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DAMAGE BOOK INIVERSAI JBRARY OU_1 66655 JNIVERSA JBRARY Osmania University Library C.IIHo53<> Accession No. Author This book should be returned on or before the date last marked below. - PREFACE Physics by Haliday and 'Resnick "Has been in use for numerous undergraduate and engineering courses all over the world for over 2 quarter of a century. The book is so ambitious that for want o! space a sizable amount of material, which, in the traditional approach, is normally worked out in the text, appears at the end o! each chapter as questions and problems and it is expected that the teachers and students .would work them out as supplemental") material for the text. Of the numerous merits which the textbook enjoys, probably the selection of problems is the most outstanding The problems have been selected with the purpose of illustratinf the underlying physical principles and have a variety which range? from the "plugin" type to the sophisticated bordering on "brain teasers.'* Furthermore, the textbook is a rich source of problems and i< ideally suited for setting examination papers at various levels Students must, therefore, get acquainted m with the techniques fo -S? 1v5 .ag luc proDlems. To this end these solutions to all the prob Jems, (about 1450 in number) from the Physics, of Resnick atu Halliday and Halliday and Resnick, Parts I and II, respectively have been prepared. While detailed solutions have been provide* for most of the problems, for a few alternative solutions have al&<. been given. An attempt has beep made to retain the terminology of the tex )k as the solutions are likely to be used by readers who ar< v dy in its p$$sskta. {Solutions are given in the same units a* in the^foWems. As the textbook does demand a prerequisite e in calculus, problems have been freely using calculu; 4 fcr.,uOds. A few problems have warranted the use of non-calcului and the alternative solutions have been given. Solutions to additio nal supplementary problems given at the end of Part II have bcci presented at end of the corresponding chapters. We ho|e that the Solutions will meet the requirements of both the teacheh and the students. A A AHMED ANWAR KAMJA* ofjboth /A* CONTENTS 26. CHARGE AND MATTER 1 27. THE ELECTRIC FIELD 15 28. GAUSS'S LAW 41 29. ELECTRIC POTENTIAL 37 30. CAPACITORS AND DIELECTRICS 83 31. CURRENT AND RESISTANCE 105 jfi). ELECTROMOTIVE FORCE AND CIRCUITS 120 33. THE MAGNETIC FIELD 145 3^. AMPERE'S LAW 166 35. FARADAY'S LAW 192 36. INDUCTANCE 206 37. MAGNETIC PROPERTIES OF MATTER 222 38. ELECTROMAGNETIC OSCILLATIONS 234 39. ELECTROMAGNETIC WAVES 254 40. NATURE AND PROPAGATION OF LIGHT 274 41. REFLECTION AND REFRACTION PLANE WAVES AND **f ANE SUKhACES 284 42. REFLECTION AND REFRACTION SPHERICAL WAVES AND SPHERICAL SURFACES 302 43. INTERFERENCE 324 44. DIFFRACTION 339 45. GRATINGS AND SPECTRA 348 46. POLARIZATION 367 47. LIGHT AND QUANTUM PHYSICS 375 48. WAVtr AND PARTICLES 391 26 CHARGE AND MATTER 26.1. Number of protons/meter j -sec falling over earth's surface, ft =0.1 5xlO= 1500 Number of protons/sec received by the entire earth's surface tfrMictf n Where R is earth's radius. JV=(4*0 (6.4 x 10 e meter) 2 (1500) per sec =7.717x10" per sec Since each proton carries charge q= 1.6x10"" coulomb, total current received by earth i=jty=(7.717x 10 M ) (1.6 X 10~ M ) amp =0.1 23 amp. 26.2. The mangitude of the force on each charge is (9X I0*nt-m/coul) (0.12 meter) 2 -2.8 nt The force is attractive. 26.3. Consider one of the balls which is in equilibrium under the joint action of three forces, the weight of the ball mg y the coulomb's repulsive force F, and the tension in the string T. (Fig. 26.3) Balancing the vertical component of forces F* =mg -.-(I) Balancing the horizontal component of forces -F ...(2) Dividing (2) by (1) tani)~ Ffmg 2 Solutions to H and R Physics II f*=mg tan 6e*m sine pk_ mgx f W=^ " . But F== a *= T - /7J/ or >: 8rrr whence x : That is, * : /" 2n O V / l(r 12 couiVDt-m a )(0 ! 01_kg)(9-8m/s > K0.05ni) > _ 1.2 meter = 2.39 x 10~ 8 coulomb. 1/3 26.4. .... , Initial relative speed, v= - = . - dt 2 / /g 8 Y/J ^^2^ rfg "3qr\ 2ne a mg / dr~~lq * dt " f couj/scc _ 3 (2 ^1.4 mm/sec. 265 The repulsive force ^^ on /I due to charge 5 acts in the diiection BA and is represented by AD. Similarly, th> repulsive fMcrFcx on -4 due to charge C acts in the direction CA and is K presented 4>y ^. Resolve the forces BA and Fc^ along two nu:! 1 ; illy perpendicular directions BC and A/L. It is seen that along hC. ihe cumponentb of F^.-c and VCA being equal in magnitude but Charge and Matter 3 7 Fig. 26.5 opposite in direction get cancelled out On the other hand, the components along A/ A get added up. Thus the net force on A is given by but where F=FBA cos FBA FCA 0.1 meter. cos 30 0) -.(2) Consider ihc ball /j which is in equilibrium under the joint action of three forces, repulsive force F due to B and C, tension of the thread Tand the weight of the ball tng. These three forces are in 4 the same plane. Balancing the vertical components ...(3) where 6 is the angle made by the thread with the vertical. Balancing the horizontal components 4 Solutions to H and R Physics II Divide (4) by (3) tan 9=F/mg Fmg tan 6 ...(5) 0.1 From the geometry of the figure we find OA= ^~ meter. sin 6= (0.1/V3)/i.O= j~ -(6) Since 6 is small, tan B^sin 6 C 7 ) Combining (1), (2), (5) (6) and (7) F=mg tan 6=mg sin -=2 FBA cos 30= V3 ^/M 4*6,,** ,* = 4*e,; t 2 mg sin 8 _(0.1 meter) 2 (0.01 kg) (9.8 m/s 2 ) _0.1 or ^"~ V3 "" (9xlO i nt-m 2 /coul 2 ) -36.3xlO" le coul 2 #== 6X 10"" 8 coul. 26.6. Choose the origin at the charge 1 2 in the lower left corner of the square. The ~ force F lt dye to charge 2 on 1 has com- ponents ; The force F lt due to charge 3 on 1 has components 3 -o C 4 Fig. 26.6 (13) - F lt sin 45 sin 45 4nt.(V2a) 2 4*t. S 2 The force F u due to charge 4 on 1 has components Charge and Matter 5 Net x-component of force is given by n* nt (H) =< 4ne. V2 a 2 O' 7 coul) 2 (9xlO nt-m a /coul') (1+4V2) V 2 (0.05 meter) 2 =0.17nt Net ^-component of force is given by ^ _ (1 .0 x 10" 7 coul) 2 (9 x 10' nt-m/coul')(2V2 1) V2 (0.05 meter) 1 = -0.046 nt 26.7. (a) The force due to Q 2 on Q x is repulsive and will be directed along the diagonal QzQi of the square of side a. * Q 2 fQQ ^ 4T^VY0)> --- x * 2 V os " r) 8n7a* The forces due to q l and -#2 on J , Q l are attractive and are directed / ^^2 along the sides of the square as *p shown in Fig 26.7. Resolve the ^ forces due to the changes q l and Fig 26.7 qt along the diagonal QyQi and perpendicular to it. Along the perpendicular direction the components get cancelled. On the other hand along the diagonal QQ the components get added up. Due to two charges of magnitude q, !Fccos45 =- 4*e,V 2 a 1 If the resultant electric force on Q is zero, then we must have FQQ =2 FQQ cos 45 _ . a* 4nc,V2a*~ 6 Solutions tottandR Physics It whence Q^-2<f2q ...(1) (6) If the resultant force on q is to be zero, then the condition would be q=-2S2 Q ... ( 2) Obviously both (1) and (2) cannot be satisfied simultaneously. Therefore, no matter how q is chosen the resultant force on every charge cannot be zero. 26.8. Let the protons each of charge q coul, be at distance d apart. Then the electrical repulsive force on either one is F= - -mg by Problem, J== r =-(1.6xlO" lf coul) V 4ne fi mj? ThatU A/ 9Xl0^nt- n r/oniP V (1.66X I(T 17 kg) < 9.8 meter/sec r ) ==0.119 meter---! 1.9 cm 26.9. U/) Let a charge +Q be placed on earth and an equal amount on moon. By Problem, electrical repulsive force gravitational attractive force. Q 1 GMm where M and m are the masses of earth and moon respectively and d is the distance of separation. [^nf Yr nt^/^ V 5.74xlO oul. (9xiO f nt-m 2 /coul 2 ) (ft) No, since the distance d gets cancelled. (c) Number of protons N each ofcnarge q required to produce the charge Q is Q 5.74 XIQ" coul, ^"9-1.6x10-" coul 3>t> 1U Neglecting the mass of electron, mass of hydrogen required =(6X10 gmKl.l x lQ- ton/gm) 660 ton Charge and Matter 7 i6.10. Coulomb force between two parts q and Q q placed at a distance d apart is Holding Q and d as constant differei liatc F with respect to q to get dF Q2v For maximum force, set - That is, <7= Thus, the charge Q must be divided equally in order to get maxi- mum repulsion. We can test whether it is actuary a maximum by (} 2 F () l F ' finding the sign of .;-,-. We get, - - -= - --* . Since the sign o'<T <^7 3 2nt t j s is negative it is a maximum. 26. It. Let the charges be q and (5xi(T 6 ..... q) coul. </i</2 _ ?(5x 10"*-?)(9X 10* nt-rnVcoui'i ^" 4K Gc J a ~~ """" (2.0 meter) 1 " -l^nt Simplifying, 9x iO f ^- 45 X 10 4 <?+4-0 Solution of the quadratic equation yields q= 1.2x10 " 6 coul and 3.8x10 "* coul, 26.12. Let the test charge +q be placed at C a distance r on the bisector of the line AB joining Q l and Q?.. The force /LV of C^i n q is directed along AC and is represented by CF. Similarly, the force FQIQ due to (? t on ? is directed along BC ;:nd is represented by CD. Since Gi^ ?-"=(? nd the sides ^C- 5C, these forces are equal and consequently CD^CF. Complete the parallelogram CDEF which is actually a rhombus. The resultant is given by the diagonal CE which lies on the bisector of AB ^ Qq __ Qq 8 Solutions to Hand R Physics tt E Fig. 26.12 Resultant force on q is given by FFo 19 cos cos = _ 6? cos But COS = F= For F to be maximum, set -~ =0 =o whence (r j +a)/ 2 ~3r(r a +a a ) 1 /=0 or r+a~3r l =0 (b) The direction of force is along the bisector of the line joining the two original charges and away from the line. 26.13. (a) Let us first calculate the x-com- $ 7 ponent on charge 1 due to charges 2, 3, 4, 5, 6, 7 and 8. Choose origin at charge i . 4* __ 4we.(V"2~o) ' V2 Fig. 26 13 Charge and Matter 9 - -14-* 14 4lfCe fl , ^16=0 _ _ 3 ~ 4ne tf (/3 a) ~ 12/3*6, a 2 where we have used the fact that the direction cosine of the body diagonal with x-axis is T-: , =F cos 45= The x-component of net force is then =0.1512 --. e e a a By symmetry F y and ^_aiso have the same magnitude. /r= JiV+FS+FJ'** 'f 3 F. =(1.732)(0.1512) -^-j= ^o; The force is directed along the body diagonal. 26.14. Assuming 500 cm 8 of water, (m= 500 gm), number of watei molecules. where N. is the Avogadro's number and M is the molecular weight. _(6.03xlO)(SOOgm) (18 gm) 1. 67 X10 W molecules. Since each water molecule has 10 protons, number of proton* io the sample =1 .67 X 10* x 10= 1 .67 x 10*. iO Solutions to ti and R Physics- tl Now each proton carries +1-6X 10" 1 * coul charge. Hence, positive charge in the glass of water -( 1.6 X l<r* coul) (! ,67 X 10") 26.15. (a) Since the penny is originally neutral, the charge Q asso- ciated with n, the number of slectrons, that are to be removed is equal to 1(T 7 cou!. If q is the charge carried by each electron, then n _ . .__ - 1(T* coui n ~ q -7:63ri0^^coT-"' 6 - 25xl (b) The number N of copper atoms in a penny is found from - *~* .=. where N 9 is the Avogadro's number, m the ma*s o r the coin and A the atomic weight of copper. Assuming Mia- ?*r-3.i gtn, x ! M aioiiii; -ok) (3.1 gin -- ..... - - -, - - - - 64.gii>, u;-o:c atoms. As there arc 29 electrons in each atcni.fv :o,ip;, total number of electrons in the penny is tfive-i by n^&N^CM) (J^x-lO*)-^. X 10" electrons vu f . r n 6/25 X1O 11 ^ . The fraction /- -- g^ - /-4 A ,. 4 26. 16 Mass of copper pipm ^f == "^ -/^^ where A is the atomic weight of copper ^nd AV is Avpg;Hro*s number A/=(64 gm/mo!e)/($.03 x lO 13 atoms/m^k) Volume of copper nucleus, T a meter) 1 Charge and Matter i 1 As mass of the nucleus is approximately that of the atom, Nuclear density, -^-=(1 .06 x l(T 2i kg)/(2.87 X 10"" meter' = 3.7 X10 19 kg/meter 8 . The answer is reasonable since ordinary density of matcii.tls is of the order of 10* kg/meter 8 , and because nuclei have radii which are smaller by a factor of 10 6 compared to atomic radii, the nuclear volume is smaller by a factor of (10 5 ) 3 or 10 16 , and nuclear densities are expected to be larger by a similar factor i.e.. would be of the order of (10 15 ) (10* kg/meter 3 ) or 10 lf kg/meter 3 which is of the right order of magnitude. 26.17. (a) The coulomb force is given by where Zi and Z a are the atomic numbers of a-particle and Thorium nucleus and r is the distance of separation. F- (2) (9Q)(1.6xlO" 19 coul)-(9x 10 nt-m a /coul*_) " "(9x i'J" ia meter) 3 = 512 nt (ft) Mass of a-particle A/ - 6.69 x 10 " kg Acceleration, a= ^ - '^xTo^' kg" " ;7 ' 7x ' 28 mcter/cc SUPPLEMENTARY PROBLEMS S.26.1. For equilibrium, it is necessary that the third charge Q be placed on the line joining the other two charges. Let Q be located 3 / ft ' f l Fig. 26.14 at distance x from -hy. The electric force /^ on Q due to 4 q has magnitude. The electric force F n on Q due to charge +4q has magnitude * F * 4n. (1-x? Solutions to H and R Physics 11 Condition for equilibrium of Q is e After cancelling the obvious common terms, 1 _ 4 which yields the solution x=//3 Next, consider the equilibrium of +q. The force F 2l due to charge +4q on +q has magnitude. / 2 The force F 31 due to g on+</ has magnitude As the force due to +4q on +q is repulsive/, that due to Q on +q should be attractive so that +q may be in equilibrium. We must then have ^si ""^! i Qq_L V Putting x=//3 and solving for Q, we find o 4? fi- T" S.26.2 By Problem 26.3 the distance of separation between the charged balls is given by 2*e rwg For 7=120 cm, ,v==5 cm, w=10 gm, we find q~ :2.4x 10~ 8 coul. When one of the balls gets discharged they come into contact, share the charge of the other ball equally so that each of them will now carry charge i^= + 1.2x 10~ 8 coul and will mutually repel each other. For the new charge \q> formula (I) gives the new distance x* , s Dividing (2) by (1), x' =0.63x^(0.63) (5 cm)=3.15 cm. Charge and Matter 1 3 S.26.3 Let initial charges on the spheres be Q and q. When they are separated by distance r they attract each other with a force F Qq *~ 4ns, r 2 . (0.5 meter)* (0.108 nt) = 4nt r*F= -- 9xl0 9 nt-mVcoul a = - 3 X 10-" (coul) 2 -0) When the spheres are connected by a wire then the net charge (Q q) is shared equally by the two spheres since they are identical. Thus, each of them now carries charge \(Qq). They now repel by a force given by (O ^- W) (0.016 nt) (0.5 meter)" ' ' (Q V - =4xlO" M (coul) a e-^=2XlQ- (2) Solving (1) and (2), C=3XlO~ coul. ^= : FlXl S.26.4. (a) The coulomb force is Fc= The centripetal force is Fc=: _L Qi. ...(i) Equating the coulomb force to the centripetal force, \ r / 2w \ ==m _ V T ) (b) The gravitational force between two particles of mass m and M is GmM r 1 The centripetal force is as given by (2). Equating the gravitational force to the centripetal force, -**,-,--' 2 * v . GMT 1 14 Solutions to H and R Physics II S.26.5. Let the electron be projected with initial speed v at infinite distance from proton. Let it acquire a speed v at distance r from the proton. The gain in kinetic energy AA'-imv^-imv. 2 ...(1) Initially, the electron being at infinite distance has potential energy t/V^O. At distance r the magnitude of potential energy is Loss in potential energy is ...(3) By work-energy priniciple, gain in kinetic energy is equal to loss in potential energy, Using (1) and (3) in (4), By Problem, v=2v ...(6) Using (6) in (5) and solving for r, e 2 \ (1.6X i(T A coul) 8 (9X 10nt-m 2 /coul 2 ) / (9.1 .1 X fir 11 kg) (3.24 x 10 6 meter/sec) 2 = 1.6lXl(r 9 meter. S.26.6. (a) Force exerted on the left dipole __1 -L+ J I (R 2a) 2 /? T (R+ 2a)* J ^ R* & (b) The above result can be rewritten as F 2re a L I/? 1 4O 2 ) 1 /? 1 J Now, R > a Neglecting the second term in the numerator in comparison with the first one, and approximating Af a 4a a by 7? 2 in t \\ A /\tmr\ rr\ i r i ^ f\. the denominator, </ 12/? 2 a 2 __ (3) " with = 27 THE ELECTRIC FIELD 27.1. Fig. 27.1 2T.1. (a) Lines of force due to equal charges placed at A, B and C, the vertices of equilateral triangle are shown in Fig, 27,2. (fc) The \est charge is kept at D, the center of the triangle. Let AD~a. Take the origin at D and let the plane ABC be the xy plane. It canrbe shown that near D the potential is where all the charges are assumed to be equal to unity, it is easily deduced from the above formula that the potential at D is not a minimum for all directions in space. As we move away from D in directions lying in the plane ABC, the potential increases, on the other hand, the potential decreases as we move away in a direction 16 Solutions to H and R Physics II perpendicular to this plane. Thus, the test charge at D, although in stable equilibrium in the plane ABC, is in unstable equilibrium off this plane. Fig. 27.2 27.3. From the upper charge, initially the lines of force tend to be projected radially outward with an angular seperation determined by the electric field strength. In the absence of the lower charge the same angular separation should have been maintained at a large distance. However, actually in the presence of the second charge which is equal in magnitude, the relative field strength at any distant point must be doubled as now the lines offeree are arising from two charges rather tluuTaT single charge, thereby reducing the angular separation to half of its previous value. Hence, if the angle between the tangents to any two lines of force leaving the upper charge is 9, it becomes i6 at great distance. 27.4. Let the electrical field intensity at a distance r from a point charge q be given by 1 <7 where n^2\ the field direction will be radially outward from the c barge and for any point of a spherical surface of radius r concen- tric with the isolated point charge, the direction of the field will be perpendicular to the spherical surface. If A is the surface area then Thf Electric Field 17 the total number of lines threading through the sphere is therefore, ..n-a This shows that the number of lines is dependent on the radius of the sphere, from which it follows that the same number of lines do not cross every sphere concentric with the charge. This then means that some of the lines of force may originate or terminate in the ipace surrounding the sphere so that the assumed continuity of lines of force will be violated. Qt is only in the case of n=2 that N is independent o r.) 27.5. Let the magnitude of the point charge chosen be # . The electric field (0.5 meter) i (2.0 nt/coul) crvy , A -ii , = /nw 1AO ' T. Siv = 5.5X10 u coul. (9 X 10 9 nt-m a /coul") 27.6. (a) Consider a positive test charge placed midway between the given charges. The test charge will be attracted by the negative charge and repelled by the positive one so that the electric field E will be directed towards the negative charge. 2q __ (2)(2Xlcncoul)(9xl0 9 nt-mVcoul 1 ) ~ (6.075 "mctcr>" f r, ^T" =6.4 xlO 5 nt/coul towards the negative charge. (b) Force on electron, F=Ee=(6AX 10 5 nt/coul)(1.6x 10" li coul) { =1.02 X 10"" 18 nt toward the+ve charge. 27.7. E is calculated from Fig. 27.7 shows the plot of E versus x. 18 So 'utions to H and R Physics II o o c UJ 5xiO -30 -20 -10 L i i^ 20 30 x (cm) -5X10 - Fig. 27.7 27.8. The force F+ due to +Q on +q and F~ due to Q on -f^ are equal in magnitude, and are indicated in Fig. 27.8. The angle sub- Fig. 27.8 The Electric Field 19 tended between F* and F~ is 120. The resultant force, F will make an angle 60 with F~. Hence, the direction of force on +q is parallel to the line joining +Q and Q. 27.9. (ft) Let <7!=+2.0x l<r coul and <?,= 4-8.5 x 10~ coul. Force on each charge is given by (9x 10* nMnVcoul i )(2x 10"* co'ul)(8.S X 10"' coul) ~ (0.1 2 meter) 1 = 1.06xlO~*nt (o) Electric field produced by q v at the site of q t is Electric field produced by q % at at the site of q v is 27.10. (a) Electric field at P is given by =+ 4- ~ Where E+ is the field due to charge +q and ~ is the field due to q, 4n. If r > a, then 0* can be neglected in comparison * with r a in the denominator. 4 r ^ E= ^7, P, g . 27.10 By definition, the dipole moment p= 2aq. E __ e_ C 2*c.r (6) Direction of E is parallel to P 27.11. (o) E= The vector sum of EI and E, points along the perpendicular bisector joining the charges. 20 Solutions to H and R Physics II Fig. 27.11 From the geometry of the figure we have COS 0= z Used) and (3) in (2) to find iff 2*7 d * ...(3) 2qr If r>tf, then a* can be neglected in comparison with r 1 in the denominator. *- l 2* J& - ~~~1 " i~ './ 4c r| (6) E points radially away from the charge axis and lies in the median plane. (c) At great distances the two charges of the same sign behave like a monopole (single charge) for which the electric field is expected to vary as r"" 1 . On the other hand for the dipole with charges q and q the field is expected to vary as r"" 8 . 27.12. (a) Let the point at which the electric field is zero be located at distance x on the charge axis on the right side of +2q. The intensity is then given by whence 2(x+fl)-5x=0 = ; (1.72)(50 cm)=86 cm. the Metric field 21 Fig. 27.12 27.13. The electric field at the centre Pdue to the dipole +q, q is given by ?, cos 6 where, EI= .-(2) ...(3) 22 Solutions to If and R Physics ft Use (2) and (3) in (1) to get 2 V2 4 rv * Fig. 27.13 ...(4) ' points downward. 1 he electric field E" due to the dipole +2(?, ~ 2<7 is given by simply replacing q by 2q in (4) ^ _2/2U2L 11 4wc.n ( ' It points up. The resultant field due to the two dipoles is given by E'-E+E"-- 2 2 4ns w a ne a = (2 /2) (9x 10 9 nt-m 2 /coul) (1.0 x 10~ 8 coul)/(0.05m) = 1.02xl0 6 nt/coul. It points up. 27.14. (a) Let the charges q l and q? be placed at a distance d apart. Let the point P be located at a distance x from # 2 and away from q l on the charge axis, where E=0. Assume th^t q l and q 2 are of opposite sign. Fig. 27.14 4KC.X 1 or or*-- ..(1) Thus, charges must be of opposite sign. The nearer charge g t mutt be less in magnitude than the farther charge o, since x must be positive. The distance from 9, is given by (1). The Electric Field 23 (b) As x has to be positive, the second solution for the quadratic equation is unacceptable. Further, at any point away from the charge axis aad other than infinity E is always finite. Therefore, no other solution is possible. 27.15. Consider a length dx of the rod at distance x from the center. Then the charge associated with the length dx is q -~ . The distance of P from dx is W-fx 1 . At P the electric field due to dx is The ^-component of the field along the perpendicular bisector of the rod is given by j cos 6= Total perpendicular component of the field is given by 112 qydx Fl. 27.1$ 24 Solutions to ti and R Physics tt Set x~y tan 9 dxy sec* 6 d& _. Then v ' o =7] 08 e d e= s -^ 1/2 Since the x-componcnt of the field would vanish upon integration, we have F - i = /i > = Since -y ==A, the linear charge density (charge per unit length) in he limit /-*oo, we get 27.16. Let the coordinate system be located at the center of the circle. Choose the j-axis so that it divides the bent rod into two equal parts. Consider a segment ds of the rod subtended between 6 and 6+d6. The charge in ds is given by, where s is half the length of the 'rod. . <W . d6 dg^ w =2q j- The field at O, the center of the circle due to dq is given by The ^-component of the field due to dq is given by Jm jwf A ^ COS rf8 dE>=dE cos 9 The Electric field 25 i> I Since the x-component of the field would vanish upon integration, have COS 49=* Consider an element of area ds in the form of a circular strip symmetrically placed over the inner surface of the hemisphere. The radius of the strip is a sin 9 and its width is (a dV) where 6 is the polar angle. sin 6) sin 6 pig rt.n 26 Sofaions toft and R Physics It As the area of the hemisphere is f ^=2na f , the charge spread over this strip is The electric field acting at O> the center of the hemisphere is The ^-component of the field is </,=</ cos 6 As the x-component of the field vanishes upon integration the total field E is given by ^ " ~~\ dE > ** 4^*| 8i COS b i sin 6 It points along the axis of symmetry and away from the hemi~ spere. 27.18. The combined electric field due to the charge +q, 2q 9 +q at distance (r-^a); r and (r+a) respectively, is given by 1 J Since f > a, a^-a^Sr 1 and r f Also, 6 27.19. The electric field on the axis of the charged ring is given by 1 qx where a is the radios of the ring and x is the distance of the electron from the center of the ring along the axis. For x < a, we can neglect x* in comparison with a* in the deno- minator. The Electric Field 27 The force F acting on the electron is _._ _eqx A i F eqx o .. Acceleration, a= ~ - A 2 - =<o 2 x * m TCE,, ma As the acceleration is directly proportional to the displacement but oppositely directed, the motion is simple harmonic, with angular frequency >= /\/ ^ ma , 27.20. The electric field is given by E= __ I* _______ i jrt Differentiating (1) with respect to .x and setting T- =0, to find the maximum value of E, d (a 2 +x-) 3/2 - x. 2x - (aH-x 8 ) __ = , =o Factoring (a 2 + x-)^ 2 (aM -x- 3x a )i=0 Since (a 2 + x l! )i^0, we have a 2 2x 2 =0 a or x= ? - 2 . Thus, the maximum value of E occurs at x=a/V2. 27.21. Consider an element of the ring of length ds located at the top of the ring shown in the textbook Fig. 27.10. The element of charge associated with it is As dq^q, - where q l is the charge in the upper half of the circumference. The differential electric field at P is given by l <fyi_ \ (q\ds \ _ 1 l ~- 4ne. H"T4ne V ~a Ja*-r where x is the distance of P. from the center of the ring whilst r is the radial distance of/Mrom the circumference of the ring. Note that for a fixed point P t x has the same value for all charge elements and is not a variable. 28 Solutions to ti and R Physics It (a) The x-component (along the axis) of the field is given by 4*, (wa) (a'+x*?* **** But ldsna, half of the circumference of the ring, so that r (x w 2i* Cl v ' 4e. (a-f x a )/ Similarly, due to charge q t in the lower half of the ring, * w ~ 4e. (aHkx^i Field along the axis due to the entire ring containing charge 7, is since E l (x) and t (jc) point in the same direction. 7- axis /-axis (5) The transverse component (perpendicular to the axis) of the field due to ft is given by The Electfic Field 29 t (T)=!dE 1 sin 6 _ f J__ q\ds ~"]~4lW. WOG^ + 4e, (no) (a +*)" Since ,. v-> (W x 1 )" 1 * ds ^ (r)= -4^7(S^ Let ds be defined by the azimuthal angles <j> and <f>+d+. The angle ^ being measured with respect to the z-axis. The transverse component will lie in the direction OA. Hence, its projection on the z-axis averaged over the azimuth angle .gives the z-component. j w/2 W2 E l (z)=E l (T) Jces ^ 4+ = -1^ (70 sin 4 ! 7C For the other half of the circumference, q n will contribute to the field in the opposite direction. E t (*)=- Net^component of electric field perpendicular to the axis in a fixed direction is 21.22. Consider a ring of radius x and width dx t concentric with the disk. The charge in the ring is The electric field at P at distance r along the axis from the center of the flisk, due to element of charge dq is given by , __ dq [__ _ (2nx dx)o _ ox dx dh ~ " The component of field along the axis due to dq is r, a ox dx r arxdx =aE cos w=- - - The component of the field in the direction perpendicular to the axis vanishes upon integration. 30 Solutions to H and R Physics II Fig. 27.22 The electric field on the axis of the disk at distance r is given by 8/2 Set The integral x*=r tan 8 x=r sec 2 6 </6 Therefore, *=- 27.23. The field due to charge +g at P is The x-component of the field is Similarly, due to g, i cos 8 1 =: 1 gx Neglecting the term a 1 in the denominators, (x) J E 1 (x)+, (jc) 1 1 V"T 2^ ^ Tl *+?) J The Electric Rdd 6q xya where we have retained terms linear in a and ignored higher order terms. Setting p=2aq m = 4. (*+/)/ The /-component of field due to charge ]+q is Similarly, due to charge 0, the /-component is ly Fig. 27.23 Neglecting o a in the denominators, E(y)=E 1 (y)+E t (y) _ 2 Where we have neglected terms involving a* and higher order terms Setting/' =2aq 32 Solutions to H and R Physics II 27.24. Weight of electron, F=mg=(9.1 x 10~ l g) (9.8 meter/sec) =8.92xlO-nt Electron charge, e = 1 .6 x 10~" coul F R 0? V 10~"*0 nt Electron field E = y = ^~ fr~ T =5.6 X 10~ l * nt/coul in the downward direction (6) weight of a-particle F=mg=(6.68x 10""*' kg) (9.8 meter/sec 1 ) =6.55xiO-nt ' Charge of a-particle, q=2 e=3.2 X 10~ coul Electric field, E= ^ =^ffi=vT =2 X in the, upward direction 17.25. (a) E V 1.5X10' nt/coul (6) F=<?=(L5 X 10 nt/coul) (1.6 X 10"" coul) =2.4xlO-"nt(up) (c) Gravitational force, F=m=(1.67X iQ-n kg) (9.8 meter/sec 2 ) = 1.6xKT 2e nt (d) Ratio of the electricjto the gravitational force, g _2.4XlQ-^nt^ 15xlol0 F Ee 27.26. (a) Acceleration, a = nt nt ( 10* nt/coul) (1 .6 X KT 1 * coul) t t A17 . , ^ xldrttk ^ =i8x 10 17 meter/sec (b) Initial velocity v =0 Final velocity v=0. lc==( 3 X 10 8 meter/sec j =3 x 10 7 m/sec QA a 1. 8 xlO 1 (c) High speeds attained by the particles owing to intense electric fields limit the applicability of Newtonian mechanics. 27.27. (a) Force, F=*=(rOx 1(T 8 nt/coul) (1.6 X 1(T 10 coul) 1.6xl<r ie nt Acceleration, a- ==t ^I-^XIO 14 meter/sec v.* v' (5 x 10* meter/sec) 1 2 a ~(2)( 1.76; XlO" =7.1 X 10"* meter=^7.1 cm. _ s ~ 2 a ~(2)( 1.76; XlO" meter/sec*) The Electric Field 33 (6) v=v at ,- v o- v (5 x 10* meter/sec) -0 , fivlft _ 8 _ /= -- = . -r ^, tn \A - : / =- 2.8X10 8 sec. a 1. 76 XlO 14 meter/sec* (c) v 2 =v<* 2 <M = (5x 10" meter/sec) 2 -2 (1.76X 10 U meter/sec") (0.8 X 10~* meter)=22.18 X 10" (meter /sec)* v=4 7 x!0 meter/sec Fraction of kinetic energy lost K. 5.0 x 10 meter/sec) J* = 1 -[(4.7 X 10 meter/secM =0.1 16 or 11.6%. ' 27.28. (a) Acceleration, a= s =~- ml m (2.0 x 10 nt/coul)(1.6x 10 18 coul) (9.1 x 10-" kg) =3.5 XlO 14 meter/sec 2 =3.5 X 10" cm/sec 2 Equation of the trajectory is y=(tar\ 6.) x - zr- ax 2 -ct 4W - L cm-uan ^ ) x 2(v cos e o ) a J3.5 XlO 18 cm/sec 2 ) -v 2 2 (6.0 X 10 8 cm/sec) 2 (cos 45) ? Simplifying, 3.5x 2 ~36x+72=-0 ...(1) As the discriminant (ft 2 40c) of the above quadratic equation is positive, the roots will be real. Hence, the electron will strike the upper plate. (6) Eq. (1) yields the roots JC=2,7 cm and 7.6 cm. The electron will strike the upper plate at a distance 2.7 cm from the left edge Here the second solution (x=7.6 cm) is not realized. 27.29. The field at A is Ess ?. n - ! 4 TCE O L 2 2 (/ z where / is the distance between the two charges (Fig 27.29 Differentiating E with respect to z + / > dE 2q Set 34 Sc Buttons to H and R Physics II _. dE __ %q 1 hen. -7 rz dz T*e / 8 Now, the magnitude of the force on an electric dipole moment placed in a non-uniform electric field E is given by the relation Fp where/? is the dipole moment. But, at z= is non- zero. Hence, force Would be exerted on a dipole placed at z=-^- notwithstanding the fact that =0 at this point. 17.30. (a) Balancing the electric force by the weight of the oil drop, Eq^mg m=yir/? 8 p=yn f 1.64 x 10~ 6 meter) 8 (851 kg/meter 8 ) = 1. 57X10-" kg = "*g (1.S7X IP"" kg) (9.8 meter/sec 2 ) q E~~ 1.92x 10 5 nt/coul =8.0 xlO- 19 coul. But q~ne where e is the charge of the electron and n is the number of electrons. Therefore, - <7 SxlO" 19 coul ~e """ 1.6 xlO" 19 coul ~ (A) First, electrons cannot be seen. Secondly, in order to balance, the fields to be employed would be too small. 27.31. Taking the differences between various measured charges arranged in the ascending order, we have in units of 10~ 19 coul, 1.6374 ; 3.296 ; 1.63 ; 3.35 1.600 ; 1.63 ; 3.18 ; 3.24 The above figures are seen to be in multiples of the elementary charge of about 1.6xlO~~ 19 coul. We therefore find the mean elementary charge from <<>>= J[l. 637 + 4(3. 296)+ 1.630 + ^(3.350)+ 1.600 + 1.630+ i(3.180) + U3.240)]xlO~ 19 coul = 1.63xlO~ 19 coul. 27.32. (a) Gravitational force on the sphere must be balanced by electric force. qE^mg _ mg (0.453 kg) (9.8 meter/sec 2 ) q ~~ E "~ 150 nt/coul = 0.03 coul The Electric Field 35 The charge must be positive. kg) met::? 3 ) J 4*P / "~(4ic)(2000 kg/met: =0.0378 meter The electric field on the surface of the sphere is 6 _ (9 x 10 9 ni-m 2 /coul 3 ) (0.03 couH 4*e r 2 " (0.0378 meter)- = 1.9X10 U volt/m a value which is much in excess of 3 X 10 6 volt/meter for the electri- cal breakdown in air. The sphere itself may get blown off owing to intense electric field. SUPPLEMENTARY PROBLKMS S.27.1. The electric f /rce actinc on the sphere of charge -f-r/ is * Fi-. S.27.1 Let the sphere be displaced through a sm ill angle P from the equilibrium position, the linear displacement hcinp v :ilonp tlu arc. While the sphere is attracted down due to praviLttional force /',.., it is repelled by the positive charge on the lower plate. I he mt component of force along x is F~-(F *F*) sin 8=~ (m.e gf I sm The negative sign has been introduced as the restoring force acts in the direction opposite to the displacement. As d is small sin 0-6- -\mf qE ) *- 36 Solutions to H and R Physics II Acceleration is given by with F f qE \ x 0= = I g ]-.- = m V rn ) I a l ( 9E\ r = In f I / V 1 m ) Equation (I) is that of simple harmonic motion as the accelera- tion is proportional to displacement and is oppositely directed. (frEm) ...(2) If the lower plate is negatively charged the net force due to gravi- tation and electric field would be enhanced as the forces] act in the same direction, the time period would then be 7=2n / g+(qEM .-.(3) Note that by setting either q^Q or =0 in (2) or (3), we would get the familiar formula for simple pendulum viz. T S.27.2. (a) Owing to symmetry the magnitude of forces F+ and F" actinc on the charge q will be equal (Fig.^ S.27.2). The resultant will be obtained by completing the parallelo- gram and drawing the diagonal. From the geometry of the figure, it is obvious that the resultant F points in the direction antiparallei to that of the dipole moment. (Recall that the dipole- ment is directed from negative charjre towards positive charge). (A) The direction of force on the dipole will be opposite to that of F i.e. parallel to the dipole moment. (c) Magnitude of force on the dipole is F-=5xl(T 6 nt. (b) and (r) follow from Newton's third law of motion viz., action and reaction are equal and opposite. Q g ' s - 27 - 2 S.27.3. Consider the differential element of charge dq in the ele- ment of arc ds for the upper half of the semi-circle. The electric field atfP due to this elementary charge points along the the radius vector and is indicated by dE+ in Fig. S.27.3. Resolve dE + into </*+ and #.,* along the x and y-axes respec- tively ...(1) (2) JQJL S -JXL dq where use has been made of (1) ' dq ...(3) F,g. S.27.3 ..-(4) where use has been made of (3) and (2). Integrating (4), i* the downward direction. It is found that due to charge -g spread over the lower half of tht semi* circle, again in the downward direction. Electric field due to both +Q and -Q, F -E t,-t,, However, the x-components + and .~ due to the two charges cancel each other. Hence electric field at P is S.IT.4. oaidtr differential element of length 4r at 4isUae x fre Ji. The elemeat of eharg* associated with dx is rff-AAf ...(1) 38 Solutions to H and R Physics tl The magnitude of the field contribution due to charge element dq at P is 1 *~ ' *dx , VAJ dF- uL = -, -r- = where use has been made of (1). The vector dE has the components dEx dE sin 6 dE y =dEcos 8 -(3) -..(4) The minus signs indicate that dEv and dE 1 , point respectively in the negative x and y-directions. The resultant x-component of the field is obtained by integrat- ing (3). 00 1=8 ""4^1 xdx ...(5) where use has been made of (2) and the relation sin 8=x/V/P+j?. With the change of variable x=R tan 0, and dx=R sec* 6 we find sin The Electric Field 39 Similarly, we have for the y-component, With the change of variable *=/? tan 6 and dx=R sec 2 6 rf8, we find n/2 E y =*dE>~- ^~ fcose</6=~ ^ ...(8) > 4*e R J 4 nt R The angle which the vector B makes wi{h the ^-direction is given by .= l=l ...(9) / where use has been made of (6) and (8). From (9) we find 6 a =45. S.27.5. The torque acting on the dipole is given by T =PxE=/?JS:sin e=p6 ...(1) where small angles have been considered so that sin 0=0. Also, T=/=/<p ...(2) There is a restoring torque acting on the dipole which enables it to return to the equilibrium position. Comparing (1) and (2) /S--* -0) Where we have inserted the minus sign in the right side since the restoring torque acts in the direction opposite to the angular dis- placement. -- .11 .-(4) dt* II Where k is the torsional constant with, k=pE. The period of oscillation for* the torsional pendulum is given by and the frequency of oscillations is given by ^L^L [T mf IpE * ~ S.f7.6. Two forces +/! tad-/, act in the oppposiu 4irectis as shows ia Fig. S 27,6. 40 Solutions to H and R Physics 11 Net force Ft~qE where E is the field at the negative charge. As the field is varying in the vertical direction, the field at the positive charge will be E+dE=E+ ~dy F l ^=q (E+dE) Fig. S.27,6 - Set dy=2a, the charge separation distance. Then F^qE+ (2 a?) |? dy =**+% pointing upward. S.27.7. The field E due to dipole at a point a distance r, along the perpendicular bisector of the line joining the charges is given by where 2a is the distance between equal and opposite charges q of the dipole and r > a. The magnitude of field at P due to the dipole closer to P is 2 o pointing down; that due to the dipole which is farther is __ 2aq E *~ 4itc,(/M-a) pointing up. Therefore, the net field is 2a Neglect cf in comparison with J? 1 . Then 28 GAUSS'S LAW 28.1 The flux is given by *=/. dS The field E makes an angle 6 with the element of area dS which is equal to 2nR' sin 6 d& (Fig 28.1). 4te= /(E) (2*R Z sin 6 rffl) cos 6 (2 sin cos 0) sin 26 sin0 -* 42 Solutions to H and R Physics II ( J 28.2. The flux J>E can be written as the sum of three terms, an integral over (a) the lower cap, (b) the ft^ _ cylindrical surface and (c) the upper cap (Fig. 28. 2). fa=/E.dS = \E.dS + [E.dS j J (a) ^J (6) ^ Now, for the caps, :.dS = fE.rfS ^A (a) J (c) U because 6=90, E.</S=0 for all points on the caps. For part (6), the curved surface may be ^ divided about a plane perpendicular to E 8 and dividing the cylinder into two halves, right and left. For right halfO < 90 whilst for the left half 90 < 6 <180*~'so that for reasons of symmetry the contribution to the intergral for the entire curved surface vanishes. Consequently ^=0. 28.3. fa=/E.dS = /"cos 6dS =EA cos 6 where we have taken cos 6 outside the integral, as 6 is constant. 28.4. Flux and the net charge q enclosed by the Gaussian surface are given by the relation -2 = ~ 1.0xlO~ 6 coul = 1.1 xlO*nt-m 2 /coul 8.9xl(T 12 coul 2 /nt-m a ?i> 5 8 , 5 6 is positive, ^^ is also 28.5. As the net charge enclosed by positive. As the net charge enclosed by 5, is negative, ^E is negative. As the net charge enclosed by 5 4 is zero, <f>E is zero. 28.6. Describe a sphere to represent the Gaussian surface enclosing the mass m. The angle between the field direction and the element of area on the sphere is 180. Thus 1 1 ^ , 1 Gauss's Law 43 or G m A mass AM placed in the gravitational field g due to m would experience a force given by The negative sign signifies that the force is attractive. 28.7. Electric field at point a at distance r from the charge is given by q = (l.OX 1(T 7 coul) (9X 10* nt-m 2 /coul 2 ) 4rce r 2 U.SxlO"" 2 meter v- =F=$X leWnt/coul^ At the point 6, the electric field since the point is^within the conductor. ^ 28.8. (a) Consider a point charge q located at the center of an uncharged thin metallic surface (shell) of radius R. Let the point P lie on an element of surface ds within the shell, Fig. 28. 8 (a), on the surface of a sphere of radius r concentric with the shell. Since the net 'charge enclosed within this Gaussian surface is q y accor- dingly to the Gauss' theorem, the flux at P is given by Fig 28.8 (a) But here the* entire field is normal to the surface and points out from it, i.e. 8=0. =jtfs cos 6 44 Solutions to U and R Physics It (b) As the shell is conductor, negative charge q will be induced on the inside and +q will be induced on the outside of the shell. Choose a point P outside the shell at distance r from the center of the shell and draw a spherical Gaussian surface of radius r and concentric with the conducting shell. The surface is indicated in Fig. 28.8 (b) by the dotted lines. The field vector emerges every- where normal to the Gaussian surface. Further, the field has Fig 21.8 (b) constant value over the surface. By Gauss' theorem whence /? = - t v c) It is seen from the results of (a) and (b) that the shell has ao effect on the field. (d) Yes, negative charge is induced on the inside surface and positive charge on the outside surface of the shell. (e) Yes (/)No (g) No 28.9. The small rectangle in Fig. 28.9 (a) is a side view of a closed surface, shaped like a pillbox. Its ends, of area dA, are perpendicular to the figure, one of them lying within the sheet the other in the field. Lines of force crossing the surface of the pillbox is EdA where is the electric intensity. The charge within the pillbox is adA. Then from Gauss* law *dA or (a) On the left of the sheets, Fig 28.9 (6) electric intensity E l due to sheet 1 of charge on the left hand is directed toward the left and its magnitude is a/2e t . Then intensity E\ due to sheet 2 of charge on the right hand side is also toward the left and its magnitude is also o/2*,. Gauss's Law 45 The resultant intensity is therefore, = 4-= (6) Between the sheets EI and 2 are in opposite directions and their resultant is zero. (c) On the right hand side of the sheets, \ and , again add up and the magnitude of the resultant is a/e,, directed towards right. (a) 4- (b) (c) Fig.28.9 (b) p +. - 4- EI E} f ? f, E 2 (a) (c) - 4- - 4- (b) Fig.2g.10 28.10. (a) and (c). On the left side as well as on the right side of the sheets the intensity components E l and E 2 arc each of magnitude cr/2e but are oppositely directed so that their resultant is zero (Fig 28.10). (6) At any point between the plates the field components arc in the same direction and their resultant is a/e and is directed towards left. 28.11. By Problem 28.10, the electric field intensity E between the plates is given by But, the charge density e~qlA =-(55 nt/coul)(8.9xlO- 12 coul*/nt-m a ) (1.0 meter 2 ) =4.9xHT l0 coul. 46 Solutions to H and R Physics II 28.12. (a) As no charge is enclosed within the charged metal sphere, E for a point inside the sphere is zero. (b) In calculating the field external to the spherical charge distri- bution, the charge may be considered as concentrated at the center. _ q (9 X 10 9 nt-mVcoul 2 ) (2 x 10~ 7 coul) E (c) E 2 (0.25 meter) 2 =2.9xi0 4 nt/coul* q ^ (9 X 10 9 nt-m 2 /coul 2 ) (2 x 1CT 7 coul) ~ ~ (3 meter) 2 = 200 nt/coul. 28.13. =---- Force on the electron, / r =Ee= Where e is the electron charge. Let the electron be fired from a distance x meters so as to just miss striking the plate. Then work done Setting, W =K, the initial electron kinetic energy <*ex ae ^( 1 00 ev) ( 1 .6 X I 0~ 19 joule/ev) (8.9 x IP" 12 coul 2 /nt-m 2 ) (-2x 10"* coul/meter 2 ) ( 1.6 x 10~ 19 coul) > =0.44 x 10~ 3 meter=0.44 mm. 28.14. Describe a Gaussian surface in the form of a right cylinder of radius r coaxial with the given cylinder and of length b Fig 28.14 (b). Let r < R. The charge within this Gaussian surface is As the caps of the cylinder are perpendicular to E, they do not contribute to the field inten- sity. The only surface which is of consequence is that of the cylinder's curved surface of area According to Gauss' law As E is normal to the elementary surface dS and is constant, we have CoE Fig. 28.14 (a) Gauss's q pr - ^ \' ^""**/ For r > R, again construct the closed Gaussian surface in the form of a right cylinder of radius r and length 6 coaxial with the given cylinder. No lines of force cross the ends of the cylinder. The lines of force cross outward normal, to the curved surface as before. We have 1 1- - t b I ^, h hr ^> - 1 __ ^ ^T> U_ Fig. 28.14 (b) 2e t) r 28.15. Forr < a ; "=0 ; E- -,- e a r 2 3 For a < r * ~ ^ , q 4 71(6* a* For r > b: E7- 2 ^= 47Sc r- 3 47ie r 2 Fig. 28.15 shows the plot of E versus r t 37.V* o ?ooo - 30 Mcrr Fig. 2K 15 48 Solutions to H and R Physics II 28.16. Case (i) r > R. Construct a Gaussian surface in the form of a right cylinder of radius r and of length 6, coaxial with the metal tube. As the lines of force do not cross the ends of the cylinder, the only surface that matters is the curved surface of the cylinder through which lines of force cross in an outward direction normal to the surface. The quantity of charge within cylinder is Afc. According to Guass' theorem. 4 -" r .-i- * __ or (r>R) Fig. 28.16 (a) Case (ii) r < R. Since no charge resides within the tube, the field =0 (r < R) Fig. 28.16 (6) shows the plot of E versus r. D O c E K 4 3 Fig. 28.16 (b) (cm) 28.17. (a) For r > b the point is outside both the cylinders and the Gaussian surface drawn at radial distance r would enclose a net charge equal to zero since the two cylinders carry equal and oppo- site charge. Hence, =0 Again, for r < a, the Gaussian surface does not enclose any charge. Hence =0. (6) Between the cylinders, we have a < r < b. Desctibe a Gaussian surface in the form of a right cylinder of radius r and length L coaxial with the given cylinders. Then the charge enclosed by the Gauss's Law 49 Gaussian surface is XL and the curved surface through which E is projected outward and is normal to the surface has area A^=2nrL. The ends of the cylinder do not contribute to the field intensity. By Gauss' theorem (t E) (2*rL)=AL or =, 28.18. -~ 2n? v r Electric force acting on the positron Equating the electric force to the centripetal force r Whence, the kinetic energy of positron =(9x 10 9 nt-m'Vcoul 2 ) (3x 1(T 8 coul/meter) (1.6xl(r w coul) =432 Xl(r lf joules =*(432xlO" l9 joules)/(1.6xlO" 19 joules/ev)=270ev. 28.19. (a) Describe a Gaussian surface in the form of a right cylinder of radius r and length 6. The area of the curved surface winch alone contributes to the field intensity is given by A = 2itrh. The net charge enclosed by the Gaussian surface is 2q-\-q or q. By Gauss's law (t E) (2nrb)^ q whence br The negative sign shows that E is directed inward. (M A charge q will be distributed on the inside surface of the shell and a charge q will reside on the outside surface. (r) Between the cylinders, the Gaussian surface (again cylindrical in shape) will enclose a net charge +q so that The field being radially outward Assumptions made are (i) 1 he cvlipder is sufficiently long so that only radial component :>f the field exists. 50 Solutions to H and R Physics II (//) Fringing field near the ends of the cylinder is not present. (///') The charge is uniformly distributed. 28.20. (a) Construct a Gaussian surface in the form of a sphere of radius r, concentric with the spherical shells. Since no charge is enclosed by the Gaussian surface with r < a, E^Q. (b) Here the net charge enclosed by the Gaussian surface is <?<, As E is normal to the spherical surface by Gauss's law or = - *- Gaussian surface Gaussfan r 'face Fig 28.20 (a) Fte 28.20 (h) (<) Here the net charge enclosed by the Gaussian surface is * \ if*, and E is normal to the spherical surface. By Gauss's law or Fin 2* <7ui<N.\'.N IMW 51 28.21. For the conducting S'KVI I'M? field \ gi\en **\ -i ...ill The electric force acting on the sphere IN ,J cr F^E<i^' ^ ...(2) The sphere is held in equilibrium IKKKY the jnint action of three force: if; weight nig acting tl--\Mi. I ft \ r (11) electric foue F acting hoii/oniallv. ami j|j (//i) tension in the thread acting at an angle r A ah ;{ the vertical. !*i fr;^ H k ? We have from Fig. 2S2I I tan i* mg Using (2) and (3) ian ...(3) HK.2i.2l "i* W tan t> " </ R.9 in-HcouP'nt-m-sd > l<> H- 0.8m -V Man 30* 2.0-. in cinil) 2.5" 1 - 10"" 28.22. C'f. For .* sphere o '/'Jrer r . -.inc- hi' siirf.ico .'n-.i l i plictv is 4nr-. 28.23. The -pnrticle^ i*i at .1 i!- f i:ui' ^R fn-in sin eriiti-r of the goh! nucleus if radius R I > Mi '' m f r. ' ><i: n! rini! a- particle to be .1 point charge, n e 1 1- .trie I'orti- IN where Z,f and Z f are :h=j char^.^ .-i 'lie /-p n t-ir .: i '. : ( iUI nucleus respectively. (2x |.6x IP" 1 * .ii!iM7*J ^ I fi in t is 1 - / >(i' ?.: , *191 nt. Acceleration. ** ^ 2 ' 85y l<)a nicli ' f / 28.24. fa) Consider i cm 2 of face arc, t Th* .n the v. iun, of li.c gold foi!. ? v 10 - cm thick, is i- ?x 10" cmx I cm x I cm 3^ i cm' 1 52 Solutions to H and R Physics H Number of gold nuclei per cm* is where N is the Avogadro's number, p the density and A the atomic weight of the material. Af=(6x 10 2S atoms/gin atom) (19.3 gm/cm 8 )/197 = 5.88 xlO 22 atoms/cm* Number of gold atoms in volume v is //== :tfv=(5.88 x 10 2 * atoms/cm 1 ) (3x 10'* cm 8 ) = 1. 76xl0 18 If a is the area of each gold nucleus, then the total area arising from n nuclei is S==HO If R is the radius of gold nucleus, then a-Tc/^-K(6.9x 1(T 13 cm) 2 - 1.5x KT-'cnr S- (1.76x 10 18 ) U-5 x i(T 24 cm-)-2.6x I(T 6 cm 2 .". Fraction of surface area "blocked out" by gold nuclei ..2J*10j< m x 1 .0 cm 2 (fe) Volume occupied by each gold nucleus ..8xl( G cm Volume occupied by A r nuclei per cm 3 of foil is M'=-(5.88x 10- atoms/cm 3 ) (1.38X 10~ 36 cm 3 ) -8.lxl(n<cm 3 Fraction of volume of the foil occupied by the nuclei is S.lxKr^cm 3 , , 14 . r . -8.1 x u 1 .0 cm 3 (r) Rest of the spuce is tilled vuth electrons. But, a major part of the space remains empt> . 28.25. (a) The flux is completely determined In the .t-component of the field as E y - f;.= 0. In-flux, 6,n ^ fx-j-7) <J a-- Out-flux, >.,< . a-b v/2a - S2hu* Net outward flux. 6 ^u/ 6 aM ' \ 2 Gauss's Law 53 -l) (800 nt/coul-m*) (0.1 meter)*' 2 1.05 nt-meter a /coul (8:9 X 10" ls coul 2 /nt-m a ) (1.05 nt-m a /coul) 9.3xKT ia coul SUPPLEMENTARY PROBLEMS S.28.1. Consider a cube of side 0=100 meter which encloses a charge q, the upper and lower surfaces of the cube being at 300 meter and 200 meter >i 1 ^F^coWme'er altitude. By Gauss theorem the flux . E is given by J t_ S ^ \ (Q) - -'OOV/rneter The flux can be written as id* the sum of six terms; (a) integral over the bottom sur- Fig S.28.1 face (6) integral over the top surface and (c) integral over the four vertical faces. JE.rfS (tf) (b) (c) In (a), E and dS point in the same direction so that the angle 6 between these to vectors is zero. 1 .rfS= j E l cos dS^E^ f dS^ (<*) where S=a 2 , is the area of the face. In (6), E, and dS point in the opposite direction so that 0=180. f E r </!5= [ 2 cos 180 dS=* -E 2 J rfS= -25 (^) In (c), E and dS point at right angles so that 6=90. For each vertical face, f E.rfS= f (a) E cos 90 54 Solutions to H and R Physics II <jr=eo (E^ E 2 ) a 1 =-(8.85 xlO" 1 * coul 2 /nt-m 2 ) ( 100-- 60 -^iL VIQ meter) 2 \ meter meter/ -3.54xlO" 6 coul. 5.28.2. Electric field extends from higher potential to a lower one. In order that the electric field may have constant direction the charge in the region must be uniformly distributed in planes perpendicular to E. The fact that field is decreasing in strength in the direction of /;" implies that the charge is negative. 5.28.3. Potential difference between the concentric spherical shells of radii a and b is 4rcE \ a I 1 1 -; n- -- . 145 meter 0.207 meter/ -1115 volt Kinetic energy gained by electron of charge e in falling through a potential difference of V volts is eV /2ev^ 1(2) ( 1.6xlO~i* coul) (11 15 volt) v v'w V (9.1XlO~ 81 kg) =--1.98 XlO 7 meter/sec. S.28.4. (a) The electric force between the two spheres would be as if the charge is concentrated at their centers. By Coulmb's law where R is the distance between the centers of the spheres. (b) If the charges are like, then force will be less than in (a) and for unlike charges the force will be greater than in (a). the two cases must be distinguished. (i) The two spheres have like charges. In this case owing to coulumb's repulsion, the charges on cacih sphere instead of remaining uniformly distributed with the center of charges coinciding with the centers of the spheres are now displaced towards the rear surfaces of the spheres, resulting in a greater value for R, the distance of separation of the centers of charges. This has the consequence of reducing the force. (/O The* two spheres have unlike charges. *S Law 55 In this case the coulomb's attraction causes the charges to be pulled towards the front surface of the spheres, leading to a reduc- tion in the effective value of R. This has the consequence of increasing the force. S.28.5. The field due to the point charge Q at the center at a distance r is The charge between the spheres of radii r and a is r r r q^ fp(47W 2 ) rfr= ( ^-(4r 2 ) </r=4,4 f rdr a a a By Gauss theorem _ A + Q-2KAa* Total field, If is to remain constant in the region, a < r < b. for any value of r, then the numerator of the second term of the right side in the above expression must vanish. ^ *~ Hence, S.28.6. (a) Consider a Gaussian surface of radius r. By Gauss' Theorem, Jrj if\ M^. 0* m E -* r 4Ke,r 2 '" (l) where ar is the unit vector pointing toward P from the center of the sphere, and q is the total amount of charge within the Gaussian surface. For uniform charge density, p is constant and 4 Solutions to H and R Physics II Using (2) in (I) _ pr_ 3 e ff "3 (b) The electric field at the center of the cavity due to the remain- ing portion of the sphere is where R is the radius of the cavity. The electric field due to a sphere of radius R corresponding to the volume of the cavity is Using superposition principle the electric field at any point with- in the cavity for uniform field is 29 ELECTRIC POTENTIAL .. TKe M* =5-6xlOnt/coul Let the equ {potentials be A* meters apart. As E is constant, or -r = 1- 5.6 X10 1 nt/coul 0.89 mm. 29.2. (a) The potential KB at the point B is given by ra TA Setting r=oo; K^ co a R co For the first integral, we find using Gauss' theorem 4 wr 2 *=?' where q' is the charge enclosed within a radius r. Since q f is proportional to the volume we have . r The first integral is then evaluated as follows: 4ntJP~ 8 R a 58 Solutions to H and R Physics II For the second integral, Gauss' theorem gives 4 xr** E=q E ^ <1 4 rce r 2 and the integral becomes R R 4 itsj* 4ne, R 00 00 '" 8*e, /JV " / T 4iw.JT-8ne.* (6) Yes. Actually, F=0 at oo. 29.3. The potential is given- by q (l(r coul) (9 X 10 Dt-mVcoul 2 ) nnn V ^4l^Tr (0.1 meter) -900 volt. ^ A . , q ( 1 .5 X 10- coul) (9 X 10 n>m"/ooul) 29.4. (a) A = ^V~ (30 volt) = 4.5 meter. or = Clearly, A/? depends on /?. Hence surfaces whose potentials differ by a constant amount are not evenly spaced. 29.5. (a) (/) Let T^Q at distance x from +q, between the charges, then, , .(rf x) -JC-3x 100 cm ^ c =25 cm A *J 4 4 (//) Let=0 at distance x from +9 outside the charges. Then Electric Potential 59 d 100cm <A or x= =50 cm. (6) Let =0 at distance x from +q outside the charges. Then _ ?...._ 3? x 4 we or Set = 1.0 meter. Then 2x-l=0 l-f3 = 1.37 meter= 1 37 cm. Between the charges E cannot be zero since the forces due to +q and 3<7 would act in the same direction. 29.6. 29.7. 4TSS 4 rce * id we, (o+J) " qd qd 4 ne. a(a+d) .'. VAVB .- 4 rce a(0+a) 4 we a a(fl+(/) 2 KC<, a(fl+^) When rf=0, A and # coincide and therefore, VAVB-+VA- which is the expected result. Also, when q=Q, VA~VB~Q. Hence, VA Va=Q is again an expected result. 29.8. (a) VA= - 60 Solutions to H and R Physics ft where TA and rj are respectively distances of A and B from q. (9X W nt.nWcoul)(1.0x lO-'coul) ( 2 ; ^ ~ f^ = -4500 volts H - - 45 29.9. The center of negative charge lies at 0, the oxygen nucleus. On the other hand the center of positive charge of the hydrogen atoms lies at P, midway between the two protons. The distance of separation of the center of posi- tive charge and center of negative charge denoted by a can be calcu- latcd from ^the triangle OPH *=(OH) cos 52 =(0.96 X 1(T* meter) (0.616)=0.59 X 10~* meter Dipole moment arises due to the separation of +2e and 2e charges of hydrogen atoms by distance a. Dipole n^oment p=(2 e) (a) =2 (1.6 X 10" lf coul) (0.59 X lO' 1 ' meter)= 1 .9x 10* w coul-meter. This value is to be compared with the figure of 0.6 X 10~ M coul- meter quoted in the text which is lower but correct. The discrepency is to be attributed to the oversimplified model. 29.10. Potential at P is given by the sum of potentials due to the charges +q, +q and q at distance (r a), r and (r+a) respec- tively. q _ + 3 _ 4 e r 4 nej (r a) 1 / q 2ga \ V 4e V r V-fl/ Since r > a, r* o*s*r l 1 (q ,2qa\ K ~4.Vr ^ r* ) Right hand side is nothing but the potential arising due to an isolated charge plus a dipole at distance r. Electric Potential 61 29.11. Energy released is Kqy=(3Q coul) (10 9 volts) =3 XlO w joules =(3 X 10 10 joules)/(4.18 joules/cal) = 7.2xlO'cal Heat required to melt m gms of ice at C is mL, where L is the latent heat of ice. mL=7.2xlO*cal 7,2xiO*cal =9 X 10* kg- 3 8 ^90 tons. 6 10 3 kg/ton 29.12. (a) The electric potential is ? (9x 10* nt-mVcoul 2 ) (1.6X 10 ~ 19 coul) ~~ ~~ (5.3 X 10" 11 meter) =27.1 volts (b) The electric potential energy of the atom, is '/= 'pK= 27.1 eV (r) Equating centripetal force to the Hcctrostatic force mv 2 e* r "~ 47ie r 2 e* or mi' 2 - - =27,1 ev 4ne r Kinetic energy=.l mv 3 , A (27.1 ev)^13.6 ev ((/) Total energy = kinetic energy ^-potential energy = 13.6ev-27.1 ev = -13.5 ev Hence, energy required to ionize the hydrogen atom is 13.5 cv. 29.13. The electric potential energy of the charge eonfiguialion of the textbook Fig. 29.7 is ~~Sla r a T a 2n a By Problem, ft= + 1.0 X I9 M> coul ; </'-'= -2.0 x 10" coul; <7i=-f 3.0 x 10" f coul; ^ 4 -^-f2.()x 10"^ coul ; and 01.0 meter 62 Solutions to H and R Physics^ II +(1) (2)+(-2) (3)+ ( ~^ 2 (2) -+(3) (2)]x 10-" 6.4 XlO~ 7 joules. _ j? (9 x 10 9 nt-m 2 /coul 2 ) (3 x IP" 6 coul) r ~4ne.K~ (500 volt) S4 meters (b) Assuming that the drops are incompressible, the volume of new drop will be twice that of the small drop. JL 3__ A r ;u 3 3 .*. Radius of new drop, /?=2 1/B r. ...(2) The charge of the new drop, Q^2q ...(3) The potential at the surface of new drop is w,__ Q 2 <y ...(4) where we have used (1), (2) and (3). F =2 2 " (500 volts)=794 volts. 29.15. (a) Total charge on earth's surface where e is electron charge and r is earth's radius. The poten- tial is _ g itie__ er_ (JL6^<U)" lf coul) (6.4 x 10* meter) _ n . f . . (8.9 x 10-* couiVnt-m 3 ] - --i 5 volt. (b) The electric field due to the earth just outside its surface is q ^ 4^^ e = -(1.6x 10-" coul (8.9X 10" u coulVnt-m 1 ) == 1.8x 10"* nt-mVcoul. The negative sign shows that the electric field points radially inward. Electric Potential 63 29.16. Under the assumption of constant density the volume of each fragment is half of the U* 38 nucleus. y r* -K . ,. r u < . A 8x 10"" 15 meter . . Radius of each fragment, r=rj-^= The distance between the centers of the fragments rf=2r = 2(8x 10~ 15 meterj^'^l^x 1(T 14 meter The charge on each fragment is ?i=?i= =r 4- (a) Force acting on each fragment is (9xl0 9 nt-m 2 /coul 2 ) (46 X 1.6x lO' ( i. 27 xl0^ 14r meter) 2 -3020 nt. (6) Mutual electric potential energy of the two fragments is 0^nt^n_ 2 /coul 2 ) (46x 1.6 x 10~ 19 coul) 2 "" (L27x 10" 14 meter) = 3.8 XlO" 11 joules. 29.17. Potential difference between the plates L>V=- L = (1.92x 10* nt/coul)(0.015 meter) = 2880 volts. 29.18. (a) The potential at the point P on the ring of charge radius a can be computed from Pig. 29.18 64 Solutions to H and R Physics- II where r is the distance of P from the differential element of charge. Since r 2 Where x is the distance of P from the center of the ring along the axis. (6) The field is given by 3K 8 q "~ dx^^dx***. an expression which is in agreement with that obtained by diixv, calculation of E in Example 5, Chapter 27. 29.19. ^VH7 -r) * -l] If r > o, expression (1) may be re-written as \ where we have expanded the radical by binomial theorem. Now, the total charge q is given by Electric Potential 65 This is the expected result for the field of a point charge. This is reasonable since at longer distances the disk appears as a point. (b) If r=0, expression (1) reduces to *>; ' This expression is identical with the field of a charged sheet of infinite extension. Very near the disk, the conditions of an extensive sheet are fulfilled. 29.20. (a) For a dipole r 1 where p is the dipole moment. ' dV L^ a _ / L. /> cos Lr dr ' dr \ 4* e ; r a ~ -A P cos B ^r .T cos 6 (b) Er is zero for 6-90 or 270i 29.21. Field on the surface of the sphere is r-JL _ (4 X 10~ coul)(9X 2 (0.1 meter) 2 ==3.6xl0 6 nt/coul This value exceeds the dielectric strength of 3x 10 6 volts/meter or 3X 10 nt/coul. 29.22. T*he field E at a distance y from an infinite line of charge density A is given by -(I) * r* f~ 'i t- I' Fig. 29.22 66 SolvhJM to H and R Physics II The potential difference between points 1 and 2 is given by = 2tu, J >> 2Tce /?, 2JIC, 1?! R* where /?j is the radius of the wire and /? a the radius of the cylinder. Use (2) in (1) to get "= Set F=850 volts J? t =0.0025 in -0.00635 cm J?,= 1.0 cm -.. r 850 volts 850 volts 168 ... . Then, =-- , -- -- --- y volt/meter () Electric field strength at the surface of the wire is zvnv 168 168 VOltS *i^tn ^,)-^- 6 :35xlO-T7ne^r 2 - 6X 10 (6) Electric field strength at the surface of the cylinder is 29.23. (a) The charge is assumed to be located at the respective centers of the sphere. The point midway between the centers of the sphere is 1.0 meter from either center. The potental at the mid point is py-_gt_ _i_ ?* But /v Electric Potential 67 (9 X iOnt-m a /coul a )(l.x 10~*coul 3X 10~coul) (1.0 meter) ==180 volts. (b) The potential at the surface of sphere 1 is due to its own charge <?i plus that of q* of sphere 2 acting at distance r. =(9xl0 9 nt-mVcoul 1 ) V 0.03 meter 2.0 meters / =2865 volts The potential at the surface of sphere 2 is due to its own charge q^ plus that of qi of sphere 1 acting at distance r. 4ice0jR 4ice r / ^ V 1 n~~8f*nii I 1 V in~*i/niil \ = (9 x 10'nt-mVcoul*) ( ** , + V-jyLJS. ) \ 0.03 meter 2.0 meter / = 8955 volts. 29.24. Let the total charge qqi+q* (!) After the spheres are connected the potentials of the two spheres are equal. Also, -^ ...(3) <?i ^2 where ? t is the charge on the sphere of radius RI and q % is the charge on the sphere of radius J? a . The surface charge densities for the spheres are given by "- and "- By Problem, Hi =1.0 cm ; R 2 =2.Q cm and <?=^2.0x 10" 7 coul. (a) Eliminate g a between (1) and (3), 1.0 cm __ 1^ ^^2.0 cm ~ 2 whence ^=6.7 X 10~ a coul (small sphere). Abo, # t = 1 .3 X 10"^ coul (large sphere). 68 Solutions to H qnd R Physics II (ft) Using (4), " =5>3 X 10 ~* coul /m 2 (small sphere) (c) Using (2), F=(9xlOnt. m */coul*) 0.01 meter =6xi0 4 volts -60 kv. 29.25. There will be a greater density of-charge in regions of large curvature and a lower charge density on surfaces of small curvature. Since the electric field intensity near a point charge is proportional to the charge, electric field will be largest near points where the charge density is greatest. Accordingly, lines of force may be drawn by spacing them more closely in places the charge density is larger. As the surface of the conductor is an equipotential surface, the lines of force are normaHo the surface. These are shown 'solid lines pointing inward in Fig. 29.25. The intersection of the equipotential surfaces, wirh the plane of figure shown as dashed lines are every where normal to the lines offeree. Fig. 29.25 29.26. Let the charges +q,-q and +q be at the vertices of an isosceles triangle of sides 2a, 2a and a (Fig. 29.26). The potential energy of this configuration is '""' a 2a 2a '~ Electric Potential 69 t M Fig. 29.26 29.27. The potential energy of the configuration taking charges in pairs, is ^J- (-+_-_+ ...-\ 475e fl \ a *f1a a a V2a a ) where potential energy at oo is taken as zero. This is the work re- quired to put the four charges together. 29.28* When the a-particle just touches the surface of gold nucleus, the original kinetic energy is completely transformed into electric potential .energy. ] 4716. r C-t n < ")"' OCl i/j ^c, 9i= s 79e, where e is the proton charge, and r=5x!0" 18 meter (2x 1.6x lQ-*coul)(79 x 1 .6 x lQ- 1> coul)(9x I0*r (5 x I0 r " meter) =7.3x10-" joules =(7.3 X 10-" joules)/( 1 .6 x I0' u ^~ -45.6 Mev 70 Solutions to If and R Physics II The a-particles used in the experiments of Rutherford stayec well outside the radius of gold nucleus. Rutherford could only put an upper limit for the estimation of nuclear radius of gold nucleus since "anamolous scattering" did not show up with oc-particles oi 5 Mev energy. 29.29. The potential gradient at distance r is given by q _ (79 X 1.6 X 10~* 9 coul)(9 X 10 9 nt-m a /coul a ) 4ne, r 2 (10~ 12 meter)* = 1.14 X 10 17 volts/meter The potential grdaient at the surface of gold nucleus of radius /?=5xlO~ 15 meter is __ ^ (79xl.6x 10~ 19 coul)(9x 19 9 nt-m 2 /coul a ) 4ic c * (5xl(T l5 meter) a =4.55 X 10 M volts/meter 29.30. E^ _ g *~~ 4ne, JJ t 2 ^ -<li *' " ^i ?t^i 8 But g ^~ 29.31. (a) Energy acquired by an electron in falling through a potential difference of V volts is Set v=c TH v- 1 *- 1 (9>1 x l(T 8l kg)(3 X 1 meter/sec) a men, K- 2 e c-- 2 (i.6xiO- = 2.6 x 10 s volts i*V ,.13 !l+ ^ s=sl+ T^T Electric Potential whence ~ = y ~^=0.745 v=0.745 f=(0.745)(3x 10' meter/sec) = 2.236X10* meter/sec. 29.32. (/) r</? 1 As no charge is enclosed inside the sphere of radius R l9 we conclude that ==0. Due to smaller sphere,^ = ^ JKm and due to larger sphere, ^2= 7 ^ Due to smaller sphere,K 1 =j~^ and due to larger sphere, K f = Fig. Z9.32 () I 4 r (mtttrl Fig 29,32 (b) (0) Due to smaller sphere, V l - 72 Solutions to H and R Physics II t and due to larger sphere, K,==~- t Figure 29.32(b) shows that plots of E (r) and V(r) from r=0 to r=4.0 meters for ^=0.5 meters, ,= 1.0 meter, ^ coul and ft= + 1 .0 x 10~ e coul. 29.33. Power delivered to the belt, P=rate of energy transfer =(?/unit time)(F) ==(3 X l(T a coul/sec)(3 X 10 e volts) =9000 watts =9.0 kilowatts. 29.34. (a) y=~ e, r Set, =4e r V r = 1 .0 meter and K= 1 .0 X 10~* volts, _ (1.0 meter)(l .0 X 10 6 volts) , , w IA a 4 *nroUTAiT * - Ti - ^ - '= 1.1X10 * COUl (9x 10 9 ntm 2 /coul 2 ) For r=J.O cm = l.ftxlO"" 1 meter, (1.0xlO~2meterKl.OxiO volts)_ 9 ~ 9xi0 9 nt~mVcoul 2 ) 1.1X10 coul (b) Owing to a larger value of E on the surface of a smaller sphere, charge would leak out rapidly. 29.35. (a) Kinetic energy acquired by the a-particle is /k= 9 j/=2<*M2)(1.6x 10~ 19 coul)(l.OX 10 volts) =3.2xlO~ w joules (A) Kinetic energy acquired by the proton is * a 9 K=eK=(l,6x 10* li coulXl.O X lO* volt) = 1.6x10-* joules (c) From (a) and (b) we find eV _L 2 Electric Potential 73 /L \2 2 A/P M2 '1 That is, vp > va The proton has greater speed than a-particle. SUPPLEMENTARY PROBLEMS 5.29.1. The charge on the surface of the conducting sphere of radius r is q~4it* 9 rV where V is the potential. a Ve Surface charge density, a= r-*- =- e ' 4 rcr 2 r _ (200 volt) (8.9 X IP" 12 couWnt-m 2 ) ~ (0.15 meter) = 1.2x 10~ 8 coul/meter 2 5.29.2. As the conducting spheres are far apart (10 meters), we can ignore the influence of one sphere on the other in altering the potential. The potential on individual spheres would be caused by the charge residing on a particular sphere. On the sphere with K= + 1500 volt, q=4Ke Vr _U500 volt) (0.15_ meter), _ _ - (9 X 10* nt-mVcoul 2 ) On the sphere with K= 1500 volt, =-2.5xlO~ 8 coul. xl(rcoul X 1U C UL S.29.3. (a) If the spheres are connected by a conducting wire, charge will flow from the smaller sphere (higher potential) to the larger sphere (lower potential) until the spheres acquire the same common potential. Let charge q be transfered from the smaller to the larger sphere. The smaller sphere will now have final charge tfitfo? and the large sphere will have final charge qt~ where q, is the initial charge on either sphere. Let the common potential be V. 4 itc. r t 6.0cm __!_ 12.0cm~ 2 74 Solutions to H and R Physics // whence - A.'XlO^oul^^ (6) The final charge on the smaller sphere is ?i=?o-?<,=(3xl(T 8 -l xiO' 8 ) coul=2xiO"" coul. The final charge on the larger sphere is ?j=9*+?=(3 X 1(T 8 -H X 1(T*) coul=4x 10"' coul The common potential finally reached is ~~ (9X 10 nt~mVcoul*) (2x 10~ coul) _ AA f = - ^7777^7 - , v - =3000 volts. (0.06 meter) S.29.4. Let the sides of the rectangle be a=5 cm and 6=15 cm. (a) The electric potential at corner B is v _ . a " 6 - -7.8X10* volt The electric potential at corner A is v gi ,-gj. _ V f-JJ + J\ 47ie 6 4 ice a 4 T5 \ ^ a ) ,n tnft . </ n\ /" 5xlO"" coul 2xlO"" 6 coul \ =(9xlO* nt-m f /coul*) ( ~~ - - - -u --^- - ^~- I 7 V 15X10^ meter ^5 X 10" 1 meter/ =6X10* volts. (fr) AP=^-PjB=6xl0 4 volt-(-7.8X!0 volt) 8.4 X10> volt Work done, ^^ A^=(3x 10^ coul) (8.4 x 10 volt) =2.52 joules (c) External wprk is converted into electrostatic potential 'energy since positive charge is moving from lower to higher potential. S.29.5. Let the charges be q t <7 t , q 3 and q 4 each being equal to q. The distance between any pair of charges is the same being equal to a.. The potential energy for the given configuration is Set Electric Potential 75 S.29.6. The potential at A due to the charges at B and C is K , ' (i + i 4 jcsA a a 2? where 0=1 meter. Let the charges at B and C be fixed and the remaining one be moving from A to D. At the mid-point D of BC, _ 1 D ~~4n&_ *q "~~~~ A Potential difference, 1 meter 1 meter Fig. S 29.6 2q ( \ a (9 x v 4rc e,, a -1.8X (LO meter) Work done in taking the charge from A to D is jf= g A F-(0.1 coul) (1.8 XlO' volt) == 1. 8 Xl0 g joules Energy supplied is 1 k\V = 1000 watts=1000 joules/sec. .". Time taken to move the charge from A to D is __ ___ "rate of supply of energy 1.8 xlO* Joules 1000 joules/sec = 50 hours. = l.8xl0 5 sec S.29.7. Suppose the density of lines of force increases in the trans- verse direction i.e. upward (along >--axis) in Fig. S 29.7. Consider a closed path in the form of the rectangle ABCD. Let the density of lines along BC be aj and that along AD be <r r The electric intensity z- 1 - and that along /</> will be .= J o * * , along J9C will then be . The potential difference between B and C is VicBid "~~ , where d is the distance BC. Similarly, the potential difference between D and A is . L c t us take a test charge q along 76 Solutions to tf and R Physics It the indicated path. Along AB and CD no work is done as the paths lie on equipotential surfaces. Work done in moving the charge from B to C is e D Fig. S.29.7 Similarly, work done in taking the charge from D to A is Therefore, the work done in the round trip ABCDA, is But a l > a 2 , by our postulate. Hence, Jf^O. However, because of the conservative character of the field W should be zero. We, therefore, conclude that our assumption is wrong. Thus, the den&fty of lines at right angles to the lines of force cannot change fo;/ an electric field in which all the lines of force are striaght parallel lines. S.29.8. Electric field near a long line of positive charge is The potential at a point P at distance r^ is given by r -lice, 1 where C is a constant of integration. The potential with the origin O is given by K.= ^ In a+C 2ne, -(D ...(2) -(3) The absolute potential V v at any field point P is given by the potential difference between P due to positive charge and the origin O at zero potential. Subtracting (2) from (3), F.-K- dln -(4) Electric Potential 77 The superscript refers to the positive line of charge. Similarly, for the line of negative charge where r 2 is the distance of P from the negative line of charge. The net potential at the field point P is given by the algebraic sum of the two potentials, Fj> + and V P ~. Hence, _, -^ J_l/~ A . T2 y^v 2 rce<, r l Since A and e n are constants, we obtain the equation of an equipotential surface by assigning some value to F, either positive or negative. Re-writing (6), we get, -= e ...(7) where C is a constant for any fixed value of V v . Now, the locus of points with a constant ratio of the distances to two lines is an equation to a cylinder. We, therefore, conclude that the cquipoten- tial surfaces in this field are a series of cylinders along each line of charge However, the cylinders are not concentric. Further, an equal negative potential Ogives rise to a cylinder of the same size but surrounding the negative rather than positive line of charge. Fig. 8,29.8 shows some of the equipotential surfaces in the xy plane. Fig. S.29.8 S.29.9. In the circular orbit of radius r lf the kinetic energy is and the potential energy t/ 1 = total energy is 4* ice. Hence, the .-(1) 78 Solution* to H and R Physics II As the centripetal force is provided by the coulomb force, we have Qq 4 ne r\ = CL- ... (2 ) 4 *e. /*! Using (2) in (1), and simplifying, Similarly, for the circular orbit of radius r a , we have 2 = L & The work W that must be done by an external agent on the second particle in order to increase the radius of the circular orbit from r l to r, is S.29.10. Loss of potential energy in moving from ^ to r 2 is r, r, "V Gain in kinetic energy is Gain in kinetic energy=Loss in potential energy =J-M 2 <2)(9 X 10 9 nt-mVcoul^Q. 1 X 1(T 6 coul) a (25 X 10~ 4 in -9 X IQ^m (2x I0" 6 kg)(9x 10" 4 meter j (25 X 10"" 4 meter) *=2.48x 10 8 meter/sec 5*29.11. (a) As both, the projected particle and the nucleus, are positively charged, the electrical forces are repulsive. If the aim is perfect then the particle will proceed head-ron towards the nucleus. As it does so it loses kinetic energy and gains potential energy. Distance of closest approach corresponds to the situation where the particle momentarily comes to stop. In that case the initial Electric Potential 79 kinetic energy is completely transformed into coulomb potential energy. When the particle has initial kinetic energy, its potential energy is zero as the particle is at infinity. Conservation of energy demands that /! -0) where r, is the distance of closest approach. vQ ^ _ q T Jg*t ' (b) This is the case of glancing collision. Nowhere does the kinetic energy vanish. Let v be the speed of the particle at P, the distance of closest approach at P from the center of the nucleus being R. Conservation of energy implies that ^r -.,(2) But by Problem, R=*2r 9 ...(3) Combining (1), (2) and (3), we get whence e <*A <* c" u * *u potential difference S.29.I2. Fjeld strength -jr-- -r distance moved Fig S. 29.11 * 1 o -H5 410 + 5 46 x ^ (meters) Fig. S.29.U 80 Solutions to H and R Physics II AK As -.(I) The minus sign is used in thd above definition so that is positive when AF/As is negative. Formula (1) is used to find cenetered around various intervals of distance. The calculated fields are plotted in Fig S.29.12 as a histogram. S.29.13. (a) Consider a differential element ds of the segment at distance s from the end closer to P. The charge associated with the clement ds is dq= A ds ...(1) The differential potential at P due to the charge element dq is </?__ - y (y+s) 4rce, (y+s) where use has been made of (1). The potential at P is obtained by integrating (2). ...(2) T jy^ _!_ " l l ' J (y + j) .s=0 ds J Fig S.29.13 A i / i , L \ -. In ( H -- I 4ne, V y I dy * yOH-D S.29.14. (a) Consider an element of length dx at distance x from 0, along the length of rod. The charge in dx is dq=Xdx=kxdx -U) dx -..(2) Electric Potential 81 The potential at P is given by dg ___ = k__ t *</*_ 4ne e VV 8 +x* 4 * e . J Jy*+x z Set z*=y*+x* and zdz=xdx in (3) F ^ fc f ?* 4e. J z (b) _1 1 J Allter ^=dE cos -(3) But </= +- * L 1 FigS.29.14 dg and Using (1), (6) and (7) in (5) JE- kyxdx /i ...(4; ...(5) ...(6) ...(7) 82 Sofa 'ons to H and R Physics II Integrating, L xdx Set ^+x f =z a xdx^zdz zdz _ _ Jky JL Z 9 4ire z L (c) F involves ^ alone. 30 CAPACITORS AND DIELECTRICS 30.1. (a) The equivalent capacitance of two capacitors in series is given b, C -.CiC. _(2xiO-/)(8xlO-f). Ci+C, (2xl(r/+8xl(r f) '- OXJU / The magnitude ^ of the charge on r =;>,,* r o . each plate must be the same. ' ^ ^=8^1 q=CV=*( 1.6X 10~ a f )(300V) | - 1 1 - 1 | - =4.8xlO~ 4 coul. The potential difference across 2 j capacitor is 10lkoul)/(2x IO- -240 volt Fig. 30.1 The potential difference across 8 /uf capacitor is K,=(jr/C 2 = (4.8 X 1(T 4 coul/(8 X \Q~* f )=60 volt (b) Total charge, Q=*q+q = 2q=2x4.& x 10' 4 coul = 9.6X10~* coul The equivalent capacitance of the Capacitors in parallel is ci\ n by C=C,+C,=2V+8|!if-10 ^f- 10-* f .'. The potential difference for each is F=0/C=9.6xlO- coul/10- f- 96 volts. The charge on 2 nf capacitor is ft =C,r=(2xlO-f)(96P) = 1.9x10 4 i->ul The charge on 8 ^f capacitor is ? 2 =C,K=--(8xlO- 6 f)(96 F)=-- 7.7x10 ! ., u! V) The charge is neutralized. <7i~7t=0 Also, F,-^^, 30.2. The potential is given by v " 84 Solutions to H and R Physics II Capacity of earth is Q 6.4x10* meter C= ~V - 4 * e '*-9 x 10 nt-m*/coul 2 =711XlO-f=7ll & 30.3. Q is charged and then connected to C a by closing switch S. The measured potential difference drops from V, to V. v V ' C * v cT+c, r C 1 (V -V) (50 volt -35 volt) whence, <?,=-* r? -(100 ^() T7ir 35 volt =43 30.4. Textbook Eq. 30 7 is C=- MKS units of e u are farad/meter. u t Farad coul __ __ cou\ Meter ~(volt)( meter) """" (joule/coul)(raeter) __ c _9}^ 2 __ C 5P]! ~~ joul-meter "" nt-meter^ 30.5. The capacity of upper capacitor is given by and that of lower one is given by For the series combi nation of capacitors,the equivalent capacitance is given by C C Using (1) and (2) in (3) we find, / ^ 1ZZJ' t, // ^ I / / t> ^ / s^% But, rfj Fig. 30.5 Capacitors and Dielectrics 85 a - 30.6. Since the effective area of the capacitor is that of interleaved portion of the plates only, the maximum effective area of each plate will be A. The neighbouring plates constitute a parallel-plate e A capacitor of capacitance, C=-^ ; and as ( D P^tes in parallel make up the variable capacitor the capacitor has the maximum (n-\)e A capacitance, C= - -3 - 30.7. The outer spherical plate is invariably grounded and contact is made with the inner plate through a small hole in the outer one (Fig. 30.7). The field at point P is caused entirely by the charge Q on the inner sphere and has the value The potential difference between the two spheres is given by Fig. 30.7 b Q(b -a) whence, C= ^ __ 30.8. Imagine that the capauior i> divided into differential strips which are practically parallel. Consider a strip at distance x (Fig. 30.8) of length a perpendiculur to the plane of paper and of width dx in the plane of paper, the area of the strip being <//<=< At the distance x, the separation of the plates is seen to be The capacitance due to the differential strio facing each plate it 86 Solutions to H and R Physics It Fig. 30.8 The capacitance is given by a a c = f dc = ( < =ea [ J* ] ac j(d+xQ) ' a }(d+x6) 39.9. (a) For the two concentric spherical shells of radii a and 6, the capacitance is If 6 a, than oft eio* and ba=d where we have set ,4 =4* a 1 , the surface area of the sphere. Capacitors and Dielectrics 87 30.10. Capacitance of the parallel-plate capacitor is ,,_ *^L * (**'). Charge, C=Cr= _ (8.9 X IP' 12 coulVnt- m a ) (0.08 meter)' (100 volt)* ~ (1.0XKT 3 meter) = 1.8xl(T 8 coul 30.11. As C\ and C, are in parallel, their equivalent is C t fC,. Now C s is in series with C t +C 8 . Hence, the equivalent capacitance of the combination is given by 1 1 "Crf C 3 (C 1 +C 8 ) (10 30.12. (6) The charge on C 8 is equal to that across the combination of Ci and C a in parallel. where F 8 and K! arc the corresponding potential difference. v ^il^^<li(5^i I5 v n\ K ^ C 3 " ^~" 4 '/rf "T Kl '"' Also, ^+^3- 100. -(2) Solving (1) and (2), y^2l volts. When the capacitor C 3 breaks down, the voltage across C l be- comes 100 volts.* Hence, change in potential difference is AKH100T-21)=79 volts. (a) Change in charge is (10 X 10^0(79 volts) 7.9xlO~ 4 coul 30.13. Let the effective capacitance between points x and y be fr Apply a potential difference V between x and y and let the effective capacitance be charged to q. Let the charge across C, and C, be ?i 88 Solutions to tt and R Physics It and <?j respectively. The charges across various capacitors are shown in Fig. 30.13. Fig. 30.13 The potential drop across C l plus that across C,j must be equal to the potential drop across C t plus that across C 8 .j But,' by Problem, Cj=C,=C 4 ==C 6 Multiply (1) through by C\ to get Add (2) and (3) to End .-(2) .-.(3) 30.14. (a) Five 2.5 fif capacitors in series would provide an equivalent capacitance of 0.4 ^f. At the same time each will be able to withstand 200 volts without breakdown, Fig 30.4 (a). . (b) Three arrays, each consisting of five 2.0 /if capacitors in series give the equivalent capacitance of 1.2 /if. At the same time each will be able to with- , ib) HI II IMMh-i 1000 v HMMHMH kMMHMM HMMHMH Fig 30.14 stand 200 volts without breadown, Fig 30.14 (6). 30.15. The equivalent capacitance of C\ and C, in series fa Capacitors and Dielectrics 89 The equivalent capacitance of C and C 3 in parallel is r.r. C=C +C = (10 /if) (5 30.16. (a) The equivalent capacitance of C, and C 3 in series is ^IS /~* i ^ /i r i ^ r\ v'-> A*i Q+C 3 (1 fif-f 3 txf) Similarly, the equivalent capacitance of 2 and C 4 in series is ^ _ QC 4 (2 4 ~ (2 The combination of C 13 and C 24 in parallel is C=C 13 +C, 4 =0.75 /f+ 1.33 ^f-2.08 /if Charge on the equivalent capacitor is 08xlO- f) (12 volts)-25xl(T 9 coul. The charge on C l and C 3 will be equal and is <7i=<7 3 =C 13 K = (0.75x 1Q- 6 f) (12 volt)-9x lO' 4 coul The charge on C t and C 4 will be equal and is <?*=9^C 4 K-( 1.33 x 1(T 8 f) ( 12 volt)= 16x 10' 8 coul. (6) The equivalent of C t and C t in parallel is The equivalent of C, and C 4 in parallel is As C lt and C 34 arc in series, the equivalent capacitor is given .by f)_ ? . f ~ M- The charge on the equivalent capacitor is C=CF=(2.1 X 1CT f ) ( 12 volt)=- 25.2 X blO' 8 coul. The potential difference across C l or C t is v v Q 25.2xl(>~ - e . , 3x|0 - n -8.4 volt. The potential difference across C, or C 4 is .. .. Q 25.2 X1Q-* coul ,, K,=F 4 =^= 7xio -_.. 3 . .'. The charges are Ci=C l !' l --<lxl(r t f) (8.4 voltj^g^xlO" 1 coul. 90 Solutions to If and R Physics - // <?a^C 2 K a =(2x 10~ 6 f) (8.4 volt) - 16.8 X 1(T 6 coul G 8 =C a fV=(3x 10~ 6 f) (3.6 voh)-10.8x iO" 6 coul C4=C f 4^ 4 =(4x 1(T 6 f) (3.6 volt)- 14.4 x 10" 8 coul 30.17. The given parallel plate capacitor is equivalent to a combina- tion of two parallel plate capacitors each of area \ A and dielectrics A^ and K 2 . The capacitance is then given by C/T t /-i AI \2 ^* / I **2 ^o \2 '*/ = C 1 +C 2 = = 7 1 ~f Limiting cases: / j\ IS YS Jt^ \l) A| A| A e, ^# rf () ^=^2=1 (for air) c= e ^ These are the expected results. 30.18. The parallel-plate capacitor may be thought of as two capaci- tors with dielectrics K and K 2 im series, their capacitance beirfg given by The combined capacitance is then given by Using (1) and (2) in (3) r= 2e ' A Limiting cases: (i) *,=*, Capacitors and Dielectrics 9l (ii) *!=*,= ! (for air N , A These are the expected results. 30.19. The parallel-plate capacitor may be thought of as an arrange- ment of two capacitators, one consisting of a dielectric K with thickness b and area A y and another with air gap (</ b) and area A. The combined capacitance is then given by With AK (for the dielectric) ...(2) ,= : 4 (for air) ...(3) o Using (2) and (3) in (1) Setting /l==IOOx 10~ 4 meter 2 ; </=1.0x 10" 2 meter 6=0.5 x 10~ 2 meter, and AT=7.0 r (7) (8.9 x IQ-^ coul 2 /nt-m 2 ) (100 X 10"'' meter') (7) (l.Ox 10~ a meter)-(0.5x 10' 2 meter)(7-l) = 15.6xl(T la f This is in agreement with the rounded off value 16 ppf obtained Example 5 of the textbook. When 6=0, C= e -^p, When A"=1,C=^ When ^b=d,C=^~ These are the expected results. 30.20. Before the slab is introduced the capacitance of the parallel- plate capacitor is given by the usual formula 92 Solutions to H and R Physics II After the slab of copper is introduced, the original capacitor if reduced to two in series each having a gap of | (db). Each has capacitance r r-r A 2 *o A Cl c *~R^r7=&~ The combination of these two capacitors in series has the capaci- tance 30.21. For a parallel-plate capacitor of dielectric K. Since to and ^4 are constant it is sufficient to find the ratio Kid in order to estimate the relative magnitudes of C. For mica, K/d~ 6/0.1 mm=60/mm For glass, K/d=7/2.Q mm = 3.5/mm For paraffin, K/d==2/lQ mm=0.2/mm. Clearly, to obtain the largest capacitance, we must place the mica sheet. 30.22. (d) Before the dielectric slab is introduced, the capacitance \^0 T" a _ (8.9 x IP"" coulVnt-m)(lQ- a meter 8 ) ____ =8.9 M" Thee capacitance with the slab in place is given by _ v , ._ 8.9xlO Kd-b(K\) __ (7)(8.9XlO-"coulVnt.m 8 KlQ- meter') (7)(10-> raeter)-(0.5x 10~ a meter)(7-l) - 15.6 Xl0-"f= 15.6 MMf- (a) Charge on the capacitor before the slab is introduced fi.=C.K=(8.9 x 10-"f )(100 volt) =8.9xlO~ w coul Capacitors and Dielectrics 93 Charge on the capacitor after the slab is introduced is l(T 12 f )(100V)=1.6X 1(T' coul. (6) The electric field in the gap before the slab is introduced is C, ^ __ 8.9 X10~ 10 coul __ * t A (8.9 x 10-" coul 2 /nt-m 2 )(1.0x 10~ 2 meter 8 ) = 1.0xl0 4 volts/meter The electric field in the gap after the slab is introduced is = J? = _ 1 6 X IP" 1 * coul _ ~~* A\t (8.9 X 10~ 12 cou! 2 /nt-m 2 )(1.0 X KT 2 meter 2 ) = 1.8x10* volts/meter. (c) The electric field in the slab is given by ., #o' 1.8 X I O 4 volts/meter ^x^, u/ . K *** - 7 ---- ^ volts/meter. 30.23. Assume If =5.4 for mica (V) The free charge on the plates is 9 -CK=(100x I0~ la f )(50 volts)=5x 10^ coul (a) The electric field in the mica is _ 5X10~ coul _ "" (5.4)(8.9 X KT 12 coulVnt.m f )(1.0x 10~ 2 meter 8 ) = 1.04X 10 4 volts/meter. (c) The induced surface charge is =(5xlO-coul) 1-5- =4.lXlO-coul The induced charge of 4.1 x 10~* coul appears next to the positive plate. 30.34, (a) Electric field E=- -~-r- 94 Solutions to H and R Physics -II Dielectric constant, AW -JL A t. EA (8.9 xlO"? coul) (8.9X l<r u coulVnt-m*)(1.4xlO volts/meterJd.OxlO" 1 meter 2 ) =7.14 (b) Magnitude of the charge induced on each dielectric surface is -=7.7X10" 7 coul. 30.25. The capacitance of a parallel-plate capacitor is Dielectric strength == 18 X 10 6 volts/meter. Electric field strength =4000 volts/meter. Setting dielectric strength equal to the electric field strength, we have strength we have, a </= ,4^? A V fl olt f , - -2.22 x 10~ 4 meter 6 f 18X10 6 volts/meter From (1) we have ^ C d (7xlO"xlO- 8 f M2.22X10" 4 meter) e K^' "'"("8TxlO-"c6uiVnt-m a )T2.'8)"~ =0.62 meter 2 . 30.26. The electric field for a cylinder capacitor is The energy density (energy/unit volume), where use has been made of (1). The energy stored between the coaxial cylinders of length and radius R and a is R U^iudv lu(2*rl)dr - (2) a Capacitors and Dielectrics 95 where </v=(2nr dr) /, is the volume element. Using (l)in (2) R n* u -jf- ( dr __ I* . ~ . I * , 111 " ' 4* e / J r 4n e, / a Similarly, the energy stored between the coaxial cylinders ol radii b and a is Therefore,' i ^^ , or In =2 In - =In fl a ^ a* and or a a~ 30.27. Radius of the metal sphere r ^5cm=0.05 meter Electric field at the surface of the sphere is 4Tr fl r* The energy density at the surface is 1 ^> <r But, v C=4n c, r q 4n t f Vr 1 e,^'_ 1(8.9 x IP"" coul'/nt-m^fgOOO volts)* "~ r 2 r (0.05 meter ) 0.114 joule/meter* 96 Solutions to H and R Physics II 30.28. (a) Capacity, C= -*~ Initial potential difference, V l ^ ^-= q v^ A New potential difference, e A t (b) Initial stored energy '~ 2 l Final stored energy (c) Work required to separate the plates ii ~ 2a 30.29. The energy on the parallel-plate capacitor with plate separa- tion x is ^ 2 C i (c. If the plate separation is increased to xdx 9 then the energy becomes Therefore, work to be done to increase the separation of plates through dx is - 2e A a 1 :. Force F=~^ A 2 e A 30.30. In textbook Example 5, ^ = lxlO- meter 1 , rf=*1.0 meter, 6=0.5 X 10"" 1 meter, /fT=7.0 and K =100 volts 1 1 / V \ f For the air-gap energy density, 110== y eo ^o* 235 y o ^ -^r I Energy in the air gap, t/o*t*o x . , where \dA is the volume corresponding to the air gap. Capacitors and Dielectrics 97 r; _*0 AV* t/0= TV- 4d _ 1 (8.9 x 10~ u coul/nt-m)g x 10"' meter*)(100 volts)* 4 (1.0X10-* meter) = 2.2X10-* joule For the dielectric, energy density, =-pru . K. Since volume is the same as that of air gap, Energy, tf = j- /=( y ) (2.2 x 10'* joules)=3 X 10~ joules (a) Percentage energy stored in the air gap is JOO tf 0== ( 100X2.2 XlO- joules) 0/ U+U (0.3 x 10-* joule + 2.2 x 10- joule) /0 (6) Percentage energy stored in the slab is 100 U_ (100)(0.3 X IP"* joule) . J9/ U+U (0.3 x 10~ joule + 2.2 x 10" 8 joule) /0 30.31. (a) Energy stored is C/=|CF-K100X10- U 0(50 volts)* - 1. 25 xlO' 7 joule. (b) t ( V \* Energy density, "o^-^f ~j ) Since the gap d is not known, M O cannot be found out. 30.32 (a) /!= i C,JV=i(2 x lQ- f ) (240 volO'==5.8 x 10~ joule. C/ a -i CjK^-KSxlO'^ f ) (60 volt) 4 = 1. 4 X 10~* joule. (b) f/!=i C 1 P' 1 =-i(2xlO-f ) (96 volts)* =9.2X10-' joule. l/=* C,V=4(8 X 10' 8 f ) (96 volt) 2 -37 X 10~ joule. (c) As ^, = ^,=0, C/i=C/,=0 In (a) the capacitors are in series and the energy stored is maximum. In (b) the capacitors are in parallel and the energy stored is less. In (c) the charges are neutralized and no energy is stored in the capacitors. The energy is used up in heating up the connecting wires. 30.33. The electric energy stored in the soap buble is r/ = iV_ ." _ 2 C ~2 98 Sot tions to Hand R Physics II where we have used the formula C=4 ne /? for the capacitance of a spherical capacitor. Due to mutual repulsion of the charged surface, the radius increases to R leading to a decrease in energy. The new energy is U-, "' ' 8 0a / j J\ .'. Decrease in energy, A U^l^ U^^ I p - } o Ke \/c RJ or Now, the work done in expanding the soap bubble at constant pressure p is where dV is the change in volume. dV^ ~ Equating, Simplifying, we find q= y - (8.9 X ID' 1B coul'/nt-m*) (1.01 3 X 10* nt/m 2 }(0.02 m)x (0.021 m)|(0.02 m)-f (0.02 m)(0.021 m)+ (0.021 m) ==7.lxlO"coul. 30.34- The equivalent capacity of two capacitors in parallel is Total energy in the system is f/=i CF*= J(6X 10- (300 volt)=0.27 joule 30.35 The equivalent capacity of n capacitors each of capacitance C in parallel is C-C -(2000)(5.0xiQ- i f)O f 01 f Capacitors and Dielectrics 99 Energy l/= J CK*=* (0.01 f) (50,000 volt)= 12.5 X 10 6 joule cost of charging is 2c/k W-hr, or 2e/3.6 x 10" joule .'. Cost for charging to the extent of 12. 5 x 10 joule is (2c) (12.5 X 10 joule/3.6 x 10 joule) = 7 c. 30.36. Q- K-=100 volts The equivalent capaci- tance of the arrangement in which C 8 is in series with the combination of Cj and C 3 in parallel is 1 C2 (4 f *f)(lO>f^5 M Q ,uf+5 . M Fig. 30.36 The total charge, 0=CF=(3.16x i(T f) (100 vo!t) = 3.16X 10" coul. Charge across C 3 is # 3 ^3.2X 10~~ 4 coul. ' (b) The potential difference across C 3 is It follows that ^-^,=(100-79) volt=21 volt. (a) ft^QK^dOX 10~ 8 0(21 volt)=2.1 X 10~* coul (5X10-' 1 f ) (21 volt)=1.05xlO t * coul ~ 4 coul =\(\Q x 10~ f )(21) I =2.2X10- 8 joule 3 -|(5X 10' f )(21) 2 =l.l X 10~ 3 joule lO- 9 f )(79) 2 = 1.25 xlO"" joule 30.37. C 2 =5 K=100 volt The capacitor C 3 is in parallel to C l and C t in series. The equivalent capacitance is QC2 /f ^ *'*-** =7.3 Mf. The total charge is (7.3 X 10- f )( 100 volt) 7.3X10"* coul. C 2. Fig. 30.37 100 Solutions to Hand R Physics II (a) ? 3 =C,K=(4 x 10~ f )(100 volt)=4 X 10" coul 9l ^ 2 =0-0 9 =(7.3x 10--4x KT) coul = 3.3 XlO~ coul lh \ v -i _.3.3xlQ-coul_ 33 u (6) K I- IoxTo-f - 33volt v _ <? 3.3 X1Q-* coul Vz ~C^ 5xlO-f 6 K 8 = 100 volt (c) t/! = i C,JV=i (10X 10~ f ) (33 volt) 1 - 5.4 X10~ joule V z =\ C t Vf=\ (5x 10~ f ) (67 volt)=l.l x 10~ joule 1/3= i C 3 K 3 =i (4X80~ f ) (100 volt)=2x 10- joule. SUPPLEMENTARY PROBLEMS S.30.1. Capacitance is given by r 3 C ~~ V F< r an isolated sphere of radius r, 4ne ft r C=4 ier Let the spheres be oppositely charged. ,,-._..- 4 7te,,r Potential difference for the system of two spheres y = Y+V-=~- q - 2 Tte r The corresponding capacitance is then C'= . =2e r= -i-C. S.30.2. (a) Let fi=fia+y .-(I) where the subscripts a and 6 refer to spheres of radii a and b, respectively. The charge will be shared in such a way that the spheres acquire common potential, Capacitors and Dielectrics 101 whence, bQ 9 =*aQi>. ...(3) Solving (1) and (3), ...(3) (fr) From (2) we have Ca=4e F* ..(5) Q6=4ne, Va ...(6) Adding (5) and (6) S.30.3. Let ^ be the charge on each small drop of radius r. Then the potential of each small drop is o 4nc r The charge on the large drop of radius R is, Q=Nq. The potential of the large drop is --(I) Since the large as well as the small drop will have the same den- ity, the volume of the former must be N times as large as the volume of the latter. Dividing (2) by (1) ^r- where use hat been made of (3). 102 Solutions to If and R Physics II 8*30.4. The charges on the two capacitors before and after the switch is closed are shown in Figures S.30.4 (a) and (&) respectively. As no charge flows through the capacitors, the charge on the plates must be the same before and after the switch is closed. (a) (b) Fig. S-30.4 / / /t \ <7 a 9t= = ^2 9i \A; __ / , /<y\ Further, in the absence of an emf, the voltage across the combi- nation of the capacitors must be zero. +!-+ =0 *"1 *- C 3 Solving (1), (2) and (3 for q^ . q^ and q 3 ', we have (3) -C 8 C 1<?3 ' _ (CC,+C 3 C.) S -C 3 C 1 .? 2 - S.30.5. Let 9 ? be the charge on C t when the switch is thrown to the left. 0o=C,F ...(1) When the switch is thrown to the right as in Fig. S.30.5 the ini- tial charge q is shared among the capacitors such that .-(2) c, JJ Capacitors and Dielectrics 103 Further, the potential difference across Q is equal to that across the combination of C 2 and C 3 , Solving (2), (3) and (4) for <jr lf <? 2 and <jf 3 and using (1), S.30.6. If the dielectric is present, Gauss' taw gives S=?-V=-r ...(1) where q' is the induced surface charge, q the free charge, and K is the dielectric constant. Construct a Gaussian surface in the form of a coaxial cylinder of radius r and length /, closed by plane caps. Applying (1), The surface contributing to the integral being only the curved surface, and not the end caps. From (2) we find, E= ? ...(3) 2 The potential difference between the central rod and the surroun- ding tube is given by b b b q dr =- f E.</r = 2 a _ "~2 IK a The capacitance is given by r -i 2 *U O /A: ~ " S.30.7. (a) If /< is the area of the plates and d the distance of ration, then the capacitance before the slab is inserted is found from 104 Solutions tottandR Physics It /j\ i- v o 120 volt 1rt4 ... . (A *o= - = -. =10* volt/meter (e) . 2xlo -. meter volt/meter *t.o =(10 volt'meter)ro.012 meter -0.004 meterf 1 88.3 volt. ... _ q C K /1A .(120 volt) , (6) C= t""V" (10Bj (88.3volt) =1 (c) Before the slab is inserted, ^CoKo^UO farad) (120 volt)=1200 e coul. After the slab is inserted, .6 farad) (88.3 volt)=1200 e coul. (10 farad) (120 volt)-J(13.6 c farad)(88.3 volt)* . 9 XlO joules. 31 CURRENT AND RESISTANCE 31.1. (a) Charge flowing =< (current) (time) q=it =(5 amp) (240 sec) => (200 coul (6) Set q~ne> where =1.6xlO~~ 19 coul/electron. where n is the number of electrons flowing, each of charge e. g _ 1200 coul _ TC^.A*! , n= - = rTTTTiv^itt ---- 1 71 I ==7 -5 X 10" electrons. e 1 .6 x 10^ * 9 coul/electron 31.2. Since positive ions and electrons (negative) flow in opposite directions, the effective current in the direction of the flow of positive ions is given by the addition of both the components. t t = (1.1x10^+3.1 =0.67 amp 31.3. (a) The resistivity p may be written as VII >=ilA For iron and copper wires, V, i and / are the same. A p But X^wr 1 where r is the radius of cross-section. .'. r ( iron ) / _P (iron)~_ /I r (copper) V p (copper) V 1 . x m> M 2 4 .7 x I0t~ ohm-m VII Since K and / are the same, but p being different for iron and copper wires, the current densities cannot be made the same with any choice of radii. 106 Solutions to ti and R Physics tt 31.4. " ' *i Fig. 31.4 31.5. Let the charge density be a (coul/meter 1 ). Charge conveyed by the belt to the sphere per second is /=- = y-(area of belt)=( a ) (width)(length)/time =( a ) (width)(speed of belt) / _ _ __ 10" 4 amp _ ~" 0== (width) (speed) ~" (0.5 meter)(30 meter/sec) =6.7xlO~coul/m. 31.6. (a) Resistance of aluminium rod, P/ _ (2.8X1Q-" ohm-m)( 1.0 meter) _ R -~A ~T5^xlO ;: ^eTe7jS -- L1 X 10 ohm (fr) For the circular copper rod, A D" 1 , where D is the diameter. n - R ~ A Set R 1 . 1 x 10~ ohm, then D= f 1 P 7 _ f W (1.7 X 10~ ohm-m V KR V' ()(UxlO- J =4.4 X 10~* mcter=4.4 mm. -meterXl.O meter) ohm) Current and Resistance 107 31.7. The resistance of the original wire is ^.eA.W-ii ... (1 ) 1 ^i ^1/1 v where v is the volume of wire. Since density of the material of wire is constant, it follows that v is also constant. The resistance of new wire is jib-ei 1 ...( Dividing (2) by (1) -{4 > *M '1 But /,-S/! ohm)=54 ohm i 'i 31.8. Resistance of copper wire is _ Pi/ _(1.7xlQ- ohm-m)(10 meter) ^J Ml^xTo^meterT 2 '" ~ 5 ' 4X 10 hm ' Resistance of iron wire is ff _Pi / _(.0x 10^ 7 ohro-mHlO meter) . |A _, ^" -- ~- ~~" s3I - 8x|0 ohin - (c) The potential difference across copper wire is y ^^J^i^ _ (100 volt)(5.4xlQ- ohm) 1 /?!+/?, (5.4X10~* ohmT3 1.8X10-* ohm) = 14.5 volt The potential difference across iron wire is V = K/?I = (lQOvolt)(31.8xl(Tohin) /?,+/?, (5.4xlO-*obm+31. 8x10-* ohm) =85.5 volt. (a) Electric field for copper is r Kj _ 14.5 volt f ._ . . l== T "loiter 1 - 45 volt/meter. Electric field for iron is r K t ,=,85.5 volt ef . . i "7T loiter -8-55 volt/meter. 108 Solutions to H and R Physics It (b) Current flowing through the composite wire is ,. = _ 100 volt ,,. R (5.4 x 1(T ohm+31.8 X 1<T ohm)"* 2 ' 67 amp =8.5 xlO 7 amp/m*. = 1(0.55 Xl0^meter)(2.87xl0- f ohm) ___ _ _ =6.8 X1(T ohm-meter The materials is nickel (Textbook Table 31.1). (a) Resistance, *=4 ,= (6 - 8x 1(r> ohm-meterKl.Ox 10- meter) wr a *( 1 .0 X 10"" 1 meter)' = 2.17 Xl(T 7 ohm 31.10. Cross-sectional area 4=7.1 in 1 =4.5 8 XlO"" 1 meter 1 Length of the track, /= 10 miles =1.61 X 10 4 meter . A n p/ (6 x 10" 7 ohm- meterX 1.61 X 10 4 meter) Resistance, /?=- -^ = - j-^ 1/ ._/ -- ^ - 4 4.58 X 10 8 meter 1 =2.1 ohm. 31.11. (a) The resistance U t at any temperature T is found from the relation where /? is the resistance at 0*C and <x is the temperature coefficient of resistivity. From textbook Table 31.1 we find for copper, = 3.9 X10~ B . By Problem, /?=2/? At temperature 256C, the resistance is doubled. (b) As <x is independent of shape and size, the result of (a) is valid for all copper conductors. 31.12. (a) For iron 20 C is P 1.0 x I0~t ohm-in and -5 x HT* we have, e Current and Resistance 109 The resistivity of iron at 0C is p 1 .0 X 1(T 7 ohm-m - rt1 ^ f _ - SS5 - 91 x 10 For carbon at 20C p '=3.5xlO- ohm-m and <x' = ~-5x 1(T 4 . The resistivity of carbon at 0C is 9' 3.5X10-' ohm-m _ 354x - 1(ri ohm ~ p < ~ ''^ 1-20X5X1CT* - 3 ' 54X1 ohm m If the resistance of iron is R and that of carbon /?' then the resis- tance of the composite conductor is where / and /' are the total thickness of iron and carbon disks, respectively. Change in resistance MP , J^^ __ AP = P / Ap' Po'a'AT Po'' _ (0.91 X 10~ 7 ohm-m)(5 X KT 8 ) __ ^ 9 fi , a (3.54 xlO-'ohm-mX- SxlQ-*) ^ OX1U This is also the ratio of thickness of individual disks as the number of disks of either material is the same. (b) As the current is the same in carbon and iron disks, the ratio of the rate of joule-heating in a carbon disk to that in an iron disk is (3.5xlO~*ohm~m)(2.6xlO~ 2 ) __ i*R pl/A ~ p/ ~ ( 1.0 x 10-' ohm-m) 31.13. The temperature coefficient of resistivity for copper is 3.9 XlO-*/C. /. Percentage change in R is (3.9 X 10~) (100)=0.39%. Coefficient of linear expansion of copper is 1 .7 X 10~ /C. .*. Percentage change in length / is (1.7xlO-)(100)=0.0017% Percentage change in area A is (2) ( 1.7 X 10-*)(100) =0.0034%. From the above it is seen that percentage changes in / and A are by far less than that in resistance with the variation in temperature. 110 Solutions to H and R Physics II We, therefore, conclude that for the calculations of variation of resistance with temperature the changes in dimensions can be safely- ignored. 31.14. If /? 20 , /? and RT are resistance at 20C, 0C and TC res- pectively, then *n=*o(l+20) ...(1) RT=R*(l+T*) -.(2) Dividing (2) by (1) 1 +20 a For copper, the temperature coefficient of resistivity is 58 ohm 1 +(3.9x 10~ 3 ) /? 2 o 50 ohm 1 +20 x 3.9 X 10" Solving, we find the temperature, r=64C. 31.15. The resistance is found from the relation, R y The plot of R against V for the vacuum tube is shown in Fig. 31.15 (a) and for the termister in Fig 31.15 (&). (Q) \ 20 100 200 300 V-*(volts) Fijj. 31.15 31.16. The drift velocity is given by 30 ne Where j is the current density, n is the number of electrons/cm* and the electron charge. Current and Resistance 111 If # is Avogadro's number, d the density and M the atomic weight, then number of free electrons/cm 8 , (6x 10 aa atoms'moleMg.O gm/cm)(l electron/atom) _ M ~~ 64 gm/mole =8,4X10" electrons/cm 3 . JL ___ 1.97X10"" 1 amp/cm 2 r ne (8.4 X 10" electrons 'cm 3 ) ( 1 .7 X 10~ w coul) = 1.5 xlO" 11 cm/sec. 31.17. Rate of energy transfer is, _ P (100 watts) . . . R -f =OvA --- ci = 1 i ohms. j 2 (3.0 ampj* 31. 18.- Length, /=100 ft=30.48 meter Radius, r=0.02 in =5.1 x 10~ meter Cross-section area, A=ttr z 'n (5.1 x 10~* meter)* = 8.17 X10~ 7 meter 3 Resistivity for copper, p=1.7x 10~ 8 ohm-m. n ?i (l.7xlO~ 8 ohm-m)(30.48 meter) /?= 1 -, = ---- (8.17x10'' meter) -- ==0.63 ohm. V 1 .0 volt , , Current, ,- -^ = __-!. 6 amp. '^ ^ t ... ^ 16 amp (6) The current density, j- - - -- (c) Electric field strength, JS== = 1 .96 x 10 e amp/meter 2 V 1.0 volt / 30.48 meter 3.3 XlO~ a volt/meter r a do volt) 8 (d) The rate of joule-heating, P=~r = ^'^ ' ^t- 6 watts * /( 0.63 ohm 31.19. Radius of wire r^O.05 in1.27x 10~ 8 meter. Cross-section area of wire, v4=*r 2 = (1.27 X10~ S meter) 1 = 5.06xlO"" 6 meter 2 . Length of the wire, /=1000 ft=304,8 1 12 Solutions to H and R Physics II (a) Current density, 7=^=^7^ rf mp =4.94XlOamp/meter a (A) Electric field strength, E=pj =(1.7X 1(T ohm-m)(4.94x 10 amp/meter 11 ) = 8.4 xl(T 2 volt/meter (c) Potential difference, V=El =(8.4 X 10~ volt/meter)(304.8 meter) =25.6 volt (d) Ra^te of joule heating, P=Vi (25.6 volt) (25 amp) =640 watts , 31.20. (a) Heat absorbed per second=(500 joules)(80%) =400 joules _ 400 joules _ 95 - = - A v * i 7 i "7J.U v*l 4.186 joulcs/cal Heat required to raise the temperature of water from 20C to 100C is H^mc AT -(2000 gm) (1.0 cai/gm) (100C-20 C)-16x 10 4 cal. Time required to bring the water to boiling temperature Heat absorbed by water ~~ Rate of absorption of heat = ^/r ;n~ =1670 secs=28 minutes. 95.6 carsec \ (b) Heat required to boil half the water i.e., 1 liter or 1000 gm, H=mL=(mQ gm) (540 cal/gra) = 54x10* cal. Time taken to boil half the water away _ Heat required for vapourization ~~ Rate of absorption of heat 54Xl0 4 cal .. AQ A . . 4 = ?^~z n =5648 sec=94 minutes. 95.6 cal/sec 31.21. Power, P=^ V* 500 watt V* (HOvolt^ 1 Resistance at 800 C, * ioo = = 24 ' 2 hm Current and Resistance 113 Resistance at 0C _ /?8oo 24.2 ohm . ~l+ Ar ~ H-(4xlO-*/C)(800C) " I8 ' JJ onm Resistance at 200C, Rw-R* (l+* A T) -=(18.33 ohm) [l+(4xlO-/CX200C)]-l9.8 ohm Power dissipated at 200C, F _(UOvolt) 1 == 31.22. (a) Current, i=/te, Where rt is the number of deutrons of charge e striking the block. - Tk vxiA-ift ~-=9.4x 10 18 deutrons per sec. 1.6X10 lf coul f If the deutrons are completely absorbed in the block then the energy dissipated per sec=nK, where K is the kinetic energy of each deutron. Total energy available per sec =nK = (9.4XlO u/ sec)(16Mev) = 1.5X10" Mev/sec - 1 .5 x 10" Mev/sec .*. Heat evolved/sec=(1.5 x 10" eV/sec)(1.6 x 10~ w joule/eV) =240 joule/sec =240 watts. 31.23. (a) Power, P= ^ J\ .: Resistance, *--" V '" =26.45 ohm Percentage drop in power, (6) With the drop in power, temperature as well as resistance would decrease. Therefore, the actual heat output ?' will be larger and so will be smaller than that calculated in (a). 1 14 Solutions to H and R Physics II . I 31.24. Power P = Power per unit volume, ., = A = iV/d-(!p P V IA ~ A* where VIA is the volume. E . . E Also p= i.e. j J P SUPPLEMENTARY PROBLEMS S.31.1. (a) Let N a-particles strike the plane surface in time t sees. Then the current, .__Ng_NOe) ,___ __ it _(0.25X lQ- amp)(3 sees) la 2* (2)(1.6xl<r"coul) - Z - JX1U Speed of a-particles, v= f^ = f (2) (20X 10" eV)(1.6x A/ m V 6.68X10"* 7 kg =^3.1X10 7 meter/sec. Number of a-particles crossing a given plane area per second, ___J_^ 0.25 XlO" amp q (2)(l.6xlO-coul) =7.8 XlO 11 a-particls/sec. Number of a-particles per meter length of the beam Hence number of a-particles per 0.2 meter length of the beam =(2.5 X 10 4 per meter)(0.2 meter)=5tXX) a-particles. (c) By definition A particle of charge ze in falling through a poten- tial difference of 1 volt gains energy equal to z electron volt. For a-patricle z=>2, hence, the potential required to accelerate clcs to 20 MeV is 20xlOeV 1nf . V= ~ = jo volts, Current and Resistance { 1 5 U 3 | O S *5 di a 3 ^rt o w 3 rt"^ JH S 4- S o o "*5 c ,2 x! s 04*3 E |j ^ x o *Q t- 1 tJ IM S D 1? 1 feU|i 1 1 fi 4> Q *^ t> V* QL 4^* ^ r^ ^rt C O ' ~ O U J3 h C rj C3 si ctl ri_ ^ *" ^S o ^_u^>^c 00 ^n o J> |s/.22.2c?/ a ^ 0*5 c^ S *s.o-^O^ i -o *O a. tg M ^c c 53 S * E i I* 1 o S g IS D C3 *- II S ^ o S3 B i O (X ^ -^ ^ 2 1 53 * <u a "T3 * nj CD *rt O ^ --^ Z* O *3 i -- i> !T 2. S + 3-S 5 .S i 35 e J2 O .P"~ (^ ^\ % A> Co n*^ </) ** o| 1 o f> 2l U S ^ 5 co U ^ Oft JS t 3s a >! .- * *** i> *2 fS ^ i IM 50 I o o C C3 "3 o o w< || 1-4 2 -3 ^ S S ^ S 3 ^ a| -s Z3 C3 - <L> -fr a o v- ^* S"2 Ji ^ o XS o^3. ^3- c^ }! hi 1 *i 1 ! ^ .-2 rt V .^ - e & 8Si> X^SS c -9 a 2 jil ^ III a I OU V B 41 *o ll W O ' C3 o ^ a B ^ 9 o 6 ^ =* go ^ g & | o^ Ona-s Ot2 rt> ^>,og .2 a ^^S3>, If 1 I? 11 i 111 wt5 P& ^qa o3u U{J3<&8 3 S ^i ^ s^ i> >S5 1 16 Solutions to H and R Physics 11 s U 93 u* O 1 u, cy on O E S i.t: g o cu ex o 31 - a s lis x: ^ ^ 2 t> - <u CO J- 05 *o a cd CO o cc: oo O O- Cu .ti TD to -" 3 o t^: .iu^ > o o " 00 -~ CO 2 '^5 I te-o ^ 5 S *g 2 ^ 5 *> u- ^o.H s s 8 i- O O <u o - w a c S 87, bo g? CL-0 <L> ^> a. a a Uj o c <t> g 2 O o : p - ^ to .ti o - X / .22 C -o 11 P g ^* . o <^ 2 O '5 o - II pj o 00 Current and Resistance 1 1 7 5.31.3. As the conduction electrons get accelerated, sooner or later, they suffer loss of energy following collisions with the ions in the lattice of the metal. The kinetic energy of the electrons is dissipated into heat by way of imparting energy to the lattice vibrations. The conduction electrons being part of the conductor can only exert internal force on the conductor and therefore this can not give rise to a resultant force on the conductor. 5.31.4. (a) Consider an elementary disk of radius r and of thickness dx at distance x from the truncated end and symmetrical about the axis. Then, The resistance for this volume-element is ...(1) where use has been made of (1). 1 P/Vx * [la+(b-a)x]* o Set, y=la+(b a)x dy(ba)dx So that, bt at =__ f d y = -L(\- - it(ft-tf) J y* x(b-a)\al Simplifying, -> (6) For a==b, (3) becomes J?- p/ p/ *"n^ s ~A with 4=ica 2 . S.31.5. (a) The resistivities of the two wires are given by RA A (40 ohmKO.l meter 2 ) _ n P/i= : = - -TT ---- - ------ =01 ohm-meter. / (40 meter) (20 ohm KO.l meter 2 ) Anc , = - 7-177 - - : - =0.05 ohm-meter. (40 meter) 1 18 Solutions to tt and R Physics It (b) and (cf) : The resistance of the two wires in series is 4Q ohm+20 ohm=60 ohm. Current, /=- = T==1.0 amp R 6U ohm The potential difference across wire A is =(\ amp)(40 ohm)=40 volt. Therefore, the field is ^===r==~--==1.0 volt/meter / 40 meter ' The potential difference across wire B is Kfl= //?*=( 1.0 amp)(20 ohm) =20 volt Therefore, the field is !B= T^M - l"^ volt/meter. (r) The current density in each wire is 1.0 amp IA >-* " oT^te^ " 10 S.31.6. (a) Power, />=Ki r- . P 125 watt in 0-7 . . Current, , = _- - - . =10.87 amp. (6) Resistance, fl=~= J 15 volt =10.6 ohm / lu.o/ amp (c) Power, P=1250 watt=1.25 kilo watt. In 1 hour 1.25 kw-hr energy produced is 1.25 kw-hr=(1.25 kw-hr) ----lOTS kcal (1 kw-hr) S.31.7. At equilibrium temperature, Rate of heat radiated= Joule heating a,r 4 -r o )(2Tcr/) = i''/? ...(1) Where T is the temperature of iron wire, r =(273-f 27)=300 K that of surroundings, 2itr/=surface area of wire (neglecting the area of cross-section of wire at the ends), i== 10 amp is the current, R the resistance, s=5.67x 10~* watt/(meter a )(K 4 ), and r=0.5X HP 1 meter is the radius of wire. ...(3) Current and Resistance At20 e C, p (ro+20 ) e =Po(l+20) -.(4) Divinding (3) by (4), p r P(r +20) 1 +20a Pr = Combining (1), (2) and (5), and using ^ 10 '' ohm-m and =5x 10~/C and solving for T, we find 7=670K =(670-273)C =400*C. 23 ELECTROMOTIVE FORCE AND CIRCUITS *4 * 11 4 D v 6.0 volt t ^ u 32.1. Resistance /?= = r7; =1.2 ohm i 5.0 amp Chemical energy is reduced due to joule-heating and is given by Pt=i*Rt*=(5 amp) 2 (1.2 ohm) (360 sees) = 1.08X10* joules 32.2. Let the resistance of the original circuit be R and a potential difference Kbe applied. Then ^=1=5 amp With the additional resistance of 2 ohm the current drops to 4 amp ,. ,=4 amp Dividing the two equations, 5 . B . = - , whence /?=8 ohm. 32.3. Potential difference, K^AMS amp) (8 ohm)=40 volt The new resistance of circuit is /?,== (8 +0.05) ohm or 8.05 ohm. The current drops to V 40 volt , n , A /2==~^ = Q Ag .-^ 4 -969 amp. 2 R 2 8.05 ohm v :. Change in current A/=/ 2 /i= (4.969 5 '0)= 0.031 amp V. Percent change in current, 4^-xiOO=(^ 31 amp )xlOO /! V 5.0 amp / = -0.62%. TC 32.4. (a) Current, /= p Set =2.0 volt and r= 100 ohms, Fig. 32.4 (a). Then Electromotive Force and Circuits 121 Fig. 32.4 (a) Curve (0) in Fig. 32.4 (b) shows the plot of current / as a function of H over the range to 500 ohms. (b) The potential difference across the resistor R is 2 /{ where use has been made of (1). Curve (b) in Fig. 32.4 (6) shows the plot of potential K across the resistor R as a function of R. (c) Curve (c) in Fig 32.4 (b) shows the plot of the product P~Vi as a function of R. This plot shows the variation of power P with the external resistance R. 2X10 1X10 2-0 1X10 - 1-0 "o > > 100 200 300 400 500 R (ohms) 100 200 300 400 500 P (ohms) Fig. 32.4 (b) Curve (c) in Fig. 32.4 (6) corresponds to the plot of power P delivered to the resistor R as a function of R. It is seen that P has maximum value at /?=100 ohms, a value which is identical with r, the internal resistance of the battery. 32.5. (a] Current, '= -(I) Power delivered, ...(2) For maximum power set ^ir=0. oK ' ' (R+rV (R+r)* whence, R=r (b) Maximum power is obtained by putting R=r in (2). 32.6. (a) Current, /= - Power p r =/"/?/>-/? =[(1.5 volt) 1 (0.1 ohm)/(10 watts)] 1 '* -0.1 ohm =0.05 ohm. (b) Potential difference across the resistor, watt) (0.1 ohm)=l volt. 32.7. (a) E! and 2 are in opposition. Effective emf is given by =,-! _ . Current, ,= 0.001 amp= ~~ K (R+6) ohm whence, -R=994 ohm. (b) Rate of joule heating in R is P=,-/?=(0.001 amp) 1 (994 ohms) =9.94 X 10~ watts. 32.8. (a) Current, i= j^ Power developed in the resistance (R+r) is 2 - watt. Electromotive force and Circuits 123 Energy transferred from chemical to electrical form is /V=(-J- watt J(120 sec)=80 joules. (b) Joule beating in the wire is ohm+1 ohm) ~ 9 Total energy that appears in the wire as joule heat is />/ = f -5- watt J(120 sec) =66.7 joule. (c) The difference in energy (8066.7) or 13.3 joules is to be attributed to the joule heating of the battery owing to its internal resistance. 32.9. J? t =4 volt a =l volt J?! /tj=lO ohm jR 8 =5 ohm Traversing the right loop in the clockwise direction, ^i+^i+'i^t^O ...(1) Traversing the left loop in the counter clockwise direction, i-'i*i + 'VRs=fO .'..(2) Traversing the path 6 a d c b in the counter clockwise sense, i-/i*i-/A-,=0 ...(3) The junction theorem yields /i+'t-ft-O ...(4) It is obvious that (3) is not an independent equation as it can be obtained by substracting (1) from (2). Substituting the numerical values in (1), (2) and (4) we have 10i 8 +5/ t =-l (1)' 10 tt-5 iV=4 (2)' *!-/.+' =0 (4)' Solving (I)', (2)' and (4)' /,=0.025 amp Kj/ ft Jl l (0.02S amp) ( 10 obm)0.25 volt. Traversing various loops in opposite sense would net yield any new information. 124 Solutions to H and R Physics H 32.10. ^=2 volt 2 =4 volt ri =l ohm fa 2 ohm R=5 ohm Start from c and traverse in the counter clockwise direction along 6 and thence to a. =(-7 H-4) volt ...(1) Applying loop theorem to the entire circuit starting from c and going in the counter clockwise direction (4-2) volt .... ==0 - 25 amp Using this value of i in (1), F a =-(7 ohm) (0.25 amp)+4 volt =2.25 volt 32.11. (a) The resistors R z , R 9 and R< are in parallel. The equiva- lent resistance /? 5 for these three resistors is given by __,_,_ ~50 +50 +75 ~ 75 /?,== 18.75 ohm The equivalent resistance R of the network is obtained by com- bining /? 5 and J?i in series R=R t +R!= 18.754- 100=1 18.75 ohm. (6) Let current / lt / 3 and / flow through resistors /J p R 2 , R a and /? 4 , respectively. Applying junction theorem at the junction of all the four resistances ii=i t +i+i t ...(1) 9 Applying the loop theorem for the path comprising /? t and /? -/i*, -/,/?a=0 ...(2) Applying the loop theorem for the path comprising /? s and 7? 4 , f/4=0 ...(3) Electromotive Force and Circuits 125 Applying the loop theorem for the path comprising R t and /? 4> -"'4^4+11^1=0 (4) Put /J^lOO ohms, jR t =H t =50 ohms, l? 4 =75 ohms and =6 volts, and solve the simultaneous equations (1), (2), (3) and (4) to obtain /! =0.0505 amps *i=/3=0.0189 amps i' 4 =0.0126 amps. We can get the result for (a) by an alternative method. Since current i\ gets distributed through various resistors of the circuit the equivalent resistance of the entire circuit is ^4 6 volts /! 0.0505 amps 32.12. (a) Let the current i x , /, and j 3 flow through R lt R t and R t respectively. Applying the junction theorem at the junction of R lt R t and R tt i 1 ^i a +i 3 ...(1) Applying the loop theorem to the left loop, E-/,^- /!/?! = () Applying the loop theorem to the right loop, ...(2) Put Ri=2 ohm, /J 2 =4 ohm, /? 8 =6 ohm and ==5 volts, and solve the simultaneous equations (1), (2) and (3) to get the current in the ammeter / 3 =0.45 amps. C f- Fig. 32.12 (a) (b) Junction theorem for the circuit in Fig. 32.12 (a) yields 126 Solutions to H and R Physics II Applying loop theorem to the left loop Applying the loop theorem to the right -15) ... (6) Using the numerical values of (a) we find, ii=0.45 amp. This represents the current indicated by the ammeter. Thus, the ammeter reading remains unchanged. 32.13. Applying the junction theorem to the junction of the three resistors, /!+/,=/ ...(1) Applying the loop theorem to ihe lower loop, Applying the loop theorem to the big loop V ...(3) Eliminating i\ and i t , we have | ,Nvr*WNN (a) Power delivered to the resistor E*R p Fig. 32.13 ...(5) Maximum value of P is obtained by setting j =0 M/C o or whence R = ~5~ (6) Use (6) io (5) to tind JV>- ...(6) Electromotive Force and Circuits 127 32.14. Let the resistances be R l and In series, H 1 +J? a =jR 8 p_ P P In parallel, /? 4 = : .-(I) where use has been made of (1). .'. RiR^RtR* ..-(2) Of the resistances 3, 4, 12 and 16 ohms the choice ^=4 ohms, J? 2 ==12 ohm when used in series or parallel would satisfy both (1) and (2) and therefore provide the equivalent resistance 16 ohm and 3 ohm respectively. When used singly they provide the resistances of 4 ohms and 12 ohms. 32.15. Suppose that a and b are at the same potential. If /\ is the current flowing through R l and / through jR*, then the potential drop over R l must be identical with that over R$. (1) Similarly, the potential drop over R^ niust be the same as that over R*. ...(2) Dividing (1) by (2), or x^ 32.16. Applying the junction theorem at b. (Fig 32.16). /,+/=/. ...CD a Applying the junction theorem at a, /, + /-/! -(2) Applying the loop theorem to the path efcbde, R Q ...(3) Applying the loop theorem to the path cabc, * ...(4) Applying the loop theorem to the path abda, ir +t*Xm -i t /J0 ...(5) Solving (1), (2), (4) and (5), R R.+R. Fig. 32.16 128 Solutions to H and R Physics 11 Eluminating / between (1) and (3) .-(7) Eliminating / between (6) and (7), and re-arranging . = _ E(R,-Rx) 1 (R+2 r) (R 9 + R 9 )+2R*R* The current z=0 if /?=*. Already, R l R^ t This result is there- fore consistent with the result of Problem 32.15 viz., 32.17. Let the resistances be J? x and R t . For the series arrangement the equivalent resistance is Joule heating is 2 2 For the parallel arrangement, the equivalent resistance is #= l ^ R l +R 2 Joule heating is P - ^_*!l*i**> ...(2) F >~ R ~ R t R t ^ } By Problem, P 8 =5P 1 Put /J x =100 ohm. We then have The solutions are R t 38 ohm or 262 ohm. 32.18. P=V*IR, Therefore, ^ 1 =F/P=(100 volt)/(100 watt) =100 ohm. X Arrangement Equivalent ff _ E t (RiR^Rt /? 4 =100ohm) resistance/? ^R (= 100 volt) Elecrtomotive Force and Circuits 1 29 25 ohm (vii) Mf Mf AA/Y V\W- 250 ohm 400 ohm 400 watt 133 watt 100 watt 75 watt 60 watt 40 watt 25 watt 32.19. (a) Let a current/ be sent at x. As the resistances in the arms xa and \b are equal, the current will divide equally, 12 in each arm. Also the potential difference between xa and xl> will also be equal so that no current flows through ab. A current equal to //2 flows through ay and i/2 through by. The outgoing current at y is therefore again /. The potential difference between x and v is Vr y ^lR where R is the equivalent resistance of the network. But V*^--V** I r,,:=(i/2) X 10+072) X 10-10/ iR^-lQ i, or /?-!() ohm. Fig. 32.19 (a) o y 130 Solutions to H and R Physics II (b) Let a potential difference V* y exist, across the points x and y. Let a current i enter at x and a current i leave at y. If R is the equivalent resistance of the net work, then iR=V 9> ...(1) , AlSO, -(2) ...(3) . .(4) -(5) or F^=30i-30/ 1 -20/ 1 The potential difference between points x and B is =Wi l +\Qi l =2Ql 1 =W (/-/!)+ 10/ 2 Combining (3) and (4), x i Fi. 32 19 (b) The potential difference between A and y is or or Also, VA, = = 1 0(i - /, - /,) + 1 0(/- 1\ - /,) K^=20(/-/ i -/ 2 ) Combining (6) and (7), 3/,+4/ 8 =2| ...(6) -(7) ...(8) Solving (5) and (8), . 2i Elecrtomotive Force and Circuits 131 -() ...(10) Using (9) and (10) in (2) and combining with (1), whence, /?=14ohms. (c) Let the potential difference between the points x and y be Vx y . Let a current / enter at x and the same current i flow out from y. Vx,=iR ...(11) where R is the equivalent resistance of the network. or AISO, , =207, 10/2 (12) or 2/ 1 -f/ 2 =/ Also, Vo y = or 2/j 3 2 =i 'I o www* -*~VSWA- -VWW WVs/t- '2 "^WWSAA Fig 32.19 (c) Solving (13) and (14; ...(13) .-(14) r / t -0 (15) 13? Solutions to H and R Physics II Using (15) in (12) and combining with (11), K a> =i/?=20/ 1 ==10/ /?=10ohms. 32.20. Applying the loop theorem to the lowei ioop or ==*!* 5 volt 11 fli" 100 ohms =0.05 amp Applying the loop theorem to the upper branch. 50 ohm 0.06 amps = -(5 1-4) volt -9 volt. 32.21. For the series connection . 2E Fig. 32.20 ...(I) For the parallel connection, the internal resistances of the two batteries are in parallel, their equivalent resistance being r/2. For the current / through R the effective cmf is E - E 2E < The ratio of current through R for the series connection to the current for the parallel connection is obtained by diving (I) by (2), -(3) " r+ (R+r) (a) R > r t For . (3) shows i, > i p (b) R < r. For Eq. (3) gives /. < i. 32.22. Applying Kirehoffs's rules to the circuit of Fig. 32.22, (1) ...(2) ...(3) Elecrtomotive Force and Circuits 133 Using the numerical values in (2) and (3) and solving the above three equations we find R -^ '3 ' R i R 2 1 1Q * * -- AfV**^ .*.* iy . 3 , _ ' 2== fsT amp ' ' " J ' '? r |, - . 8 '4=19-amp. ' f Fin 32.22 (a) The Joule heat produced in JRj is P 1 =/i a /^==(j|-amp Y(5ohm)-0.34626 watts; that produced in /? 2 is P a =/ a J? 2 =f ~ amp j (2 ohm)=0.04986 watts that produced in R$ is ?,=/, ^ 3 -f j|amp V (4 ohm)- 0.70914 watts (b) Power supplied by, /^ is i/3=(3 volt) f j 9 - amp j^ 1.25316 watts Power supplied by 2 is 2 i,=(1.0 volt)( - y J-= -0.15789 watts (c) Power supplied by E l and 2 is E^'g-f f^i, = (1.26316-0.15789) watts- 1.10527 watts. T oule heating in the three resistances is, />=/! -f ^,+^=(9.34626+0.04986+0. 70914) watts - 1.10526 watts. Thus, the Joule heating is equal to the power supplied by \ and a . The battery E, is charged and negligible Joule heat appears in the battery. 134 Solutions to H and R Physics tl 32.23. (a) The resistance R v is put in the middle of the range. Rough adjustment of current is made with R 2 (lower resistance) and fine adjustment is made with /?! (higher resistance). (6) The equivalent resistances of /?! and R 2 in parallel is Holding /? 2 constant, differentiate (1). Change in R is A/?//L= R 2 t A/?//? _ 1 Setting ^-20^ we get Thus, a small change in the resistance of the parallel combination is crossed by a large fractional change in /f lf thereby permitting fine adjustment. 32.24. The resistance Rv of the voltmeter is in parallel to the resistance R Fig. 32.23 (a) so that the effective resistance R' is given by -L=-L _L R' R+Ry ~R"R Ry 32.25. The resistance RA of the ammeter is in series with R, so that the effective resistance R' is given by or R^R'R* 32.26. Without ammeter resistance the current E 9 Ri+R,+r 5 Volt ohm Elecrtomotive Force and Circuits 135 With the inclusion of ammeter resistance the current is , / E _ 5 volt (5+4+2+0 f) ohm =-0.4504 amp. Error in current measurement is, &iii' Fractional error in current measurement, is A*_(0.4545-0.4S04) amp nno _ . 4CXC ---- = u . uu y . i 0.4545 amp /. Percentage error is 100(A*7/) := =0.009x 100=0.9% 32.27. If no current goes through the voltmeter (Rv=oo), the current . E 5 volt __ ___ __ (50 + 40-20^ ohm =45.45 XlO"" 8 amp. Potential difference accros R l is j/=//? 1 = (45.45xlO" 8 amp)(50 ohm) -2.27 volt With a finite resistance Ry for the voltmeter, the resistance of parallel combination of R l and Rv is RRv _(SO ohmKlOOO ^ R,-\~Rv (SOohm+lOOOohm) Current, I" = T; .- ---- M^/l/rr^Tr^'V^T'l ^ 46.46 x 10" 1 amp R. 2 + r--R (40 + 20 r-47,6) ohm h Potential difference across R t is r = /'/Z = (46.46x 10" 3 amp)(47.6 ohm)-2.2i volt. Fractional error in potential difference across R l is AK__K-r__(2.27-2.21)volt V 2.27 volt -0.026 .*. Percentage error in reading the potential difference is 0.026 X 100=2.6% 32 28. (a) TKe currents are shown Applying junction theorem at a, / 2 +'s == 'i (0 Applying the loop theorem to the left loop, Ei-tiRi-i l R l ~E l -i l R l =Q (2) Applying the loop theorem to the right loop, 136 Solutions to ti and A Physics 11 Putting the numerical values, (2) and (3) become (2) Solving (1), (2)' and (3)', we get ^=0.67 amp, counter-clockwise in the left loop / 2 =0.33 amp, up in the center branch / 3 =0.33 amp, counter-clockwise in the right loop. (b) The potential difference between a and b is = E /2/? 2 =(4 volt) (0.33 amp)(2 ohm) = 3.3 volts. Fig 32.28 32.29. For the RC circuit the charge q after time t is where CE is the equilibrium charge on the capacitor and the product RC is the time constant. g _ 100-1 . 1 =14 _ - <IRC CE B100 100 J_ 100 r//ZC=ln 100=4.6 Thus, time r=4.6 times time constant must elapse before a capaci- tor in an RC circuit is charged to within 1.0 per cent of its equilibrium charge. 32.30 '1*1 p. '2*2 lR 2 .*\+*2 Electromotive Force and Circuits 137 (a) '3*3 ib) / '* \. ^'^ -t \. L._ ( = Fig. 32.30 32.31. Units of /?C=(ohm)(farad) coulomb .coulomb ampere volt ampere coulomb coulomb/time 32.32. Energy of the capacitor is 138 Solutions to ft and R Physics 1 1 Rate of energy transfer is dV = Q dQ^Q dt c~ dt T'~ F '- The right hand side is nothing but Joule heating. Thus, at any instant the rate of energy transfer is completely accounted for by Joule heating. 32.33. RC=(3X 10 s ohm)(1.0x 1(T f)=3 sec. R _l.YL ,-(1.0 sec/3 sec) *i v f 1 r\n i_ \ ** dt RC = (4 volt) (3X10 ohm) =y,6xlO~ T coul/sec, (b) Energy t/=y dt = ^* C dt C C ( =iE (l-tT-' //fC )=:(9.6x 10-' amp)(4 = 1. 08 XlO~ watts. (c) Joule heating in the resistor is i/l=(9.6x 10~ 7 amp) 2 (3x 10 s ohm) = 2.76x UT" watt (d) Energy delivered by the seat of emf is /=(4 volt)(9.6x 10~ T amp)==3.84x 1(T watts. SUPPLEMENTARY PROBLEMS S.32.1. The current in the circuit is - '2 Fig. S.32.1 -d) Etecrtomotive Force and Circuits 139 The potential difference across the first-battery is where use has been made of (1). By Problem, V^E ...(3) Using (3) in (2), whence, S.32.2. (a) />=*r where K is the potential difference between A and B .. P 50 watts K= -r = r~A ---- ^ 50 VOlt. i 1.0 amp (6) The potential drop across R is >=i*/fc=(1.0 amp)(2.0 ohm)=2 volt In the absence of the internal resistance, the einf of C is 50-2 volt=48 volt. (c) As the element C is opposing the current /, B is its negative terminal. 5.32.3. Power, /V> = ; r K As for parallel connection Kis the same for bulbs of resistance r and R; the power P will be larger for r (r < R). Hence the bulb with resistance, r will be brighter. (6) Power, jP ( r)=i'V ; P(R) = i*R As for series connection i is the same for bulbs r and R; P ( R) will be larger than /V). Therefore, the bulb with R will be brighter than that with r. 5.32.4. For series connection, total resistance of the circuit is R+Nr. Total emf will be NE. NE Current, I- ...) For parallel connection, the total internal resistance will be r/N, which is in series with R. Total resistance of the circuit will then be R+(r/N). The net emf will be simply E. The current will then be . = E NE ' Rj-r/N~NR+r '" (2) By Problem, in both (1) and (2), / is identical. NE NE Hence, R +tir~NR+r 140 Solutions to H and R Physics II whence (7V-1) CR-r)=0 Since N^l, we have R=r. S.32.5. (a) Let a current 12; enter at the corner A and emerge at B. Because of the arrangement with identical resistors the current Fig. S.32.5 (a) is divided symmetrically as shown in Fig. S.32.5 (a). VM m - 12 4iR+2 iR+iR 7 12 / ~ 12 Fig.S.325(b) Elecrtvmotive Force and Circuits 141 (6) Let a current 12 i enter the corner B and 12i emerge at C. Because of the arrangement with identical, resistors, the current is divided symmetrically in various branches as shown in Fig. S.32.5 (6). i2T (c) 4iR+5iR 3 Fig. S.32.5 (c) Let a current 6i enter at A and 6f leave at C. Because of equality of resistors, the current in various branches is divided symmetri- cally as shown in Fig. S.32.5 (c). RAC =- VAC f 61 2iR+iR+2iR 5 f. = 2" /v 6j 6 S.32.6. Join one end of each of 1 ohm resistors together and convert the other ends separately to N terminals as shown in Fig. S.32.6. Fig. S.32.(i 142 Solutions to H and R Physics II S.32.7. (a) The resistance of ammeter and R are in series. (/?!+3.62) (0.317 amp)=28.1 volt ^=85 ohm (6) The resistance of voltmeter and R 2 are in parallel. The equi- valent resistance of this combination is n_ 307 R The voltage drop across /?2 is 307 * 2 i (307 /? 2 )(0.356 amp) 307+7?; ~ 23/7 Volt /? 2 =85 ohm. S.32.8. (b) For an ffC circuit, the potential difference across the capacitor after time t is v ^ Ee -t\RC ...(1) where E is the initial potential difference and T=/?C, is the time constant. Substituting the given values, 1.0 volt=(100 volt) e~WW -.(2) = 100/1. o-ioo Taking logarithms on both sides, ^ -in 100-4.6 T =/?C= IQ ~ =2.J7 sec. 4.6 (a) After 20 sees, the potential difference would be, =0.01 volt. S.32.9. (a) U^ ?0 = / 2/ C= V (2)(0.5 jouleXlO"* farad) = 10~* coulombs. (6) q = - dt RC? Elecrtomotive Force and Circuits 143 At f=0, i= Tryrr r r-7T^==-j Tr=10~ s amp (10 f ohm) (10 e farad) K Since /?C=(10 ohms)(10" e f)=l sec =100e-' volt W) Rate of Joule heating, Uj^i*R==(^\ Re~ 2 't RC \/vC/ r (io-coui) T = L /mn u \/tA->f 7\ (10 6 ohm) (10 farad) J = e~ 2/ watt S.32.10. fa) At f=0. C is to be considered closed. Applying junction theorem at the junction of/?! and /? 2> /i=-/2+/ 8 (!) Applying Kirchoff's law to the lower loop, E- /!/?!- i f * a =0 .-(2) Applying Kirchoff's law to the upper loop, / 2 /? 2 -/3/? 3 =0 ...(3) Since Rt=X 3 , i^h -.(4) Using (4) in (I), /a^i/i -(5) Using (5) in (2) and the fact that 1 /? 1 =/l a , -|^/i=^ 2 E 2 (1200 volt) /3=/ 2 =if 1 =J(l.l x 10"" 8 amp)==0.55x 10"^ amp. At f=oo, the capacitor C is fully charged and C must be consi- dered as open. In that case / 8 =0 and / 1 =/ 2 . From (2), / 1 jR 1 +i 2 /? 2 =J? Since jR a =^i 144 Solutions to H and R Physics 1 1 (6) Fig. S.32.10 (c) At <=0, i t - 0.55 x 10" amp K,=i2/? 2 =(0.55 x ID' 3 amp)(7.3x 10 5 ohms)-401 volt At /=, / 2 -0.82xi(r 3 amp V t =i t R 9 =(Q.*2x 10' 3 amp)(7.3x 10 B ohm)=599 volt. (//) The voltage drop across R% is seen to approach its final value asymtotically, and hence an infinite time is required for the voltage drop to develop to its maximum value. However, the time for the voltage to increase ,ro any stated fraction of its final value is quite definite, and for the usual values of R and C encountered in common practice the voltage grows essentially to its final value within a reasonably short time. Let time be measured in terms of time cons- tant T = /?C. r^KooO-e-'/W .,.(6) T =,C=(7.3 X 10 8 ohm)(6.5 X 10" farad) =-4.75 seer, Set /=T in (6) to find -I-*- 1 =0.63 Next set t=2* in (6) to find Thus, for the fraction F z /Foo=0.63, we have f=f=/? 2 C=4.75 sees, and for the fraction VjV^ =0.87 we have r=2T=2/?,C=9.5 sec. 33 THE MAGNETIC FIELD 33.1. (a} The beam will deflect to the east. The direction of deflec- tion is found from the rule for the vector product in the relation (b) The velocity of electron whose kinetic energy is K can be found from v= /"*?= /2X(12xTo" 3 eV)(1.6xlQ- 19 joule/eV V"m ^y 9.1XlO~ 31 kg~ =6.5 X 10 7 meter/sec Force on electron, FqvXHqvB sin 6= since, 0=90. F=z(l.6x 10" 1 ' coul)(6.5x 10 7 meter/sec) (5.5 X 10" 6 weber/meter 2 ). = 5.7xlO~ 1(l m A i F 5.7xlO~ lfl nt ,-^^14 4 / 2 . . Acceleration, == '= ~ _, - f - =6.3 X 10 14 meter/sec 2 , m 9.1 X 10 81 kg (c) Time taken to traverse 0.2 meter of horizontal distance along south- north direction, .v 0.2 meter ~ f ^ 1A _ Q /== = r^ TT^ ; ; =3.1 X 10 9 sec v 6.:>xl0 7 meter/sec Deflection in the eastern direction is given by y~\at* =1(6.3 X 10" meter/sec 2 )(3.1 X 10" 9 sec)' = 3X10~ 8 meter ==3.0 mm. 33.2. The kinetic energy of a particle of mass m and charge q, moving in a circular orbit of radius R under the influence of magnetic field B, is 146 Solutions to H and R Physics II (a) For cc-particle, q=2e and m= *-&>=! Mev For duetron, q=e and m<f=2mi > 33.3. The magnitude offeree on wire of length /, carrying current i in a uniform magnetic field B at 30 to the wire, is = (10 amp)(1.0 meter)(1.5 webers/meter 2 )(sin 30) = 7.5 nt. The force acts perpendicular to the wire and the magnetic field. 33.4. Magnetic force on the wire is F=ilxB=/7fl since B is at right angles to the displacement vector I. Setting, F=w?g, the weight of the wire, the magnitude of the current required to remove the tension in the supporting leads is ^_ 10~ 2 kg)(9.8 meter/sec 2 ) IB (0.6 meter)(0.4 weber/meter 2 ) =0.41 ampere in the direction from left to right. >33.5. Magnetic force. F=Bq\\ The dimensional formula for magnetic induction is JF] __[A/L7->] B ~ 1 Q l The flux <f>B for a magnetic field is defined by Thus, the dimensions of 4>D are equal to that of B multiplied by thnt of area. That i<, The Magnetic Field 14? 33.6. The force on the wire is F ilB Acceleration, a= = m m Velocity, v=v +af m directed away from the generator. 33.7. The force acting on the wire is mv= J Fdt^l Bit dt^Bl J idt^Blq ...(2) where use has been made of (1). Also, v = 0-0 X 10" 2 kg) yX (2K9.8 mctcr/scc 2 )(3.0 meter) " Bl (0.1 weber/meter ? )(0.2 meter)* =3.8 coul. 33.8. Resolve B into two mutually perpendicular components (0 horizontal component, B sin 6, radially outward in the plane of paper, (i/) vertical component, B cos 6, perpendicular to the ring out of the paper. Consider a pair of points, diametrically opposite . on the ring. While the vertical component is the same in direction as well as magnitude, the current has reversed its direction, resul- ting in the cancellation of force. Adding up the contributions of symmetrically situated pairs of points on the ring, the force on the ring due to vertical component vanishes altogether. On the other hand, the direction of the radially outward component having fixed orientation of 90 to the direction of current at 'each point on the ring alone makes the contribution to the force. F= (current)(circumference)(radial component of B) =(i)(2iui)(5 sin 6)=2na/ sin 8. 33.9. Magnetic force on the wire is F=iBl Setting the magnetic force equal to the frictional force, =3= : MWf ^(0.6X0^0.4536 kg)(9.8 meter/sec') //' (50 amp)(0.3048 meter) 0.0524 weber/meter* = o.0524 ~- (10* gaussj-524 gauss. The direction of the field is normal to the plane of tracks. 148 Solutions to H and R Physics II 33.10. Replace the wire by a series of steps parallel and perpendi- cular to the straight line joining a and b. Traversing along the steps we notice that in going from a to b the current flows as much in the upward direction as in the downward direction so that the net component of force due to the segments of wire in the direction perpendicular to ab is zero. On the other hand the current in the segments of wire parallel to ab is only in one direction aad the total length effectively traversed is equal to ab along a straight line, Hence, the force on the wire is the same as that on the straight wire carrying a current i directly from a to b. 33.11. The torque on the coil is where N is the number of turns, / the current, A the area of the coil, B the magnetic induction, and the angle which the normal to the plane of the loop makes with the direction of B. As the angle between the plane of the loop and the magnetic field is 30, it follows that e=90-30-=60 . /. T=(20)(0.1 ampXO.l X0.05 meter 2 )(0.5 weber/meter 2 ) sin 60 =4,3xl(T 3 nt-meter The torque vector is parallel to the 10 cm side of the coil and points down. 33.1&J The torque is given by t^NiAB sin .(1) Maximum torque is obtained when 6=90. L^IwN -.(2) where r is radius of the circular coil and N is the number of turns. Area of coil, A=nr* .-.(3) Eliminating r between (2) and (3), Using (4) in (i\ with e=-90, L* iB is maximum when AT=1. This gives The Magnetic Field 149 Imagine the flat surface enclosed by the given loop to be divided by a fine grid into a large number of rectangles. If the same current i flows around the perimeters of the small rectangles clockwise then the currents in each rectangle is cancelled by tlje currents around the perimeters of the surrounding rectangles, except at the edges of the grid. The net result is a current i flowing clockwise in the large closed loop (Fig. 33.13). Let the magnetic induction B make an angle with the normal to the surface of the loop. Now the area of the large loop is equivalent to the areas of all the rectangles into which it is divided, and since the same current / flows through the small rectangles, it follows that the relation, t=NAi B sin 0, is valid for a loop of arbitrary shape where N is the number of turns. Fig. 33.13 33.14. The torque which tends to make the cylinder roll down the incline is sin 6 The torque acting on the wire loop by the current is T' = # i AB sin 6=2NiLRB sin 6 where we have written 2LR for A. Condition that the cylinder may not slip for minimum current is T'T INiLRB sin 9=mgR sin 6 _ (0.25 kgH9.8 meter/sec 2 ) 1 _ 2NLB (2)(10)(0.1 meterXO.5 weber/meter-j 2.45 amps. 50 amp 33.15. (a) Current density, y--^- = (0 . 2 n^etcrKl.Ox 10- = 25x10* amp/meter* 150 Solutions to ff and Jk Physics II If n is the number of conduction electrons/unit volume, then drift speed of the electrons, - J (25 X 10 amp/meter 2 ) F U ~ " " "~ " ~ne (1.1 X 10 29 /meter 8 ;(1.6x 1(T 19 coul) = L4xl(T 4 meter'sec (b) Magnetic force is F=<7v<i5=(1.6x KT" coul)(1.4x 1(T 4 meter/sec)f 2 - > \ meter* =4.5xl(r* 8 nt, in the downward direction. (c) Set the electric field equal to magnetic field, volt/meter, in the downward direction. (d) Necessary voltage to produce this field is V=Eh=(2.%X 1(T 4 volt/meter)(0.02 meter) = 5.6 XlO- volt. Top voltage should be +ve and bottom ve. (e) E//=-vdXB The magnitude of EH is given by Vd=(1.4x 10~ 4 meter/sec)(2 weber/meter 2 ) =2.8xlO' 4 volt/meter, in the downward direction. 33.16. (a ne where j is the current density, B the magnetic induction and n the number of charge carriers per unit volume. E Ene where we have used the expression for resistivity ?=Elj. (6) 90 (c) Textbook Example 5 gives, 5-1.5 webers/meter 1 For copper, n=8.4x lO^/meter* p= 1.7x10"* ohm-meter The Magnetic Field 1 5 1 Therefore, -~ = E net __ 1.5 webers/metcr 3 ) (8.4X 10 28 /meter 3 )(1.6x 10~ 19 coul)(1.7x 10~ 8 ohm-meter) =0.0066. 33.17. (a) Kinetic energy of proton, Kv=eV Kinetic energy of deutron, K* =eV Kinetic energy of a-particle, K* K, : Kd : K* = l : 1 : 2 (6) Radius of curvature is r= J2mK A / I/w 1 ~qf~ V ^ LL =:A / md J? jr V m p qd But m* qd e rd=V P 2/- P =V2(10cm)=14.1 cm i = A / -^ ^ = A f -i x~~ = V 2 r P V >p ^ V 1 2 /o, = V2rj,= V"2(10cm) = 14.1 cm 33.18. R __ I jna qv^ V m qd 2 and vlqa~ or or R*Rp 33.19. (a) Speed, v= 7W = (2)(^6>110- M _coulHl.2 \vebcr./mtftcr 2 )(0.45 meter) (6.68X10-" kg =2.6xl0 7 mcter/scc. 152 Solutions to tt and R Physics It (b) Timeoeriod T- 2 " m - (2*)(6.69 X 10"" kg) W nme period, /- ^ (2x 1.6X1Q-" coul)(1.2 weber/meter" 8 ) = 1.09xl(T 7 sec. (c) Kinetic energy, i(6.68x 10~ 27 kg)(2.6x 10' meter/sec) z = 22.5X10- 13 joules -14.1 Mev. W) Set #=$K= 33.20. Kinetic energy of electron, K= 15000 eV -(1 5000 eV) ( 1 .6 x 10- 1 " j ^ ) = 2.4 X 10" 15 joule Magnetic induction, B=- 250 gauss=0.025 weber/meter 2 Radius /t-~ ^^(2X9.1 XjlOr^ kg)(2.4x 10' 15 joule)] 1 ^ ?fl (1.6x fo"- i9 coulXO.025 weber/meter 8 ) = 0.0165 meter 1.65 cm. 33.21. First we shall investigate the path of an electron starting at rest from the positive plate. Choose the origin at O. The electric field acts along the y-direction perpendicular to the plates. The positive plate is along the ;*>axis and magnetic field is applied along the z-axis perpendicular to the plane of paper. The electric force on the electron is directed along the >>-axis and since the magnetic field B is along tne z-axis, and further if the component of initial velocity parallel to B is zero, then the path of electron will be contained entirely in the xy plane. ig. 33.21 The Magnetic Field 153 Writing down the force equations, Writing for convenience, o>=^ and U= ...(3) Equations (1) and (2) can be re- written as - =co(7 coVjc (4) dt v (5) Differentiating (4) and using (5), d*Vy__ dv x _ or + 8 v,=0 ...(6) The solution of (6) with the initial conditions, Vx==v > ==0, is seen to be v v =J7sin o>r ...(7) Substituting (7) in (5) and solving, v*=UU cos a>t ...(8) The coordinates x and y can be found out by integrating separa- tely (7) and (8) with the initial conditions xy=Q. y= (1 coso>r) ...(9) x=Ut sin wt .-.(10) 01 V ' o U Setting 6=oi/ and /?= , --.(11) Equations (9) and (10) respectively, become -(12) -(13) Equations (12) and (13) are the parametric equations of cycloid, defined as the path generated by a point on the circumference of a 154 Solutions to H and R Physics tt circle of radius R which rolls along the x-axis. The maximum displacement of electron along the y-axis is equal to the diameter of the rolling circle i.e., 2R. Identifying 2/?=rf ...(14) ---. B/B_ Em 2 ~"~ >~~~~ Also, = j- ...(16) Using (16) in (15), gives Thus, the condition that no electron should strike the positive plate is B> \TZJI- V ed* 33.22. The magnitude of the magnetic deflecting force F is given by F^qvB sin 6 ...(1) where 6 is the angle between v and B. Equating the magnetic force to the centripetal force, and setting 6=90 in (1), mv p or r~~~n~n qB qB where p=mv 9 is the momentum and r is the radius of curvature. Thus, r a p. 33.23. Equating the magnetic force to the centripetal force, r K mv Required magnetic field will have maximum value when 6 is maximum i.e., 90. Putting 6=90 in the above equation, The Magnetic field (1.67X 1(T 27 kgHl.Ox 1Q 7 meter/sec) (1.6 X 10" coul)(6.4 X 10* "meter) sin 90 = 1.63 X 10~ 8 webers/meter 2 . The magnetic field must act horizontally in the direction per- pendicular to the equator. 33.24. Momentum of deutron is p~qr=^(\.6x 1(T 19 coul)(L5 webers/meter a )(2.0 meter) =4.8 XlO~ 19 kg meter/sec. Kinetic energy of duetron at the time of break-up is p* (4.8 X KT 1 * kg-m/sec) 1 a _ w f A _ n . . ^Tx^ Joule Half of this energy is carried by neutron which moves tangen- tial ly to the original path, as it is unaffected by the magnetic fielq. The remaining half of the energy is carried by proton. So |&=1J&=1.73X l<r u joule. Momentum of proton is L73X 10~ u joule)(1.67X 10~ 27 =2.40 X 10" 19 kg-meter/sec Radius of the new orbit is , p'_ _ 2 A X IP"" 19 kg-mcter/scc r ~~ qB (1.6X 10^ 19 coul)(1.5 weber/meter 1 ) = 1.0 meter. Thus, proton moves in a circular path of radius 1.0 meter. 33.25. Resolving v along B and perpendicular to it, we have A / A \/ V M 6A A . . ... \/ cos sin e where 6 is the angle between B and ?. The non-zero component of velocity albng B viz. v 11 makes the plane of the circular orbit of the pdsitroq path advance along the J3-axis. In other words, the actual path df the positron is helical. *=2keV-(2 X 10 eV)(1.6 X 10T U jnul/eV)=3,2x lO'^ joule V (2)(3. 156 Solutions to H and R Physics tt meter vj_ =v sin 89=2.7x 10 7 Vji=vcos89=4.7xl0 5 sec meter sec (2Tt)(9.1XlO~**kg) _ Bq (0.1 weber/meter a )(l .6 X 10""" coul) 3.6xKT l0 sec. Radius of the helix r=^ ^^Ax.l^ 8 JM)a7x 10^ meter/sec) Radius of the helix, r q ((U We ber/meter 2 )( 1.6 XlO- coul) = 1.5xlO~ 8 meter = 1.5 mm. Pitch, Pn T=(4.7 X 10 meter/sec)(3.6 X 10~ 10 sec) =0.17 X10" 1 meter =0.17 mm. mv 33.26. (a) Path radius, r= ag ^(9.lXlQ- 1 kg)(Q.lx3x ig^meter/sec) (0.5 weher/meter 2 )(1.6xiO" 19 coul) =3.4 x 10" 4 meter =0.34 mm. (b) Kinetic energy, T=lmv*=l(9.l X 10' 31 kg)(0.1 x 10 8 meter/sec) J =4.1 XlO~ 19 joule _ (4.1 X10-M joule) _ 25 , X10 . (1.6xlO-j^ie/ev)- 2 - 56X 10 ^ eV =2.56 kev. 33.27. Period, r _ 1 _2nm 1 v 5 OT ==- (1.29 X 10~ sec/7)(1.6X 10'" coul) (4.5 x 10~ weber/meter) =2.1 IX JO"" kg. 33.28. Kinetic energy of the ion as it is accelerated through potential difference F is K=<P ...(1) The Magnetic Field 157 As the magnetic force is equal to the centripetal force, Mv* <y#V= r or A/* But from (1), K= or M Comparing (2) and (3), -.(2) ...(3) M or 2V where we have set r=x/2. 33.29. (a) F=-qvB, Fig. 33.29. For clockwise motion the magnetic force will be directed towards the proton. As the electrical force will also be directed towards the proton, the net centripetal force will increase. Set q=e. --- .. (1) !n the presence of magnetic field, Eq. (1) shows that v would increase, leading to an increase of the angular frequency, co=vr (for a constant radius r). V (b) For counter-clockwise motion, Fig. 33.29. the centripetal force would decrease resulting in a decrease of angular frequency. 33.30. Taking into account both the possibilities of increase and decrease of vEq. (1) may be re-written as ...(2) __ Ar v='-y- Writing v=<o r, (2) upon re-organization becomes . . Be e* (3) 158 Solutions to H and R Physics II Without magnetic field, (1) is simplified to Whence, ..--__. ... (4) Using (4) in (3) we Halve; to8 Ite w _ k lit Solution of the above quadratic equation yields, Be [ V Be "I J where we have neglected the second term in the paranthesis, as compared to unity. A>=<4 d> = ^~ Lm But Aw=2Ti AV 33.31. University of Pittsburgh cyclotron has an oscillator fre- quency of 12 XlO 6 cycles/sec and a Dee radius of 21 in or 0.533 meter. The magnetic induction required to accelerate deutrons is B ^ ?.5VH _(2iO( 1 2 x lOVsec)(3.3 x 10~ g7 kg) ~q l.6X!0- lft coul = 1.6 webers/m 1 . Deutron energy. Kd= q ' 2m = ( 1.6X10^ coul)(K6 webers/meter*)* (0.533 metcr) (2)(3.3 x 10- ~~^ =2.8x)0~ lf joule. The Magnetic Field 159 -(2.8 X 10- joul.)( ,. 6x ,' - e .T j<)ule - )= I' Me, (<> K* Kp nip 1_ or *=! K*=\ (17 Mev)=8.5 Mev A> ntp 1 Bp=\ 5d=i (1.6 weber/meter z )=0.8 weber/meter 8 f c ) it-i-"--*- _d.6x IP" 1 * coul) 2 (1.6 webcr'm*) 3 (0.533 meter) 1 2m (2)(1. 67X10' 17 kg) < 5 - 57xl '"i" l "(l.6x'lO M ' V j ou l e) =34.8 Mev (d) Bq _ ^- 6 weber/m 1 ) (1.6x 10~ 19 coul) V ~"2nm ~ (2n)(1.67xlO- 7 kg) =24.4 X 10 cycle/sec=24.4 Me/sec (e) For (a) *~ =^ =-2 ^.=2A"d=(2) (17 Mev) = 34 Mev For (b)^ = ^- =1 DA jndlqt S = Bj 1 .6 weber/meter 1 . , . (2x 1.6X 10~" coul) s (1.6 weber/m 1 )*! .0.533 meter)* ror \c) A = -,~i,s x rx-..-.- -- (2)(6.68 x 10~" kg) =(5.57 x lO' 14 joule)(l Mev/1.6 x 10' 13 joule) = 34.8 Mev. F (A\ = Bq ^- 6 ^ebc^meter*) (2x1.6 x 10"" coul' ( v ~ 2nm~ (2*)(6.68 x 10'" kg) = 12.2X10* cycle/sec = 12.2Mc/scc. Solutions to H and R Physics II 33.32* Total path length traversed by a deutron is N L= where Rn is the radius of the n th orbit and AT=number of revolu- tions that the deutron makes during the acceleration process. If V is the potential between the Dees, then in each turn energy picked up by the deutron is 2 eV; the factor 2 arises due to the fact that the acceleration occurs twice for a given orbit. The energy gained after n revolutions is 2m ...(2) Using (2) in (1), N Now N _ nergy gained_ _ _K ' energy gained in each orbit ~~2eV __ 17xlOev ~(2)(8xlOiv) where A:=17Mev from textbook Example 7. (33xlO" 27 kgH8xl0 4 volts) (1.6x 10-" coul)(1.6 weber/meter 2 )*' =(106) 8/2 =232 meters. 33.33. The oscillator frequency at the beginning of the acceleration cycle, .. = 2nm and at the end of the cycle - 2itm OT The Magnetic Field 161 33.34. (a) Set Bev=*Ee D _ _ _ v JlK/m 10* volts/meter V (2)(10 4 evXl.6xl(r"joule/ev)/(9.iS<10-"kg)~ = 1 .7 x 10" 4 wcbcr/meter 1 . Field of magnetic induction must be horizontal and to the left as one observes along v. (b) yes. _, 4 E lo 4 volts/meter For proton, v= TT = r^'CTiTv^ i. / I v B 1.7X10 * weber/meter* = 5.88xl0 7 meter/sec Kinetic energy of proton, K\ Mv* = \ (1.67X10~ 27 kg)(5.88x 10 7 meter/sec) 1 IMev =(2.89 X 10~ u joule) ( - 6xlO- 15 joule 18 Mev Thus a proton of 18 Mev energy will pass through the given com- bination of electric and magnetic fields undeflected. 33.35. (a) Bev=Ee The mi aim urn speed is v= (6) _ 1500 volte/meter 0.4 weber/meter 2 3750 meter/sec. Fig. 33.35 162 Solutions to H and R Physics II SUPPLEMENTARY PROBLEMS S .33 .1. If a positive test charge q Q is fired through a point and if a force F aqt^ride ways on the moving charge and if the induction B is presenfat that point then the following relation is satisfied. Thus, the direction of F follows the rule for the vector product. The same result is obtained by the application of left-hand rule. On this basis particle 1 is positively charged as it is bending upward while B is acting into the page. Particle 2 is going straight undeviat- ed. So it must be a neutral particle. Particle 3 is bending downward and must be negatively charged. S.33.2. Speed. v= m _ (0-41 x 10~ 4 weber/meter*)(1.6x 10~ 19 coul)(6.4x 10 6 meters) "" (1.67xHT 7 kg) =2.5 XlO 10 meter/sec a value that exceeds the velocity of light. The fallacy is due to non- relativistic calculations. (b) Fig. S.33.2 S.33.3. Kinetic energy of the electron, A>1000 ev -(1000 eV)( 1.6 xlO- 19 ^~)= 1.6 x 10' 16 joule =i mv 2 [ ~2K /(2Xf 6"xlO" 1 '* Joule) V"^" V (9.1 x 10-* kg) = 1.88 XlO 7 meter/sec. The Magnetic Field 163 The electric field, E=~^ = Q 0? meter ==5Q voit / meter If the electron moves undeviated then the force due to magnetic field and electric field must be equal. Bev^Ee E 5000 volt/meter *~xxtA-4 u / * * rrrs r^ -- / - =2.66X10 4 weber/meter 2 v 1. 88 X 10 7 meter/sec S.33.4. (fl)The deflection is made zero if the electric force, FE on the positive charge is equal and opposite to the magnetic force. If the velocity vector makes an angle 6 with the induction vector S then the magnetic force exerted on charge Q is given by FB = QvB sin 0, the direction of the force is always perpendicular to the plane determined by v and B. Setting FB^FL: QvBsin 6=Q or B=-~-jr --.(I) v sin The least value of B is given by the condition that sin is maxi- mum i.e. 0=^90, so that a- E Jo V The direction of induction ought to be from east to west so that the magnetic force may act vertically down in opposition to the electric field. (b) From Eq. (1) it is obvious that by varying the angle bet- ween the velocity vector and the induction vector, the magnitude of B would accordingly change. Thus, B is not unique for a given ^et of values of E and v if is not specified. (r) Energy of protoo, #-3.1 x 10 6 ev=(3.1 x 10 5 ev)(1.6x HT" joule/ev) -4.96 xlO~ 14 joule. Velocity, v ! 2K - A /UM4.96X 10~" j V ;?i V (1.67X10-" kg) =7.7X10 meter/sec. _ E 1.9 X10 volt/meter 2 webcr/metcr*. 164 Solutions to H and R Physics-^ II (d) Equating the magnetic force to the centripetal force, mv* r (1.67 xt(T 27 kg)(7.7xlO meter/sec) or _ ^ r ~~ Be = (2.47~x 10-* weber/meter)(1.6x KT* coul) =3.25 meter. S.33 5. (a) Let an auxiliary resistance /? be connected in series with the galvanometer resistance R g so that the equivalent resistance s, /= Fig. S.33.5 (a) By V=iR l~ ~f &9 ^ 1.0 VOlt 75.3 ohm-^542 ohm then 1.62xl(T 3 amp Let the auxiliary resistance R Q be connected in parallel to R ff9 ...(2) 'l fi*\ vr^/ i ; o P Fig. S.33.5 b ting / , we get, ff _ Roig (75.3 ohm)(1.62x 10~ 3 amp) i-/,~(50xiO~* amp- 1.62x10-' amp) =2.52 ohms, The Magnetic Field 165 S.33.6. (a) Hall electric field, ,, V 1.0xlO-volt ,_ . La =J~-= 0.01 meter" = 10 '' 10- 8 volt/meter _ r 5~~1.5weber/meter J =6.7 X 10-' meter/sec=0.067 cm/sec. (6) Current density, /'= - : 7-v=3xl0 7 ' meter) The number of charge carriers per unit volume, _ J B __< 3 x * 7 amp/meter z )(1.5 weber/ meter 8 ) eEtt (1.6xlO- M coul)(iO-voJt7meter) =2.8xl0 19 /meter 3 =2.8xlOVcm 3 . See textbook Fig. 33.10 (b). For negative charge carriers, according to Fleming's left hand rule, face y will be at a lower potential than face x. gB _(1.6xlO-"coul)(10" 4 weber/metei) S.33.7. (a) v - 2wn (2"X9"l X 10~ kg) =2.8 xlO cycles/sec =?..8 Me/sec. (6) Velocity of electron, V~2K _ [ TlXFoJevX 1.6x 10"" joiile/evf ^ V "(9.1 x 10- r kg) == 5*9 XlO 6 meter/sec ,. . mv (9.1 X IP"* 1 kg)(S.9x 10 meter/sec) Radms of curvature, r= ^^(i^^berTSeter'Kl.ex 10' 1 ' coul) =0.33 meter. 34 AMPERE'S LAW 34.1. Radius of wire is /?=0.05 in =(0.05 in)( 2.54X10~ 2 5^L )=1. 27 X 1(T meter \ in. / Field B at the surface of the wire jig- o o_ =- weber/amp-m)(50 amp) " " (2iO(1.27xl(T*mete75 = 7.9X10~ 3 ,/eber/meter 2 . 34,2. Distance from the pov*er line is ft=UO ft)(o. r=20ft=UO ft)o.3048 ^6 meters At the site of the compass, jt 1 ^ (4nX 10" 7 weber/amp-m)( 1 00 amp) ~ 2nr (2)(6 meters)" = 3.3 X 10~ 6 weber/meter 2 =0.033 gauss/cm 2 . This value cannot be neglected compared to the horizontal component of earth's magnetic field of 0.2 gauss. Thus, there will be a serious interference with the compass reading. 34.3. ,Near the electron, Mo i ^(4rcX 10 7 weber/amp-m)(50 amp) 2nr (2w)(0.65 meter) ==2xlO" 4 weber/meter 2 . (a) Force on the electron is F~qv xtoqvB sin 6, where is the angle between v and B. Set 0^90, F=qvB=(l.6x 10~ 19 coul)(10 7 meter/sec)(2 X 10~ 4 weber/meter 2 ) = 3.2 X 10~ 16 nt, parallel to current. (b) Assuming that the velocity is parallel to the current, a force equal to 3.2 X 10~ 16 nt would act radially outward, (c) 0=0. Hence, F=0. 34.4. (a) Consider a rectangular Ibop abed of length / enclosing a portion of the conductor, (Fig. 34.4 urt. Apply ng Ampere's law, /B. </l=/Vo Ampere Law 16t Now, the contribution to the integral from the path ad and be is T ro as it is perpendicular to B. Furthermore, I0//I/ Also, B and d\ are parallel along the paths ba and dc and as their contributions are equal, or B- L (b) At every' field point the horizontal component of B alone will be reinforced and the vertical component will get cancelled from considerations of symmetry and the direction of B will be as indicated in Fig. 34.4 (6). B / oooooooooooo Fig. 34.4, (a) Fig, 34.4. (b) Consider an infinitesimal width dx through which a current di flows. Then rfi=m dx. The field contributed by di at P is -(1) "" 2*r which is the differential form for the magnetic induction for a long straight wire, 168 Solutions to H and R Physics 11 r= J Rsec0 ...(3) x=R tan 8 dx=R sec* e<*6 ...(4) Using (1), (3) and (4) in (2) dg-W'g 8 " 8 . ...(5) The induction 5 at point P is given by the integral, -^ j ^ -w/2 where use has been made of (5). 34.5. (aLBy-Amjjere's law, where i" is the current flowing through the body of the conductor. Now, - T - L ^ ' rr- A L. Cy\ Evaluating the integral in (1) and using (2) in the right hand side, or -o r This gives, J3= for fl=0. This is the expected result for 5 at a distance r from the center of a long cylindrical wire of radius b, where (b) The general behaviour of B (r) from r=0 to r=oo is shown in Fig 34.5. For r<a, as there is no current flowing, B will be zero. For a<v<b, B is given by (3) and for r>b, B is given by Ampere's Law 160 ,34.6. (a) The net current passing through the conductor bounded by the closed path corresponds to that flowing through the liner cylinder above. Hence, by Ampere's theorem or Here current through outer cylinder does not contribute to B. or Mo* (c) Here current through both the cylinders contribute to B. or J3=0. /"fat Jm^ffii ^ JrrH "- c *~- As the net current flowing through the closed path is zero, 3(4.7. By Ampere's theorem aoplied to the interior of the wire at a radial distance r, l7d Solutions to ti and A Physics tl Where i'==/(r/a) 2 = current inside radius r. Current i is the current in the entire wire of radius a. p/ r %_ AW J* V W; ~~ 2icr V a ) Since the surface 5 is normal to induction, the flux is given by, where / is the lefagth of the wire. f *-*? ) 2rca a 4 4-/ Hence, the magnetic flux per meter of wire is __ AW (4ic X 1Q~ 7 weber/amp-mctcr) ( 10 amp) "/ ~~4* ~ 4* 10"" weber/meter. 34.8. (a) By Problem 34.7, in the interior of each wire (r<a) 9 the flux per meter is 10~ 7 weber/amp'm)(10 amp) / "~4K """ 471 =10""* weber/mcter. For antiparallel currents, the inductions are additive Hence, the flux/meter in the region r<a, for the two wires is 2X10~ 6 weber/meter, where a==0.127 cm is the radius of the wire. For the space between the two wires the flux is calculated from, d d x a (4icXlO~ 7 weber/amp-meter)(10 amp; f / 2.0 \ _ -- In 1 1 X 10"" 6 weber/meter where d is the distance between the centers of the wires. Therefore, the flux per meter that exists in the space between the axes of the wires is (2 X HTH-1 1 X 10~)=13 x 10- weber/meter Ampere's Law 1?1 (b) Fraction of flux that lies inside the wires is - 2 X 10~ 6 weber/meter _ n ., 7 ~ 13 X 10- 6 weber/meter ~ U (c) Since the inductions get cancelled for parallel currents, the flux/meter is equal to zero. 34.9. Field of a long wise is B=<- (4rcXlO~ 7 weber/amp~mcter)(100 amp) (2w)(50 x 10" r weber/meter 2 ) = 4xlO~ 8 meter=4.0 mm. B will be zero along a line parallel to the wire and 4.0 mm dis- tant. Suppose the current is horizontal and in the direction of the observer, and the external field pointing horizontally from left to right, then the line would be directly above the wire. 34.10. The current at a out of the plane of paper produces magnetic induction B lf at P at distance r, a ._ direction perpendicular to in a r l and is indicated by the arrow in Fig. 34.10 (a). The current at b into the plane or paper pro- duces magnetic induction B 2 a) P at distance r a in a direction perpendicular to r 2 . Resolve B! and B 2 in a direction parallel to R and perpendicular to it. Be- cause Bj and B.j are equal in magnitude and are symmetrically oriented about R, the perpendicular components get cancelled and the parallel components are reinforced. Fig. 34. 10 (a) ?=/?! cos 6+J5 a cos 0=; cos 6 where we have used the fact that cos 0= Further, rf- 172 Solutions to tf and R PhysicsIt 34.11. Choose the x and y axes parallel to the sides of the square. The magnetic induction B l9 /? 2 , 5 8 and 5 4 due to currents 1, 2, 3, and 4 respectively, at P 9 the center of the square, are indicated in Fig. 34-25 (a). As the currents are equal and the point P is located at equal distance from the site of the currents, Resolve B 19 J3 a , 5 3 and B 4 along x and y axes; we note that the resultant of x-components vanishes. The ^-components reinforce and the net induction is 45! cos 6 where r x is half the diagonal. a cos 8=cos45=l/V fc 2 U) -(3) (4)(4nXl(r 7 weber-amp/meter(20 amp)(l//2) (2n)(0.2 meter)/ V^2 =8 x 10"" 5 weber/meter 1 , along y-axis. Ampere's Law 173 34.12. We wish to calculate the force per meter acting on wire 2 due to wires 1, 3 and 4. Since the currents in wires 1 and 2 are paral- lel, the force F tl per meter on wire 2 due to 1 is attractive and is direc- ^J ted up, the magnitude being given by 10~ 7 weber-amp/m)(20 (2)(0.2 meter) * 45 V / =4xl(r 4 nt. / The current in wire 2 and 3 being / ^4 anti-parallel, the force of 3 on 2 will ' be repulsive and acts towards left. Fig. 34.12 weber-amp/m)(20 amp) 2 Ina ~ (2)(0"2 meter) Similarly, the force between wires 4 and 2 will be repulsive, along the diagonal joining 4 and 2. 0^ 7 weber-amp/m)(20 amp) 2 (2Tc)(\/2)(0.2 meter) The x-component of the net force F*=F SS +F 24 cos 45=4 X 10~ 4 nt+2xlO~ 4 nt=6xlO" 4 nt The ^-component of the net force F,**F n -F n sin 45=4x 10~ 4 nt~2xlO~ 4 nt=2xlO*" 4 nt The magnitude of the force is F KT 4 nt) 8 +(2x 10^ 4 nt=2 v / lO X 1(T 4 nt. It is directed at an angle 6tan"" 1 -^-=tan' 1 n ^. 4 Jr y Z X I u nt ^tan*" 1 3=71.6 with the j^-axis towards left. 34.13. The longer sides of the loop alone contribute to the force. In the left vertical branch of the loop the current is parallel to that in the long wire. Hence, the force is attractive while in the right branch the current is anti-parallel and hence repulsive. The reiul- tant force acting on tfte loop is r __ Mo /i/2 / _ u o M 2 ' ^ Mo '1*2 [b ' 2na(a+b) i74 Solutions to H and R Physics 11 10~ 7 weber-amp/m)(30 amp)(20 ajnp)(0.3 meter)(0.08 metersli (2)(0.01 meter)(0,01 meter+0.08 meter) =3.Jb< 10~ 8 nt, directed towards the long wire. 34.14. Field on the axis of a circular loop is P^ _ Mo _ 2 (K'+x 2 ) 8 /' Let there be n turns per unit length of the solenoid. Consider an elementary length of the solenoid. The number of turns in the length dx is ndx. The flux dcnstiy at a given point on the axis set up by a current i in the element of length dx is dx With the change of independent variable 6 defined by x=R cot 6 dx= -R cosec 2 6 d& no niR z f B 2 J dx . (/? 3 +x 2 ) 3 ' 2 - 00 I (-R cosec 2 6</6) cot 2 6 1 3 / 2 2 J sm TC 5ta If In^llftirvn CPt nr at P HIIP -2-cos6 = Mo , w'f.J.w* lUViUVllV/11 SCI up al i UUC to current j flowing through one ; , of the wires at the sides of the square in the elementary length dx (Fig 34. 15) is, jf, MO id* s n 6 * N - -'- \ \ ^ < \ V. \ 4itr^ *^ ^ D '*=**+$ . a /2 sin 6= j- . ,jf V x 8 +a 8 /4 D Mo ' f a dx Q / P 2 / Jfma i 1 4* j 2 (* 8 -f-fl 8 /4)/ 8 Pig. 34.15 Ampere's Law 175 Set x=a/2 cot 9 45 45 I sin dQ^^r cos 6 135 135 Since B due to the four wires at the sides of the square is set up in the same direction, the value of B at the center is given by, Poi ' /2iw n* 34.16. (a) We first consider the induction at P due to current i flowing through the top side of the square, Fig. 34.16. As the point P is^ symmetrically situated about the square, the induction due to a tght conductor of length (top side) is 9 fa? Fig. 34.16 i cos 6 4nR 4rtJ where we have used the relation COS = 2r -(I) -.(2) Resolve B into two mutually perpendicular components, B\\ along the axis of the loop and ti \ perpendicular to it. |It is seen that B^ gets cancelled by the contribution of the current flowing through opposite side of the square. On the other hand, the com- ponents B\ are reinforced. J3,=5,cosa ...(3) Thus, the induction at the point P due to the entire loop is given by 176 Solutions to H and R Physics 11 J5(/ocP)=4J?ii=4J5 cos == ~~~cosa ...(4) where use has been made of (1) and (3). ...(5) ...(6) Using (5), (6) and (7) in (4) and simplifying, (6) Set x=0 which is identical with the result of Problem 34.15. (c) For x > a, the terms a 2 and 2a 8 in the denominator of for- mula (8) can be neglected. Then (8) reduces to *=^ .-(9) 2*x 8 On comparing (9) with formula (10) for the magnetic field at distant points along axis ...do, We conclude that the square loop behaves like a dipolc moment, with the dipole moment givfen by 34.17. (n) Consider a typical current element dy. The magnitude of the contribution dB of this element to the magnetic field at P is found from Biot-Sav'art law and is given by Since the directions of the contributions dB at point P for all elements are identical viz., at right angles into the plane of figure, the resultant field is obtained by simply integrating (1). Amepres Law 177 J JB ' 4w ] r But sin 0= r -112 jojR_ 1 Z 4 R* f * 1=^-11 f ;- J r 3 4jc J (>- + -t/2 dy t Flg34 t l7 (&) If /-*oo then the term 4/? 2 can be neglected and /? a result which is identical with that expected for a curreni m a long straight wire. 34.18. The magnetic induction at C due to a current clonum d\ . given by ,/_ Mo i sin 9dx UIJ~ ^ Q 4rcr ? (a) Here 6=0 for the left straight bn^ch and 0-- 1 KO for i ;- right straight branch. In either case B^-0. (b) Here 8=90 since the radius of the semi-circle is perpe*u.:ivi.,i.ir to current element. Setting r R 178 Solutions to H and R Physics II ~ i sin 90 4*** f , _ MO / ^ J dx ~~4R (c) Since the straight portions of the wire do not contribute to B, the value of B due to the entire wire is the same as that for the semi-circle viz., 4R 34.19. (a) The magnetic induction at the center of the circle due to current element dx in one side of the polygon is i /V sin 6 dx = --- - - Since dB' points in the same direction viz., into the plane of figure at right angles for all the current elements the contributions to the field is obtained by integrating the above expression and multiplying the result by the number of sides (Fig 34 J9). I I I Fig. 34.19 sio e2. r B'- 7/2 -112 112 _ 4 Ampere's Law 179 But dx J G r +t s ?> i= 1 . 1C ==-5 sin b* n 1 . ic /v * _ sm ~ ^o n _ n =a cos n '-*L tan-5- The magnetic induction at the center of the circle due to n sides is As w->oo, tan - --> 2a a result which is utentiaU for H of a circular loop. This is reason- able since as /->oo, the polygon->circle. 34 20. (a) Using the result of Problem 34.17 we find at the center of the rectangle, the induction __.>'/ 2/7 1 " where the factor 2 in the numerator takes care of the two longer sides of the rectangle* ouch of length /. Similarly, for the shorter sides each of width </, _ . 4. - >' fi-*i+fli- B V/ M . (ft) If / > d Hsa 2 ^^ /r +^ _ . ~ it// 180 Solutions to H and R Physics II where we have neglected d 2 under the radical. This is the expected result for the value of B mid-way between two parallel long condu- tors separated by distance d. 34.21. The field at any point P on the axis due to the left coil is 2(/? 2 +x 2 ) 8 / 2 where x is measured from the center of the coil. As the distance is being measured from P, the middle point of the separation of coils, replace x by x+(R/2) in the above expression. Similarly, for the second coil, replace x by (R/2)x. Hence, the resultant, 1 1 ___ 1 2~x) 2 ] 3 / 2 f ___ 2 [R*+(Rl2+x)*]*l* [/? 2 +(/?/2 Fig. 34.21 shows the plot of B versus x for the given data, -d) Qj 4 -O 0-25 B -5-0 + 5-0 x(cm) Fig. 34.21 34.22. The resultant field due to the two coils at any point a dis- tance x along the axis from the center of the left ceil ir " 1 -0) For the purpose of investigating the variation of B with x we can ignore the constant factors | ^iNP 2 . We then find upon differen- tiating B with respect to x, dx 3_ 2 3 2(z-x)(-l) "2 -..(2) Ampere's Law 181 Set x=R/2 and z=jR. Then we find upon substituting these values in the above expression that dB/dx=Q. Differentiating (2) once again with respect to x, JT (R*+x*)*l*- x _(S/2)(2x)(/^+Jc a ) 3 / 2 I L (R"~rX ) J +3 i Setx=/?/2 and z =--/?, then 34.23. The direction of B at both the points a and * is perpendicular to the plane of paper upwards (Fig 34.23). The field at a is contributed by there paths, I, II and III and is calculated as follows: Following the methods of Problem 34.17, 00 Rdx Mn / Q cos 8 90 00 = /!_!_ f Rdx = . _MO ' ' 4 J (x 2 +^) 3/a ^nfl m i Fig. 34.23 Following the method of Problem 34.18, D f , _ MO ' B " ~ Also, Bin 182 Solutions to Hand R Physics II _(4*x 1(T 7 weber/amp-m)(10 amp)(2*+l) (4*)(0.005 meter) 1.03 x 1(T* weber/meter 1 . The field at b is mainly contributed by the paths I and III, the contribution by the path II being zero. Hor* Here, Similarly, Bin * B 4f*4 j*o ' i 1*0*^ Mo /_(4n;X 10"" 7 weber/amp-m)(10 amp) 2w/l 2w* *A ()(0.005 meter) = 8X 10~ 4 weber/meter*. 34.24- The field due to a straight conductor carrying current i at the point P at distance R is given by, ******* ( dx 4* J -(0 It is seen that contribution to B at P from the left path (BJ and top path (Bt) will be identical. Similarly the contribution to B at P from the right path (B t ) and the bottom path ( 4 ) will be identical, the field in each case pointing in the same direction viz., into the paper perpendicular to the figure. -(2) 30/4 (3) where we have set, &*"-% Ampere's Law 183 304 A1S ' * == * V^+'iT a/4 where we have set #=3a/4. *-- M-'vs) - <6) Using (4) and (6) in (2), x 10~ 7 weber/amp-m)00 amp)(V"lO+ \/22) (3*)(0.08 meter) =2 x 10~ 4 weber/mcter 8 . 74.25. (a) Consider a ring of radius r and width dr concentric with the disk. Then the charge associated with the ring is -. q- R2 - The current is the rate at which charge passes any point on the ring and is given by i~vdq (2) where, v = 2 O)- is the rotational frequency of the disk. ^.-gL.*^ ...(4) where use has been made of (2). Using (1) and (3) in (4). f H r -J*>^-. J df ~ 2*R /L\ -A w 2qr dr taqrdr (6) i-v^-jj- -V 3p- Contribution to the magnetic moment is , .. (yqrdr) 2 . <*>qr*dr ------- 1 84 Solutions to H and R Physics II R = f </u- m~ f ] ^ /P j 34.26. For a square of side a, 4fl=/ (I) By Problem 34.15 the induction at the center of the square is , TCtf 7C/ where use has been made of (1). For a circle of radius R, "-(3) The induction at the center of the circle is ^%H-r - (4) where use has been made of (3). Dividing (2) by (3), = 8 ^ 2 _i 1C /O B ; -^2- -LI s ...p; As the right hand side is greater than unity, we conclude that Bs>Bc, i.e. the square yields a larger value for the magnetic induction at the center than the circle. 34.27. (a) The magnetic induction set up by the large loop at its center is P^/XQ i _ (4ft x 1CT 7 wcber/amp-m)(15 amp) 2R (2)(0.1 meter) =9.4 X 10*" 5 weber/meter 1 (b) Torque, T= M 5sin 6=^5, as 6=90. as 6=90 Here, /A#M(50)(1.0 amp)(0.01 meter) 1 ==1.57 x 10" 1 amp-meter 1 T=(1.57x 10" 1 amp-meter > )(9.4x 10" 1 wcbcr/mctcr 1 ) 1.48x 10~ e Dt-mcter. 34.28. Applying Ampere's theorem to the rectangular loop abed aa in Fig. 34.28. Ampere's Law 185 as no current flews. Further, assume that the magnetic field drops to zero along the side cd. Fig. 34.28 Now, the contribution of the path be and da is zero to the integral as B is perpendicular to ad and he. Also, the path cd does not contribute anything as by our assumption B is zero along cd. The only contribution to the integral then comes from the path ab. If the length ab is L, then or J3=0 along ab, which is absurd. We, therefore, conclude that our assump- tion that the field along cd is zero is wrong. SUPPLEMENTARY PROBLEMS S.34J. The wires labeled 1, 3, 6, and 8 above are within the closed path. According to Ampere's law, / B.<n=^i ...(I) Here, the direction of traverse around the loop is clockwise when one faces the loop, and the current is considered positive if its direction is away from the observer. As in the present case the direction of traverse around the loop is counter-clockwise the integral > ields a negative sign for the right band side of (1) with the current into the plane of {taper as poiiare. /W ; ft 1. -3, +6 and +8) 186 Solutions to H and R Physics ft By Problem, Thus, / B.rfi=- S.34.2. One must consider the effect of the magnetic field due to the current ti (Fig. S.34.2) on segments like A and C on the conductor carrying the current t\. The net result is that the torque tends to align the conductors with the currents running parallel. When this state of affairs is reached the conductors with parallel currents will begin to attract each other. Fig. S.34.2 S.34.3. Consider the force/meter on wire labeled 1 by 2, 3 and 4. As the currents in the wires are equal, ^ 'i==i2=/ 1 =i 4 =/. The force of 2 on 1 per 2 meter is given by . 1 I 2ifl acting along the line joining 1 and 2. Similarly, the force of 4 on 1 per meter is given by acting along the line joining 1 and 4. Fig. S.34 3 Ampere's Law As the magnitude of F n and F 4l are equal and as they act at right angles their resultant^ which is given by V/V+*4i a=5S / 2 F n , would lie along the line joining 1 and 3. Now, the force of 3 on 1 is given by along the line joining 1 and 3. Hence, the resultant is given by along the line joining 1 and 3, i.e. towards the center of the square. S.34.4. The magnetic field at P, the center of the hole, is obtained by considering it to arise due to two current densities (Fig. S.34.4), (i) a current density, j = I/IB (JR 2 radius /?, and carried by the cylinder of 07) a current density J carried by a cylinder of radius a. The resultant magnetic induction is the vector sum of the effects under (/) and (11). The contribution due to (j) is obtained by applying Ampere's law. or The induction at P due to equi- valent solid conductor of radius a carrying an assigned current oppo- site to the above is zero as no current will be contained within the path of zero radius. The resultant magnetic induc- tion at P, which is given by the summation of the two foregoing factors is then Fig. S.34.4 188 Solutions to H and Physics -It S.34.5. The resultant B at the center C of the circular loop may be considered as the superposition of fields B l and B a in the upper and lower semi-circles respectively. Now, the current i divides itself equally in the two semi-circular paths. By Problem 34.18, the con- tribution to the induction from the upper semi-circle with the current running clockwise would be J^. 2 l into the plane of this figure at right angles to the page. An equal contribution to the induction is made from the lower semi-circle viz., But since the current is running in the counter-clockwise sense, B t will be pointing out of the page at right angles to the plane of the figure. Thus, the resultant induction B^j+B^O. S.34.6. (a) According to the Biot-Savart law, dB is given in magni- tude by ... Mo/ dx sin 4n r a As the points S and P lie on the axis of the current 0=0 for all the elements of the wire and consequently, ,&,=&=(). Let us cal- culate the induction at Q. From Fig. S.34.6 (a), sin e= Fig. 8,34.6 () fttf. 34.6 (b) Ampere's Law 189 Set Then Ldx ~ L tan , B^^^^pL 4:t J L 3 sec 8 y 4nL J cos 45 ~J^ sinY f __ <f2/_*>1^2 ~4*L m Y J -4S2nL~ 8*L into the plane of paper. Finally we calculate BR. From Fig. 34.6 (ft), ^ x sin 6 sin 6= r 1=^ f Ldx 4* J (x a +L 2 )/ Set x==L tan y dx~L sec 1 y Then w 45 , *,=-' f ^'Y 4w J L 8 sec 8 45 into the plane of figure. (fc) Calling the induction at T due to the sides labeled ! 2. ...6 of the closed loop, by B lt fl a ...5,, we have 190 Solutions to H and R Physics 11 in the plane of figure. S.34.7. Using the result of Problem 34,18, the induction at C due to the inner semi-circle of radius R l is, -ff 1 ==/o741Z 1 . (current being clockwise) and that due to outer semi-circle of radius /? a is ^i^fj?' (current being counter-clockwise) The straight sectors AH and JD which upon extension pass through Cgivj >ero induction. Therefore, the resultant induction at C is, into the page. S.34.8. Due to inner arc f the induction at C is osin where a=90. Also dx^Ri d9. 1 4uR, Similarly, due to outer arc, the negative sign arises due to the fact that the current has reversed its direction. As the radial part of the path points towards C, it does not contribute to the B. Therefore, the resultant induction is S.34.9. (a) The resultant B at the center is given by the superposi- tion .of B 9 due to the straight conductor and B due to the circular path, both of them being directed out of the plane of figure. Mo/ Ampere's Law 191 "'~2R B=B.+Be 2R~ 2R out of the page. (b) As before B$ points up out of the paper. But, now Be lies in the plane of paper being in the direction of the current in the straight conductor, its magnitude remaining the same as before. At C, the resultant induction is given by Fig. S.34,9 /? B is inclined at an angle 6, given by 0=tan"" 1 ^-^ DC out of the page. 1 "" 1 ==18. 7C 35 FARADAY'S LAW 35.1. The induced emf, E=-N--* B - at (f(pg where TV is the number of turns, and ~y- is the rate of change of flux. Now, <}>B~BA where A is the area of cross-section. Change in fllux A<* for each turn of the coil is Afa=4A= (0.001 meter 2 )(2 weber/meter 2 ) =2xlO~ 8 weber. since the magnetic induction changes from 1.0 weber/meter 1 to 1.0 weber/meter 2 . The current / is given by ^JL^-IL A^B _A<? 1 R R Af A/ /. The quantity of charge A q flowing through the circuit is N A , (100)(2X10~ 8 weber) _. A ^ - (wrf-) --- = 2xlO-*coul. -35.2. The long solenoid of Example 1 has "200 turns/cm and carries a current of 1.5 amp; its diameter is 3.0 cm* The current in the solenoid is reduced to zero* and then raised to 1.5 amp in the other direction at a steady rate over 0.05 sec, Cross-section area of the solenoid, - -(0.03 meter) 8 Field B in the solenoid . -=:(4Tr x 10~ 7 weber/amp-m)(200x 100/meter)(1.5 amp) =3.8 x 10~* weber/meter 2 . Faraday's Law 193 ^=5,4 .=(3 .8 x 10~ 2 weber/meter 2 )(7 X 10~ 4 meter 2 ) =2.66 XlO" 5 weber. A^=2x2.66x 10~ 5 weber = 5.32 x 10' 5 weber 5 ' 32 * Rate change of flux, =E= A/ s 0.05 sec = 1.06x 10"" 3 weber/sec. The current in the coil is 100 A _. u , v weber/sec) ~\ (5 ohm) =2.1X10~ 2 amps. 35.3. As the loop is woi .cd, it normal rotates about the field direction at a constant angle of 30. In this process the lines of force cutting the loop would not vary. Hence, in accordance with Faraday's law no emf will be produced in the loop. 35.4. Area of cross-section of the loop, A=.]y*=2-(Q.i meter) 2 =7.85xlO~ 3 meter* Length of wire, /=^D==n(0,l meter)=0.314 meter Area of cross-section of wire, a ~?L rfi==JL(o.lx 2.54xlO" 1 m) t =5xlO~* meter 2 . Resistance of the wire, /?=p//a (1.7 XlO" 8 ohm-meter)(0.314 meter) n . A = - T= T^Z^ -- sr: -- := -' i.vj x ID *onm. (5X10 meter*) =J*=A 41 1 R R dt dB_ iR _(10 ampXl.05 XlO"* ohm) r dt A K (7.85X10-* meter 8 ) = 1.3 weber 'meter^sec. 35.5. Mass, m^al do =(nr*)(2*) do '...(I) where d is the density of copper, / the length of the wire and flakier 1 , the cross-section area of the wire. If A is the area of the loop and RQ the resistance then the induced current in the loop is ._A dB I- "-15 dt 194 Solutions to H and R Physics 11 Bu4 fl-P'-Pt 2 **)- 2 *? But *--- wr2 --,- Use (3) in (2) to eliminate R , . itRr* dB l = 2p~ A' From (1) we have, Use (5) in (4) to find -.(5) dt *' It is clear from (6) that / is independent of / and r (size of the wire) and R (size of the loop). 35.6. Radius of the wire, r=0.02 in. =0.02x2.54 XlCT 2 meter =5.08 xlO~ meter. Area of cross-section of wire, a=nr 2 =n(5.08x !0~ 4 meter)* =81xl0~ 8 meter*. B e u D P 7 (1.7X10- 8 ohm-m)(0.5 meter) Resistance of the wire, *o= -- = - CSlX lO's meter 2 ) - = 1.05x10-2 ohm. Radius of the loop, R=^- = ~ =7.96 X 1G-* meter. M Zn Area of the loop, ,4=*/? 2 =*(7.96x 10^ meter) 2 =0.02 meter 2 t^lQQ gauss/sec =10~ 2 weber/meter 8 -sec. Induced emf,=>4-=(0.02 meter 2 )(10^ weber/meter'-sec) at T 1 u . D -a* _ Joule heating, Pj=i*R Q =- lMxW =3.8 x 10""* watts. ) 39.7. As the north pole enters the coil the direction of current in the face Si of the coil is counter-clockwise, Fig. S5.7 (a). The current rises to maximum when the magnet is half /way through, and as the magnet continues to move in the saqtfe direction the Faraday's Law 195 current in the coil decreases but continues to flow in the counter- clockwise sense as viewed along the path of the magnet. Fig. 35.7 (6) shows qualitatively the plot of / the current as a function of x, the distance of the outer of magnet from the center of the loop. Joule heating is given by Pj=/ 2 jR. Fig. 35.7 (c) shows the qualita- tive plot of Pj as a function of x. Fit',. 35.7 (a > Fig. 35.7 (b) Fig. 35.7 (c) 35.8. (a) The magnitude of the emf induced in the loop may be computed from the rate of change of flux through the loop. Let the plane of the^loop make an angle a with the normal to the field. Then the flux through the loop is <f>=AB cos a (!) where A is the area of the loop. The rate of change of flux is then jj. j~ ...(2) dt The induced emf is - at sn a ...(3) where a=<fe/d/ is the angular velocity of the loop and N is the nura- qer of turns in the loop. 196 Solutions to H and R Physics II Put, aj^2n* Eq. (3) becomes E=2nv Nba B sin 2mt=E^ sin 2rcvf ...(4 where J E =27cv Nba\B is the maximum value of the induced emf. (b) From (4) we have, ^ V ^ _ __ 150 volt ___ ~~ 2*v# (27c)(60 rev/sec)(0.5 weber/meter)* ^-5/271 turn-meter 2 35.9. By Problem 35,8 we have- E~~2wNbaB sin 27cv/^ S jn 2 The amplitude of the induced voltage is where we have set ba=A, the area of the loop. Putting A~-nR-/2, the area of semi circular loop, and we have /To- rrv Amplitude of induced current is . _ __n*vR*B RM RM 35.10. The emf developed between the axis of the disk and its rim is 2 =i(1.0 weber/meter*j(2jtX30 rev/sec)(0.05 meter) 2 =0.24 volts. 35.11. E=-Slv=Sv (b-a) ' 2-nr b -<* \ Faraday's Law 197 _(4rcX 10~ 7 weber/amp-mXIOO amp)(5 meter/sec) (20 cm 2n ! ~ -3 X 10"* volt. 35.12. (a) Force acting on the wire is .*. Acceleration, a= = m m At time /, velocity is given by m The direction is from right to left. (b) By Faraday's law (c) The terminal speed of the wire is vrElBd. The wire is being resisted by a constant force F so that it is moving with constant speed v. Now, F=Bid ...(1) and the rate of working is But the induced emf is equal to the rate of change of flux through the circuit enclosed by the moving wire and the two rails. =l^=lWv r ...(3) where A is the area of the circuit. In order to provide current against the induced emf defined by (3), the battery must work at the rate Ei**Bdvri=P -.(4) where use hat been made of (2). Thus, toe entire work is done by the battery to slow down the wire to constant speed (terminal speed), the induced emf getting completely cancelled by that provided by the battery M that the net current in the circuit is zero. 198 Solutions to Hand R Physics It **** V**/ I. \ E \=*Qf =(12 X2+7) milliweber/sec r-2 =31 millivolt. (b) Direction of current through R is from left to right. 35.14. (a) The emf induced in the rod is ~J?/v=(1.0 weber/meter 2 )(0.5 meter)(8 meter/sec) =4.0 volts. From Lenz's law E must be counter-clockwise. (b) Force required to keep the rod in motion is B 2 l 2 v (1.0 weber/mctj^jnO.S jjnctcr) 3 (8 meter/sec) (0.4 ohm) -S.Ont. (c) Rate at which mechanical work is done by the force F is R Joule heating is Thus, mechanical power=electrical power. />=:/rv=(5.0 nt)(8 meter/sec)=40 watts. 35.15. (a) Gravitational force acting on the wire down the rail, F a =mg sin 6 ...(1) The component of magnetic induction normal to the plane of rails is B cos 0. The magnetic force on the wire up the rail is ...(2) For steady state velocity, the net force must be zero. That is B* cos t 9 / f v . fl or, - 5 - & Mg sin 8 Faraday's Law 199 , T . mgR sin whence, y-JL- ... (3) (6) Gravitation does work at the steady rate of P=FgV~mgv sin ...(4) where use has been made of (1). Joule heating is given by . a ,-. =/tiv sin 6 .-(5) /v where use has been made of (3). Since (4) and (5) are identical we conclude that Joule heat appears in the resistor at the same rate as that done by gravitational force a result which is consistant with the conservation of energy. (c) If B were directed down, then the force due to magnetic in- duction would have a component down the plane of rails and would reinforce the component of gravitational force and the net force would be 'cos a e/*v , . Q F( n e<> = - -- \-rng sin The wire would therefore, suffer acceleration, acquiring ever increasing speed, this being the case of non-uniform acceleration. 35.16. The induced emf is _ Nd$B L ~~ ~dT ._dq^ __ __ ^L ^D l ~~dT~~ R~~ R dt N " l N ( N ^q=- J </#=-^(# t - 35.17. Magnitude of emf developed is ., d$B A dB t <lB E=-j- A^-=nr*-j dt dt dt Electric field t at a distance r from the center is A. _ d JL J 4* 2itr dt ** 2 r dt 200 Solutions to if and R Physics 11 Force on electron is, Acceleration is, ^ F^ __ 1_ e_ dB m~ 2 m r dt At the point a y the electron experiences an acceleration, T ( ^TyS^^ )(0 ' 05 meter )< - 01 weber/meter 2 -sec) =4.4 x 10 7 meter/sec a , to the right. At the point fr, the electron has acceleration, a=0 At the point c, the electron has acceleration, 0r=4.4 X 10* meter/sec 2 , to the left 35.18. The electric field EE at any point 6 on the rod at distance r from the center is perpendicular to r as in Fig. 35.18. The magni- tude of EE is given by Pig. 15.18 Resolve EE along two mutually perpendicular directions, EL\\ the length of the rod and EE^. perpendicular to it. t'araday's Law 201 ii=Esm9=(^ rdBldt)(p}r) where p=0c= V# a -/ 2 '4 According to Faraday's law But J - 35.19. Fig 35.19 shows the plot of B (r) against r. Area under the curve is ,4=(3420)(200X 1) gauss-cm =6.84 X 10 5 gauss-cm - A 6. 84 XlO 5 gauss-cm Ol _ B= = 5-r- 2 =8143 gauss R 84 cm & Now, from the graph we note that at r=K=84 cm, Bie=4000 gauss. 2**=8000 gauss Thus, the relation 5=2 J Ba is nearly satisfied. 10 e (gduss) cm r (cm) Fig. Solutions to Hand R Physics It 35.20. Apply Faraday's law to the rectangular path abcda in Fig. 35.20. Then =/E.rfl=0 -(I) as emf is absent. Now the contribution to the integral from the horizontal paths be and da is zero as E is perpendicular to these paths. If we now suppose that along cd the ejectric field is zero, then (1) gives, where ab~L. Thus ==0 along ab, which is contrary to the pro- blem. Hence, our assumption that E drops to zero abruptly outside the parallel plate is wrong. a (j r 1 ' ^ i i i 1 I i L J c Fig. 35.20 SUPPLEMENTARY PROBLEMS S.35.1. [emfj-fclectric field][distance] =[force/charge][distance] =[force/(velocity)(cbarge)][L 2 j[r- 1 ] Thus, S.35.2. (a) The resistance of the circuit ADCB can be rendered approximately constant by choosing the resistance of AB equal to 1.2x 10~ 6 ohm and negligible resistance for the section ADCB. Induced emf, E^-d^ldt^Btl =(6x 10~* weber/m'XO.S meter/8ec)(2.0 meter) =6xlO~vo|t. Faraday's Law 203 Electric field, =>!=3 x lO" 5 volt/meter (c) Force on electron* F='t =(3xlO~ volt/meter)(1.6x 10 " coul) =4.8xlO~"nt. (j\ r- * E 6X!0~ 5 volt , W) Current, ,=- ^^.s amp (e) The force due to the induced emf must be counter balanced by an equal and opposite force. P=HB=:(5 amp)(2 meter)(6x 10" 5 weber/m 2 ) = 6xlO" 4 nt. (/) Rate of work, P=Fv-(6x 10~ 4 nt)(0.5 metcr/scc) =3xlO~ 4 watts. (g) Rate of joule heating, Pj=i*R =--(5 amp) 2 ( 1.2 x 10" r> ohm) = 3xlO~ 4 watts. S.35.3. (a) The induction along the axis of a circular current loop of radius R carrying current / at large distance x(x > JK), is given by 2 The magnetic flux, fa^BA^ where A is the area of the smaller loop. -(I) dx dt~ dx Differentiating (1) with respect to x and setting xNRin the resulting expression, xl - ~"~ 2 dx\x* x=NR 204 Solutions to H and R Physics It Since v is positive, expression (3) shows that E is positive and (4) showns that d<f>B/dx is negative i.e. the flux through the smaller loop is decreasing. The direction of current in the smaller loop will be such as to oppose the decrease in flux, i.e. the directfoii of the current will be in the same sense as in the larger loop. 5.35.4. (a) The projected area in a plane normal tg B is. A=nr 2 cos 45 = ^(0,037 meter);=3.04 X 10~ meter a The induced emf is .., d<f>B A dB dt dt /-> *A 1/x-n * o\ f 76 X I0~ s weber/m? \ sa -(3.04xlO>tncte>)(- 4<5xlo -3 S ec ) -5.13X10" 2 volt. (6) The emf produced in each of the two semi-circular loops would be equal in magnitude but in the opposite sense. The net emf in the complete loop would be zero. 5.35.5. (a) The induced emf is given by The current through the loop of wire is , JL __ 1 * m l ~ R- R dt - (1) But, i=^- ...(2) Comparing (1) and (2), dq=-\IRd<l>B Integrating, *~^ j **-- The result is independent of the manner in which B is changing. (6) Since the above result is independent of B, it is possible that B was changing in the time interval /! to / t causing current to flow in the circuit leading to joule heating. Faraday's Law 205 S.35.6. The torque is T=|Ltfl sin 0=//Jfi sin Q=iAB=ia*B ...(1) where p. is the magnetic moment of the dipole, i is the current, the area A a*, and 6=90 is the angle between B and the surface area. Bh_ Bla>r R = ~R" "* w where / is the length of the loop and R the resistance. Using (3) in (2), i=B<sta>r z ...(4) Using (4) in(l) ...(5) 36 INDUCTANCE 36.1. The induced emf is given by E=^L di L L dt Let the current change at the rate of di E 100 volt 1 "^r == T'" == "TAT -- =10 amp/sec. at L 10 henry ' 36.2* (a) Let the two coils having self-inductances L x and L 2 be connected in series, a great distance apart. The equivalent self- inductance of the network is defined as the ratio of the total induced emf between the terminals of the network, to the rate of change of current responsible for the emf. emf in coil 1 = self-induced emf - ~ l dt emf in coil 2 = L ,- ~ (it Net emf = (^ +L 2 ) From its definition, the equivalent self-inductance is L^L,+L 2 -(I) (ft) Separation should be large so that mutual-inductance M may not be present, otherwise formula (1) would be modified to -(2) The signs + or correspond to the sence of the windings being the same or opposite, respectively. 36.3. If i is the total current, / = -, -(I) at Let the inductances L x and L 2 be in parallel. Current in one inductance, say L % will be //2, that is Inductance 207 But ==! Comparing (1) and (2), we get L=LJ2 36.4. The magnetic induction between two parallel wires carrying equal currents / in opposite directions at a distance x from one wire is The flux is given by j PfiH f ( 1 i 1 \j /*= - - I ( f- -j Wx 2* J \ x </ x ) a d-a . in a 36.5. For the toroid, 2n a v/here N is the total number of turns. Let b=a+& where A is a small quantity, t / t i A \^ A =ln I 1H 1 \ @ / Q \_ 2na 208 Solutions to H and R Physics II But AA==X, ...(3) where A is the area of cross-section. 2*0^ / 4 where / is the length N=nl ..-'5 where n is the number of turns/unit length. Use (3), (4) and (5) in (2) to find an expression appropriate for the solenoid. Thus, if the solenoid is long and thin enough the equation for the jnductance of a toroid reduces to that for a solenoid. 36.6. The inductance/unit length, for the solenoid near its center is where n is the number of turns/unit length and A is the area of cross-section. n= _ 1-0 mctcr _____ ^104 (0.1 in)(2.54x ID' 2 meter/in) ^ ^=w(0.02 meter) 2 -!. 256 X10~ 8 meter 2 y-(47 C xlO'" 7 weber/amp-m)(394) 2 (l, 256 XlO- 3 meter 2 ) =0.245 XlO~ 3 h/meter =-0.245 mh/meter. 36.7. Inductance L is given by, L== hK5XlO-3 36.8. (a) Inductance of the toroidal core is b ~2T~ a where N is the number of turns, h is the height of the windings, b is the outer radius, and a is the inner radius. As the cross-section is square, h=b a=12 cm 10 cm=2 cm. Inductance 209 Diameter of each wire is J -0.04 in 0.1016 cm. 2na , _(4wx 10~ 7 webcr/amp-m)(618> 3 (0.02 meter) . J^2 L ---------- - __ , n 1Q =0.28X10 'Mi =0.28 mh (6) Perimeter of each turn for the square cross-section is 4/i=(4)(2 cm)=8 cm Therefore, length of the wire =(number of turns)(perimeter) = (618)(8 cm) =4944 cm =(4944 cm)(3.28x KT 8 ft/cm)=!62 ft Resistance of wire is T . ^ , L 0.28xlO- 3 h Time constant, r= ~=- == R 1.0 ohm =2.8xl(T 4 sec. 36.9. The current i at time t is related to steady state value /o by where T is the inductive constant (we have dropped off the subscript L for the time constant for brevity). or = 1.5 whence T= ' = ^P-lil^^.a sec. In 1.5 In 1.5 36.10. It .is required to have, i/i 1.0 0.001=0.999 From i=/ (I e~*l~ ) or c~' =0.00! or < T = //T=lnlOOO-69 210 Solutions to H and R Physics // 36.13. (a. For an L-R circuit Joule heating is where use has been made of (1). The total energy transformed to joule heat in time / T is T T 00 - ' = ^!T[I -2(1 -0.368)+ -^ (I -0.135)] =0.168 f . (fr) The energy stored in the magnetic field at time t is, Ua(t)=Li*=2 g(l-<r'/> ...(3) Set /=T and (c) Set f==-oo in (3) to find the equilibrium energy stored in the magnetic field. 36.12. /-/ (1 --'*) DiflTereotiating with respect to time, di___i tit dt~ : e , Initial rate of increase is Inductance 211 Suppose d//eft==/o/T= constant, at all times, an assumption which is incorrect. Then or f *-i !'<* . /=-T ! =T 'o 36.|3. /=; (l-< =2.78 XKT 4 sec _ L 50xlQ-h T ~ R~~ 180 ohm t 0.001 sec _ T ~2.78xlO~ 4 sec frf-, S-^--0.278 amp /{ 180 ohm rfi 0.278 amp ~2.78xlO~secc (=0.001 sec -a-=27.3 amp'sec. :, 4 T _:L_- 2 - 0h FJL*w T - , * 0.1 _ ,. "~ T 0.2 sec (a) Rate at which energy is stored in the magnetic field is dUa _ .di dt ~ U dt But, /=/o(l~<?~~' /T ) di I'D f/T -(I) ...(2) -(3) ..-(4) (5) Using (4) and (5) in (3), .. E 1 00 volt ,_ Also, ,.--- ,- - h --10ainp ...(6) ,..(7) 212 Solutions to If and R Physics II /=0.1 sec =240 joules/sec where we have used (6), (7), (1) and (2). 'ft) Joule>heat is produced at the rate ==(10 amp)(10 ohmXi-e" 6 ) a *=155 joules/sec (c) Rate at which energy is delivered by the battery =(240+ 155) joules/sec =395 joules/sec. 36.15. (a) The equilibrium current is . E 100 volt 1A 10 s n = ,7: r = 10 amp. R 10 ohm p (6) Energy stored in the magnetic field due to current / is C/*=4 L/o 2 =J(2.0 h)(10 amp) a =lOO joules. 36.16. Joule heat produced in resistor is OO 00 Uj OO 00 = f Pjdt=* f PRdt. ...(1) When the switch is thrown to b, the current decays, being govern- ed by i=he~th ...(2) Using (2) in (1), : ' 1 '* dt=i*R J ...(3) But *=L/R ...(4) Using (4) in (3) /^ = H/ Q =C/, the energy stored in the inductor. Inductance 213 36.17. The energy density in magnetic field is given by fJLQ where B is the magnetic induction. At the center of a loop of radius R the induction is Using (2) in (1) = J- w a (4*X 10~ 7 weber/amp-meterKlOO amp) 1 B 8 /? 2 ~ (8)(6.05 meter) 2 =0.63 joule/meter* 36.18. According to Example 9 of Chapter 34, in the hydrogen atom the electron circulates around the nucleus in a path of radius R of 5.1 x 10" u m at a frequency v of 6.8 X I0 i5 rev/sec. The current due to the circulating electron is /=?* = (1.6x 10-" coul)(6.8x 10" rev/sec) = l.lxlO~ 3 amp. At the center of the orbit, ., Pol _ (4* X 10~ 7 weber/amp-m)(l.l X 10""* amp> 2R (2)(5.1 X10- 11 meter) = 13.5 weber/meter 2 The magnetic energy density at the center of a circulating electron in the hydrogen atom is __ 1 ffl _ (13.5wcbcr/mcter a )* _ UB ~ V 2 MO (2)(4ic x 10" 7 weber/amp-m) % =7.3 XlO 7 joules/meter*. 36.19. Energy density stored at any point is Now, B at a distance r from the center of a long cylindrical wire of radius b, where r < fr, can be calculated by Ampere's law (see Example 1 of Chapter 34). 214 Solutions to H and R Physics -II <*> Using (2) in (1) Consider the volume element, dv(2nrdr)(l), where / is the length of wire. Then the total energy in the volume, v=ji6 2 /, of the wire is calculated from b u \ 16* o Therefore, magnetic energy per unit length stored in the wire is / 16* 36.20. By Problem 36.19, magnetic energy per unit length stored in the wire is / 16* Therefore, magnetic energy in length / of the wire is UB= & -(I) 16n But UB=\Li* (2) Comparing (1) and (2) r-Mo/ L ~^ an expression which is independent of the wire diameter. 36.21. (a) By Problem 36.19, magnetic energy of the wire is _(4* X KT* weber/amp-m)(10 amp)* 2.5 X 10~ joules/meter. (b) For the co-axial cable, in the space between the^two conduc- tors the total magnetic energy stored per unit length (See Example 5 of Chapter 36) is Inductance 215 . 6 Jn ,. (4x 10~ 7 weber/amp-m)(10 amp) 1 . 4.0 UB = ------- 4jj in -J-Q = 14 X 10~ s joules/mete. (c) *= 4-~ - (I) JL ^o The magnetic induction within the outer conductor is given by Problem 34.6 (c) and is Consider the volume element c/v within the outer cylinder symmetrical about the axis of the co-axial cable. </v=(2* rdr) I -(3) where / is the length of the cable. /*= f usdv^ { ( y )(2ic/rVr ...(4) where we have used f I) and (3). Using (2) in (4) rf f JL( c 2_ -6 2 ) 2 J r (C b c lnc/6 ? 4(i-6Vc)J 10~ T weber/amp-m)(10 amp)' 4 / la (5/4) 3-(16/2S) \ V (l-16/25) 4(1-16/25) / =0.83 X 10~ $ joules/meter. 216 Solutions to H and R Physics 11 36.22. (a) Magnetic energy density, _ fo/ a _(4rcX 10" 7 weber/ainp-m)(10 amp) 8 UB T 1 :? - J6w =2.5x 10~ 6 joule/meter 3 (6) Electrical energy density, WE= eo E 2 where E is the electric field. since /=1.0 meter. *=!.<> ohm/HJOO f,- =3.28X10"* ohm/meter =J(8.9x 10- couWnt-m 8 )(10 amp) a (3.28x 10~ 8 ohm/meter) 8 =4.8 X 10~ 15 joules/meter 8 . 36.23. The magnetic energy density is 1 5 a UB= -=- 2 /ao The electric energy density is M = Je E 2 Set W=/*B ' __ B __ _ 0.5 weber/meter 8 O~ weber/amp-m) = 1. 5 XlO volt/ meter. SUPPLEMENTARY PROBLEMS S.36.1. (a) By Problem 36.2, the equivalent self-inductance of two inductors /-i and L, in series is given by L =L 1 +L t 2M -(I) where M is the mutual inductance. Inductance 217 The positive sign before the last term in the right side of (1) is applicable for the arrangement in which the flux linking each coil, due to the current in the other is in the same-dhrcction as the flux due to the current in the coil, and the negative sign for the arrange- ment in which the flux linking each coil due to the current in the other is opposite in direction to the coirs own flux. If the inductors are far apart then M=0. Setting Li=Lj=L in (1), we find, L =2L. (b) If two closely wound coils are placed side by side then almost entire flux set by either coil would link with all the turns of the other. By definition, where NI and N^ are the number of turns of coils 1 and 2 respec- tively: 02i IS the flux linking circuit 2 due to current ; 2 in circuit 1, and 12 is the flux linking circuit 1 due to current i t in circuit 2. Now, 12 =0 2 (5) 0ji=0i (6) Hence, A/= ^- ...(7) Iv Multiplying (7) and (8) or M=L,L 2 ..,(9) where use has been made of (2) and (3). Setting !=,=! in (9), we get M=L Using these values in (1), A,=0 or 4L depending on the direction of winding. Solutions to H and R Physics~tt S.36.2. For condition (I), time f=0 (a) Applying the loop theorem to the left loop, E 10 volt 7 = _ = =2 amp 1 R l 5 ohm F (b) Applying the loop theorem to the outer loop* LJ/,__ ^Q dt 2 * Jt i But :. EL- --- ^/ a ==0 or i a =0 (c) Applying the junction theorem to the junction of RI and i=/ 1 +/ 1 =2 amp+0=2 amp W) (e) K=L-= J E=IO volt a/ where use has been made of (b). (/) L~ ? --/? 2 /a=10 volt-zero=10 volt J/ 2 10 volt ^ . ir-=^-r - =2 amp/sec. Jr 5 henry w For condition (II), time r=oo. . . 10 volt =2 amp- . 10 volt or ' = = (c) 1=^+1, r=2 amp+1 amp^S amp Inductance 219 (d) F 8 =/ 8 /?=(l amp)(10 ohm)=10 volt. () Ft-J^J where use has been made of (6). (/) $ -0 where use has been made of (6). S.36.3. (a) At /=0, the current i 3 through L will be zero. Hence -.(D Applying loop theorem to the left loop, E-i^-itR^O ...(2) Using (1) in (2), . . E 100 volt (6) At f=oo, / 3 will have a non-zero value! From junction theorem, 'f+*3 = 'i (3) Using loop theorem for the left loop, -/!/?! -~/ 2 /? a ==0 .-.(4) Using loop theorem for the right loop, But Therefore, (5) reduces to XV x,, v Or l,= -jp- ...(6) ^t Solving (3), (4) and (6), t (100 volt)(20Q+300) (10Q)(20Q)+(20Q)(30Q)+(30Q)(10Q) 220 Solutions to // an d A Physics // . ER* (100 volt)(30Q) _ ~=2.7 amp. "~(10il)(20Q)+(20I2)(30Q)+(30ti)(loa)" (c) ii-O The inductor would retain the same current i 3 when the switch is just opened. From (6), /,=(2.7 amp)- 1.8 amp. Also, from (3) with /^O, / a r=/ a =* l.8 amp. (d) A long time after the switch is opened the inductor would have lost all the current and i f 0. Also /^O as in (c). S.36.4. The magnetic energy stored in a coaxial cable is given by * ...d) where i is the current and / is the length of the cable (see Example 36.5 of Chapter 36). Also, its capacitance is given by C = -2?*i ...(2) C In (bid) V ' By Problem, the electric and magnetic energies are equal. 2 4 a where use has been made of (I). Using (2) in (3) 1 p, 2nc 7 MO/*/ , n b ~ ** 4n 1D ~a or !_ f 4 X 10" T weber/amp-m . b 2 V 8.83 x 10- coul/nt-m a -lo - ohms. ZK a Inductance 221 S.36.5. The dimensions and units of some of the electrical quanti- ties arc tabulated below for ready reference. Quantity Dimensions Unit Coulomb Q coulomb Current T~ 1 Q ampere Voltage ML*T~*Q~ l volt Resistance ML*T~ 1 Q~* ohm Inductance JfL 2 g~ 2 henry Capacitance M~ 1 L~*T*Q* farad Magnetic flux MUT^Q" 1 weber (a) [coulomb-ohm-metcr/wcber] (W (c) [c >ulomb-ampere/farad] ,.. [kilogranvvolt-mctcr 2 ] [henry-ampcrej (g)* 5 (coulomb) (e) (henry/farad) 1 / 2 - 37 MAGNETIC PROPERTIES OF MATTER 37.1. (a) Magnetic moment due to the current I through a loop is M-M -(I) where A is the area of the loop. Setting, 4=7cr 2 , were f is the radius of earth, (1) becomes or 6.4 x 10 al arnp-m 2 . 1A7 1*6.4 XIV meter)" ^X 10 amp. (b) Yes. (c) No. 37.2. Volume element in spherical polar coordinates is given by r sin dO If the electron carries charge e and its volume is (4/3) *R 3 , R being its radius, then the charge associated with the volume element is (2icr 2 rfr sin 6 dO) or 3er 2 dr sin Q -(1) The charge element dq goes around in a loop of radius r sin about the axis of rotation, the area of the loop being d/4=*(r sinO)* ...(2) If the rotational frequency of the electron is v then any charge clement dq would also be circula- ting about the axis of rotation with the same frequency ^ The contribution to the magnetic moment due to an infinitesimal current loop is Fig 37.2. Magnetic Properties of Matter 223 </ft=v dq dA _ o> 3er a dr sin 6 <*6 rcr 2 sin 2 6 2* ~ 2* s ^-TSi a" 4 * sin ' e rfe 4/1 3 The magnetic momont of the electron is found out by integrating the above expression R n 4 4R 3 5 3 5 M= ^f- -(3) The mechanical angular momenlum (spin) based on the classical model is given by where 7=(2/5) m/?-, is the rotational inertia of a sphere of mass m, rotating about an axis nassing through its center. We can then write ..,(4) Dividing (3) by (4), A =r f L 2m *-=& .-(5) m L Expression (5) is in disagreement with experiment, the observed value being e/m 37.3. (a) Electric field strength at a distance r is "^ coulK9x 10* nt-m 2 /coul 2 ) (1.0 xlO* 10 meter) 2 1.44xlO u volt/meter, 224 Solutions to H and R Physics II (b) Magnetic induction at a distance r is webcr/amp-m)(1.4xl(T 26 amp~m 2 ) (27c)(1.0xl(r 10 meter)' =2.8 x 1(T 3 weber/meter 2 . 37.4. (a) In the Rowland ring, a toroidal coil is wound around the iron specimen in the form of a ring and originally unmagneti/td With the iron core absent, a current i set up in the coil causes ;: field of induction within the toroid given by "(1) where n is the number of turns per unit length of the toroic! N is the total number of turns and r is the mean radius of the ring. From (I) we have =: (2)(5.5x 10~ meter)(2x 10"* weber/m 2 ) (4xlO~ 7 weber/amp-m)(400) =0.14 amp. (b) Because of the iron core the actual value of the induction will be B, given by B=B +B M .-(2) By Problem, Bw=800 B . ...(3) Using (3) in (2), B=801 fl =(801)(2 X 10" weber/meter 2 ) =0. 1 602 weber/meter 2 ... (4) The charge q flowing through the secondary coil is given by BN'A ... ^ R - (5) Where N' is the number of turns of the secondary coil, R the resistance of the secondary coil and A the cross-section area of the toroid. .1 = *<f-/4 ...(6) With the diameter of the cross-sectional area being </=6 cm 5 cm = 1.0 cm=0.01 meter A- * (0.01 meter)*=7.85xl(T 5 meter 2 . -(7) 4 Magnetic Properties of Matter 225 Using (4), (7) and the values #'=50 and /?=8.0 ohms, in (5), we find = (0-16 weber/meter a )(5Q)(7.85 X 10~ 5 meter 2 ) q (8.0 ohm) =7.85xlO~ 6 coul. 37.5. (a) The dipole moment of the bar is r*~Np , -U) Here N, the number of atoms of iron in volume ( v=(5 cm) (1cm 51 ) = 5 cm 3 , is given by v^^oPJL M where W =6x 10 23 is the Avagadro's number, p the density of iron and M the atomic weight of iron. We have (6xlQ 28 /mole)(7.9 gm/cm 3 )(5 cm 8 ) " (55.85) -4.2X10". Mb =(4.2x 10 M )(1.8x 10~ 23 amp-m 2 )=7.6 amp-m 2 (b) The magnitude of the torque is given by T=^J8 sin 0=(7.6 amp-m')(1.5 weber)(sin 90) = 11.4 nt-meter. 37.6. The atoms of a diamagnetic material do not have permanent magnetic moments, the individual magnetic moments of orbital electrons neutralizing each other. However, in the presence of an external magnetic field, a magnetic moment is induced in the atom. The phenomenon of diamagnetism can be explained in terms of electromagnetic induction. Consider a typical electron revolving in the xy plane as in Text- book Fig. 37.7 (a) and (b). Let an external magnetic field be switched on along the z-direction. In the absence of an external magnetic field the centripetal force which enables the electron of mass m and charge e to move around in a circular orbit of radius r is given by r F = m ^ -O) r Let the field B be increasing at a rate dBjdt. This will induce an electric field E around the path. In accordance with Faraday's law of induction, the induced emf is dt 226 Solutions to H and R Physics II The flux through the loop is *~r ...(2) The additional force acting on the electron is dv _ 1 dB ,,, m ^^ E -2 er T t '" (3) where (2) has been used and the sign has been ignored. From (3) we find a connection between the change in v and the change in B, dv= gdB ...(4) Suppose v increases. As the radius of the orbit remains constant, the increase in centripetal force is provided by the magnetic field. (The speed of electron v will increase or decrease depending on the direction of the magnetic induction B). Call Av the net change in v during the process the field attains the final value B. B > The magnetic moment arising from the circulating electron is or /*=t^ a a> Change in magnetic moment, holding r as constant, is given by AfA=i er a A w (6) But from (5) we have eB ,-. m ...(7) Using (7) in (6) r ^ B Magnetic Properties of Matter 221 Thus, the efTect of applying a magnetic field is to increase or decrease the angular velocity of electron depending on the sense of circulation. This, in turn, causes an increase or decrease of magne- tic moment. The change in the magnetic moment is in opposition to the applied Reid. 37.7. By Problem 37.6 . eB <*> eB is Set e=1.6xl(T 19 coul ; m=9.1xl(r" kg, a> =4.3x 10 8 rad/sec ; #=2.0 weber/meter 2 . We have chosen a typical value for B, and the value for corresponding to Bohr's model for hydrogen atom. (1 .6 y \ Q~ 19 coujK2 weber/meter^)_ 6 ~ (2)0.2 XlO" 31 kg)(4.3xTo 16 rad/sec) "" Thus, A a* << cu 37.8. The spin angular momentum is a vector which points in t' direction of the axis of rotation (Fig. 37.8). The direction of ,u il magnetic moment and the sense of flow of positive charge in the loop are related by the right-hand-screw rule. The spin magnetic moment is given by Angut* i Magnetic mompnt Thus, the magnetic moment is propor- tional to the angular momentum. Desig- nating spin angular momentum by the vector 5, we may^ rewrite the above relation urn The positive sign in the right hand side shows that the v urr.nt of the positive charge is in the s.ime direction as the spin motion. The spin and the magnetic moment point in the same direction 37.9. C.TW.XO where N is the number of protons in the sample and u is the magnetic moment of the proton. 228 Solution* to H and R Physics 11 Number of water molecules in 1.0 gm of sample (18 gm) Now, in a molecule of H 2 O there aro 10 protons. (2 from hydro- gen atoms and 8 from oxygen atom). Therefore, the number of protons in the sample is .3 x IC 22 J=3.3 X 10 23 * x !<T W Joule/Tcsla) (2w)(5xUT 2 'nicter)""' "" =7.5 X 10~ 6 weber/meter 2 . 37.10. v.- ...(I) or or "S- fi Vp/,5 Using (4) in (2), m vp.s 37.11. (a) As the field in the median plane of a dipolc is only half as larrc as on the axis, --- B 2 4 xlO"ber/air 8 X . (4w)(l. OxlO" 10 meter) 8 = 1.8 weber/meter 2 . (M Energy required to turn a second similar dipole end for end in this field is -=(2)(1.8xlO~ 23 amp-m 2 )(1.8 weber/metei 2 ) =6.5 xiO- 2 joules. At room temperature, the mean kinetic energy of translation is tfryArrf | Vl.38x 10"- 3 joulo/ K)(30nK) -6 x 10- joule. Magnetic Properties of Matter 229 JIKI-, 1'f^H/ 1 ) t//r Consequently, the energy exchanges in colli- sion > i HI ilcsi.'jy iiic alignment ofthodipoles with the external Held. .V7.I2. In t i. ;7J ? the line ab lies in the interface between the two media ! dMd 1. Let t and B 2 make angles 6 1 and t respectively, Ai'i-i i-... lanital i' the interface. We shall now apply Gauss' I:K\V.* n. iu ; u- ::!; x>x shown in the diagram I' i*^ --\* V '.- J A/.v-fl 1 s 1 + L LT 0- --t-i^c win rice. /J, co> i', ~ # 2 cos i 1 , s!;ouing I-.) % rch> :ii:it !:: n.-rmal component of the induction ha> the same value on c.ich >iJc of -he surface. 37.13. Ci usidor f c ,cciang,.iar path us ^iiown in F : ig. 37.13, Since the clojd! puti: eiidoaCh zer^ current, it follows that the line inte- gral of // *:r -und tiio p:ih is /CM*. / H.rfi -0 In the evaluation of !he above integral the contnhution due to the paths dc a:u! i/can be igmrcJ. Ihub, / 0- J //!.<//+ J Hr'Jl c c or or sin Thus the tangential component of // ha the same value on each side of the surface. 232 Solutions to ff and R Physics It By Problem FE=NFn Now, pB ...(2) .-(3) -(4) Combining (1), (2), (3) and (4) or o>=^ (Nl) (b) = (AT+1) = (0.427 weber/meter 2 )(1.6xlQ- M coul)(100+l) (9.1Xl(T 81 kg) =758 X10 10 rad/sec. JJe , ., to, = (0.427 weber/meter 2 )( 1.6X10^ coul)(100~l) B - 743 XlO 10 rad/sec. S.37.4. The magnet will tend to align its axis antiparallel to the direction of B but is opposed by the torque due to its own weight. The torque due to magnetic field is ta^pB cos 6 ...(1) Here 6 is the angle between the axis of the magnet and vertical and ft is the magnetic moment, given by M =2m/ ...(2) where 21 is the distance between the poles of the magnet and m is the pole strength. In Fig. S.37.4 the torque T B acts in the counterclockwise sense. The torque t g due to the weight acting at the center of the magnet is given by T,= A/g/sine ...(3) acting in the clockwise sense. Pig S.37.4 For rotational equilibrium we have the condition, Tf=T* 6 or "-gr (4) (5) Magnetic Properties of Matter 233 Thus the magnet is oriented at an angle 6 with the vertical given by (5) with the north pole moving away from the direction of B, whilst the string remains in the vertical direction. S.37.5. The reluctance S is given by ,3 = I -J-, p^A L M r where /ir is the relative permeability, I I is the flux path, / t is the gap length and A is the area of cross-section. 1 T 1.0 meter . M ^1 " (4KXlO-*weber/amp-mML 5000 rv "' "~""J The flux, 8121 , . , = -j amp/ weber ..-(I) )/l = 1.8^ weber ...(2) Magnetomotive force (mtnf)Ni ampere-turn ...(3) . <f>S (1. 8X weber) / 8121 , . ,\ _. . or I= ~AT w I j amp/weber) 1=29.2 amp. iV DvU \ y^l / 38 ELECTROMAGNETIC OSCILLATIONS 38.1. Ci=5 pf ; C 2 ~2 pf ; jL=10 mh (a) LC l combination: Resonant frequency, v 1 = -- -~- - - - l - -=- ^2n V" (10 X ~l(r h)(5 x 10^ f ) =714 cycles/sec. (6) LC 2 combination: Resonant frequency, v 2 = 2* V LC 2 1 2w V(10xl(T 3 h)(2xlO-f) = 1 126 cycles/sec (c) LC X C 2 combination with Q and C., in parallel: Equivalent capacitance of C l and C 2 is C=Cj+C 2 =5 i Resonent frequency, v 3 = - j 2tc V =602 cycles/sec ((/) LCiC 2 combination Ci and C a in series: Equivalent capacitance of C l and C 2 is M Resonent frequency, 4 At* i 2 YUOxHT' b)(1.43 x ib'* f) -=1333 cycles/sec. Electromagnetic Oscillations 235 .1. Frequency of oscillation is v = r^^- 2* </ LC C- ! =2.5 XKH 1 farad. Thus, by combining a capacitor of C=2.5x l(T n farad with the given inductance L= 1.0 mh, we can get oscillations at l.QXlO 6 cycles/sec. 38.3. Frequency range extends from v^ZxlO 5 cycles/sec to v 2 =4Xl0 6 cycles/sec. The frequency range is, =4x 10* cycles/sec 2 X 10* cycles/sec =?,X 10* cycles/sec. The angular range on the knob is A6=180. Therefore, the rotation through 1 corresponds to a frequency change of 2 XlO 5 cycles/(sec-degree) rotation. For any rotation of 6, the frequency would change by 2 x 10 6 0/180 cycles/sec. Thus, the angle 6 and v would be related by 2 x 10* 6 / 6 \ =2X 10 ' l + rcycles/sec ...(1) l 180 C=- (4u J )(2x 10 cycles/sec)(1.0x ID' 3 h)t 1 + -^-)' _ 634X10'" f . 634 or " , farad= 236 Solutions to H and R Physics II Figure 38.3 shows the plot of C as a fraction of i.nglc foi 180 n rotation. C Ifl/UfJ 600 400 200 180 e 90 Fig. 38.3 38.4. For an LCR circuit resonant frequency is given by o>' _ 1 [ 1 _/ R ' J 2* 2n V LC \ 2L 1> Squaring and re-arranging (1) -IE For the given data for v, R and /., the secoiui i'.rm on tlic -igln hand side of (2) is quite small compared to the lir*t term. Lq. (2) then reduces to 1 'LC or C r~. 1 cyclcs/sec) 2 ( 10 h ) The given LCR circuit is as good as resistanceless LC circuit. 38.5. The potential drop across the capacitor is Vc=qlC and the potential drop across the inductance is _.__<// VL-L W Electromagnetic Oscillations 237 Applying the loop theorem to the LC circuit, VC + VL -0 U. B ,,t But This is the Textbook Eq. 38.5. 38.6. Conservation of energy demands that the applied emf, E=2sm cos >" / be equal to the sum of the potential drop across L, C and R of the circuit "' L ~ {IR ^ Em cos " t - (1) ,- ... Using (2) and (3) in (1) which is in agreement with Textbook Hq. 38.12 of the textbook. 38 7. The stored energy in the capacitor is V'~-=fc - (l) ;iiicl thi* maximum stored energy in the capacitor is f//:..,,*-'! -(2) By Prohlcm, t/i .-~\Ui.*, ...(3) 7 2 . 1 fl' 2 M , 2C " 2 2C " (4) where use has been m;uic of (I) and (2) Rut the oscillation amplitude of charge is 238 Solutions to H and R Physics II RtlL q*f - whence e = T === ^ where use has been made of (5). or f= In 2=0.69r t where we have used the fact that the induction time-constant TL=L//?and In 2=0.693. 38.8. The undamped resonent frequency is given by '" The damped frequency is given by zrr* LC \2L _ By Problem, -- - - -= 1 x But .~ = i~- V Solving for R, we get, = V(8)(10x 10-8 h)(10- 4 )/(10- f) = V8 ohm=2.8 ohm. ,_ f ^-JAY -\ILC \2L) co c 5 | __ fi o co A/ 4L ...(2) Electromagnetic Oscillations 239 where use has been made of (I) and (2) and we have expended the radical binomially. whence, t*= In ~ = In 2 ...(4) CR* (In 2? or Use (5) in (3) to find, g>-g)' (In 2) 2 (0.69315)* 0.00608S (8X9.8696) n 2 2 0.0061 f^t . 38.10. q = qme- R *t 2L cos a// DifTerencitate with respect of time to obtain . dq I R RtjIL , , . RtllL \ l ^~dt^ qm \-2L C coso,/-*, sin 01 r^ j ( 2 ^coso>'/+sinov) >L> (tan ^ cos o/r+sin o>'/) where we have setr7~ / =tan <. This gives _ , Rt/2L /=- z!^ - - - ( s i n cos co'/+cos ^ sin cos Sin COS 9, But for low damping, the resistance R is small. Hence, ^ r -, -*0 and ^ ^ So, 240 Solutions to H and R Physics II 38.11. If UE, mat is the initial maximum energy and Us the energy at any time then change in energy But, amplitude of charge oscillation is =m e -Rtl2L ...(2) Using (2) in (1) Set UE, maxU, then AC/ , One cycle implies that f= = V OJ . At/ sin (o.'/-^)+a>'/? cos (o>*r-^)+ ~- sin 38.12. L+/?-f--=J? ra cosa,'r ...(1) Let q=a sin (o'r $) ...(2) Then, =ao>" cos (/-#) '...(3) -(4) ai~ Substituting (2), (3) and (4) in (1), =/Tm COS fti'f ,..(5) Electromagnetic Oscillations 241 Expanding the sine and cosine functions 0o/ a L (sin oft cos # cos o>*/ sin ^)+0a>*.R(cos *>"/ cos +sin co"/ sin #)+-^(sin <*>"t cos ^ cos o>*/ sin ^) o ==m COS to't ..-(6) Comparing the coefficients of sin o/f f a<*>"*L cos ^+0</ /? sin # + -^ cos ^=0 c/ Re-arranging the terms and cancelling the common factor a, Comparing the coefficients of cos a>*t in (6) aa>"*L sin 0+^01"^ cos ^ r sin ^=m m or -, -- p^- ( a>"*L--^ j sin #+o,*/J cos # From (7) using trignoraetric identities we easily find - (10) Substituting (9) and (10) in (8) and simplifying, Em Maximum value of qm is conditioned by setting ^y Cancelling the denominator as well the common factor Em, w' (2L 1 co*-2I,/C+^)=0 The nontrivial solution which gives q m a maximum is 240 Solutions to H and R Physics II 38.11. If /, mo* is the initial maximum energy and UE the energy at any time then change in energy But, amplitude of charge oscillation is qq m e~ Rt frL q*=q* m e~ Rt l L ...(2) Using (2) in (1) - -- 2C~ 1C -2C Set UE, max=U, then AC/_, -Rt/L l-e 1 2w One cycle implies that t = J v to At/ -..- -='-( -^g +) 38.12. L+/?--f--='cos tu '/ ...(1) Let ?=a sin (<a't~<f>) .-(2) Then, =aa,' cos (a//-^) --..(3) j u *f ** I ** \ (A\ a d i7*~~~ a<J * sin ^ *""$> --W rfr Substituting (2), (3) and (4) in (1), sin (art+)+a<'R cos ^^~0;+ sm i-/ ^) =m cos w'f ...(5) Electromagnetic Oscillations 241 Expanding the sine and cosine functions 0ai* a L (sin o/f cos ^ cos a// sin <t>)+aa>'R(cos aft cos +sin co"/ sin #)+--(sin a>*/ cos ^ cos a// sin ^) =; cos a// -..(6) Comparing the coefficients of sin oft, ~a<*>**L cos (f>+a<*>" R sin <f>+- r cos #==0 o Re-arranging the terms and cancelling the common factor a, Comparing the coefficients of cos <*>"t in (6) sin ^+flw"^? cos ^^T sin <f>m _ -_ or a j~ ^'L - sin #+>'/? cos From (7) using trignometric identities we easily find COS *" 7K^+(o,'L-l/C) " 00) Substituting (9) and (10) in (8) and simplifying, Maximum value of qm is conditioned by setting ,^7=0 a<o ___ _ rf'~ 2 "* [V a ^ Cancelling the denominator as well the common factor Em, w" (2L<o"*-2L/C+/?)=0 The nontrivial solution which gives q m a maximum is CO sss 2L* 242 Solutions to H and R Physics It 38.13. The LCR circuit will oscillate with maximum response i.e. the maximum amplitude of the current oscillations occur when the frequency <a" of the applied emf is exactly equal to the natural (undamped) frequency <a of the system. 1 The amplitude im of the current oscillations is given by Em At resonance, co /f ==o> and /m= !* By Problem, . m== Em 1 Ern Squaring and simplifying or o/X-- = or <oLC+ V3 o>' CR- \ =0 . .. 21, ^V 4L + LC These are the only possible solutions. _, ^yj^ ohm) / 3 /20 ohm \ 2 1 ^ - ^/r^vx- +J 4 V i.Qh / + (1.0h)(20xlO"f" 1.0 h) 01^=24 1.6 rad/sec co/ =207 rad/sec. cu/ 241. 6 rad/sec . f , v 1 <r = -= r =38.5 cycles/sec. o~ 207 rad/sec -, A , / _ _ _i =33.0 cycles/sec. Electromagnetic Oscillations 243 38.14. The amplitude im of the current oscillations is given by At resonance, co*=co and . _ Em Im ~lT . _ _ m ___ L /m ~V KL-1KCP + /P 2 Squaring and simplifying or > V LC V3 w' CR-1=0 The only accep^aole solutions are V3 . Subtracting the last equation from the previous one, and dropping off the primes, o co 38.15. Resonant frequency for L^C^Ri in series is 1 i~" HP^-- .-(I) Resonant frequency for Z- 2 C 2 /? 2 in series is ...(2) By Problem, eo^cog (3) that is, J&A-IiC, -(4) For the two combinations in series, the equivalent inductance is 244 Solutions to H and R Physics II Assuming that the inductances are far apart, the equivalent capacitance is The resonant frequency is then given by ...(7) where use has been made of (5) and (6). Using (4) in (7), we find upon simplification, 38.16. Example 5 is concerned with a parallel-plate capacitor with circular plates. By definition, the displacement current is given by . dE = ~ Displacement current density is . _ id _ dE J n*~ ^ e ~df 38.17. By definition, the displacement current is given by d$E d =e A -AYZ ^=f<Ld- d *L=-c dv ~ dt \ d ) d dt dt where we have used the following formulas for the parallel plate condenser. d being the distance of separation of plates, and for the capacitance, >m e A d A being the area of cross-section of plates. Electromagnetic Oscillations 245 38.18. Using the results of Problem 38.17, the displacement current is given bv ~ dt dV __ id _ 1.0 amp f/ ._ f . ' ~dT~ "c ~ TjoZiw** 10 volts / sec - Thus, the displacement current of 1.0 amp can be established by changing the potential difference at the rate of 10 e volt/sec. 38.19. Consider the parallel-plate capacitor with circular plates. By definition the displacement current is given by " dT In the region r < R In the region, r > R, <f>E=(e)(nR*) M=e --- (nR*E)=e *R* ~~ 38.20. (a) Ampere's law in its modified form is -d) In the wire conduction current i alone flows whilst within the capacitor C displacement current id alone exists. Since the con- tinuity condition says that the total current, conduction current plus displacement current, is constant in a closed circuit, it is sufficient to show that /<*=/. In the gap of the capacitor, Differentiating, dE/dt= -r- dqldt = . i d^L r , d(EA) A dE Rut irf = e T-T- //= JLJUk *u CA , A-I^^*Q !)- J^* dt at at Combining the last two equations, id = i. (b) The conduction current (Fig 38.20) flows up the walls of the cavity and the displacement down through the volume of the cavity. 246 Solutions to H and R Physics II Application of (1) to to region outside the cavity, such as the path r, shows that 5=0. Consequently, the net current enclosed is zero, the conduction current being exactly equal and opposite to the displacement current. Thus, the continuity condition of current is satisfied. Fig. 38.20 38.21. (a) =J? m sin a>* dE _ rr ~Lm co COS (at Set ai^ojj where the angular frequency for the fundamental mode as in Textbook Fig. 38.8 is given by 1.19c i T~ where c is the speed of light in free space and a is the cavity radius. (1. 19)(3X10 meter/sec) , . , m0 .. , 1 -- 2.5X10-' meter ~ ==1 ' 4X1 radlans /c (Jr \ -jf j=(10 4 volt/meter)(1.4x 10 l radians/sec). = 1.4x 10 U volt/meter-sec, Electromagnetic Oscillations 24? ~ By Problem, -^- = ^ f -^- L = x 1.4x 10 M volt/meter-sec =7 X 10 U volt/meter-sec. Set ^=0=2.5 X 10"" 1 meter Also B( 1=:- dE - ( 2>Sx 10 ~ g meter X 7x 1Q1S volt/meter-sec) ^ r j 2c a rfr (2)(3 X 10 8 meter/sec) 1 = 10"" 6 weber/meter 1 38.22. r < R r > R (a) B(r) 2*r (b) E(r) I 8 2rcor i ji? (c) dt 2r JL ^? _! _ 2 r dt 2 rrfr Maxwell's equations in differential form are (OVXH-/+ (/) V.B==0 (iv) V.D-p with ,ff==/x //, D=e We shall verify that Maxwell's equations satisfy (r)and (r) as given in the above table. (a) VXB-O; (r > R) ...(1) where k is a unit vector in the direction of ithe conductor. Using the cylindrical coordinates, and using only the radial dependent term 248 Solutions to H and R Physics 11 For r < R, n .*. V xB= For r > R, .'. VXB=0 (rf) Assuming electric field is negative in the direction and is proportional to r, E^Ar ...(3) The correctness of this assumption is proved by showing that (3) is satisfied by the magnetic field. Here A is to be determined and is independent of r, 6 and z. From Maxwell's equation (i7) VXE= 2=-k2X - r Using (4) in (3), Electromagnetic Oscillations 249 The induced electric field assumes a maximum value at a radius. R. The electric field does not drop abruptly but dies away at r > R in such a way that the electric field has no curl. To determine the electric field in the space beyond the magnetic field stretches out it is necessary to satisfy the following conditions: (i) E has no curl, (11) E is continuous with the field given by (5) at r=R 9 and (111) JE-*0 as r -> oo. x - Conditions (1) and (3) are satisfied by V- 7 -> where C is a constant. At radius R, (5) and (6) must give the same value of E. C __ R dB_ R~2dt C- -I* f ...(7, Using (7) in (6) *-- - SUPPLEMENTARY PROBLEMS S.38.1. Initial energy stored in the 900 /zf capacitor is If this energy is eventually transferred to the 100 nf capacitor, then or -^ I^L =(100 volt) x / ?0p./*f -300 volt. VC 2 \ 100 /if Thus, the 100 /*f capacitor can be charged to 300 volt, but a direct transfer of energy is not possible since the charge from C, to C t will cease to flow when the capacitors have equal potentials. One can circumvent this difficulty by first converting the electrical energy of C l into ihe magnetic energy of the inductor L by closing the switch 5, for a specified time and then immediately transfer the magnetic energy to C 2 in the form of electrical energy by opening switch 5 t and closing the switch SV It is known th;iMn an LC cir- cuit the transfer of electrical energy into magnetic energy and vice 250 Solutions io tt and R Physics It versa takes place in a time T/4 where T 2*/Zc is the time period of the electrical oscillations. The time period for the LC V circuit is calculated as 7 T 1 =2n v /LQ = (2w) V(10 h)(900 X 10~* f)=0.6 sec. Similarly, J 2 =2*v'ZQ==2*(10 h)(100x HT f)=0.2 sec. The procedure then consists of the following sequence of opera- tions: Step (/) Close S 2 and wait for time TJ4, or 0.15 sec, during which time the 900 pf capacitor is completely dis- charged and the current in the inductor is fully esta- blished producing a magnetic field in the surrounding space. Step (11) Immediately close Si and open 5 t so that now the current in the inductor begins to decrease and after a time r 2 /4 0.05 sec the magnetic field of L would have disappeared and the 100 jxf capacitor fully charged. Step (iii) Immediately after 0.05 sec open the switch S\. The 100 p,f capacitor is now charged to 300 volt. (6) The mechanical analogue consists of the oscillations of a mass m attached to two uncoupled springs of force constants k t and 2 as in Fig. S.38.1, the free ends of the spring being fixed to rigid walls and mass m is free to slide on a frictionless horizontal table. A mechanism R* analogous to switch S 2 in the electrical case, Fig. S.38.1 allows the spring k to be released when it is stretched through a given distance from the equilibrium position. The same mechanism /?, may be maneouvered to get the mass detached from the spring ki. Another mechanism R l analogous to switch S l allows the mass m to be attached to A: a . Furthermore, R l may also be used to get the spring fc a looked up when extended to the maximum. The procedure for the transfer of deformation energy of spring fc t (analogous to the electrical energy of C\) to the spring Jfc t (analog- ous to the magnetic energy of the inductor L) is as follows: Step (0 Operate R % so that the mass m is attached to & t and moved through a distance A. Then release it. Wait for a time TJ4, where Ti*=*2n<( mlki* The mass m would have required maximum kinetic energy. Electromagnetic Oscillations 251 Step (11) Immediately use mechanism R 2 so that the mass gets detached from k^ and use R so that ni is attached to fe a . Wait for a time TJ4 where r t =27cVw/Av in this line m would have momentarily come to stop when the spring k 2 is stretched to the maximum. Step (iii) Immediately after time T 2 /4 use ^ to get the spring fc a locked up. In the table below are compared quantities related to the mechanical and electrical oscillations for the Mass+Spring system and the LC Circuit. Mass + Spring System LC Circuit (a) Nature of motion Simple harmonic Simple harmonic (b) Equation of -~ mv 2 + -^ k& -z Li* -K- ^- motion (c) Frequency of L JtL L \f oscillation 2n \ m 2-x ^ LC (d) Description of xA sin cat q=Q sin cat oscillation dx_ ._dq V dt l ~dt S.38.2. (a) Let the left parallelepiped contain charge q^ and tlie right one charge q t . Applying Gauss law to the left parallelepiped, e c / 1 E.dS=<h ...(1) Simlarly, for the right parallelepiped, e c / 2 E.dS=q t ...(2) Now, e / E.dS=* E.dS +e E.dS ...(3) where 1 (NC) refers to all the surfaces of the left parallelepiped which are not common with the right side parallelepiped, and 1 (C) means common surface. Similarly, c f E.</S=e f E.dS +e j E.dS ...(4) 2 2(NC) 2(C) Adding (1) and (2) and using (3) in (4), +e c j E.dS +e f E.^S -f e { E.JS ...(5) 2(C) Solutions to ff and R Physics it The last two terms on the right hand side get cancelled together since the normal to the common surface in the fourth term is opposite to that in the third term and consequently e f E.dS=-e f E.dS 2(C) i(Q " fr+ft^tf^o J E.rfS +e f E.dS =e J E.dS l(NC) 2(NC) (Composite) This proves the self-consistency property of Maxwell's equations. (b) Proceeding along similar lines we can prove the desired pro- perty by using / B.</S=0. b c d a S.38.3. (a) f E.d\ - f E. dl + f E. dl + f E. d\ + f E. dl abcda J f J J e f c b / E.dl = f E. dl + f E. dl + f E. dl + f E. dl fcl J J J J b e f c Adding (1) and (2), b e f c $ E. d\+ f E. d\ = f E.d\ + f E. dl + f E. d\ + f E. dl abcda befcl * J J } a b e f d a c b 4 f E. dl + | E. dl + | E. dl+ I E.dl dt /ocd V dt A./. " < c b Now | E.dl= -| E.rflj ...(4) b c _ 4 . Further, ( #B\ . { <#B\ , I -3- I I -~ I \ at /j>fj \ at Jtife - [A(<H>ctl+A { brfti\ -J Electromagnetic Oscillations 253 dt Using (4) and (5) in (3), be f c da f E.rfl+ [ 'E.</I + f E.dl + f E.d\ + f E.d\ + f E.rfl / /&& E. </!=- Thus, the self-consistency of Maxwell's equations is proved. (b) By a procedure similar to the above, we can show by using / B.Jl=/+e ~~J~> the desired property. S.38.4. According to Gauss' law for electricity, e / E. </S=<7=Jp dV ...(I) According to Gauss law for magnetism / B.rfS- W ~(2) According to Ampere's law modified by Maxwell if O ,71 d<pE . = ^o e o 4- / E.</S+^ 1 J. </S ...(3) According to faraday's law of induction, /_, d^n d r n , E.rfl- --- j ^ -. - J B.JS Equations (1) through (4) are the desired Maxwell's equations. 39 ELECTROMAGNETIC WAVES 39.1. Conduction current results from the flow of positive charges towards negative charges and negative charges towards positive charges due to electrostatic attraction. This is seen to be .consistent with the direction of flow shown in Fig. 39.1. T ^ f \ ? \ f t , * ik > i ^ ^ > f i f \ r } r i k > ^ > L > ji, , L f > t 1 r ^ p } f > k J i * \ l \ ^ r > r 1 f \ f Fig. 39.1. 39.2. Assuming the dominant mode, with n=3 cm the as width of the rectangular waveguide, and c= 3 X 10 8 meter/sec, the velocity of electromagnetic waves in free space, the phase velocity is given by i c 3 X 10 meter/sec Vj>h f . . - -- . rrr: /*"= --, -, Trr - - - ------ 1 -=^=; VI -(A/2^) 2 VI -(A cm/6 cm) 2 where A is the free space wavelength. The group velocity is given by vpr-c Vr-(A/2a) 8 -=(3 X 10 8 meter/see) and the guide wavelength is given by ~(A cm/^cm) 1 * _ ___ ____ 9 Vl -(A/2*) ^ A cm Fig. 39.2. function of A. ), (b) and (c) show the plot of V*A, v ar and A a as a 5X10 8 3X10 8 Electromagnetic Waves 255 10X10 8 r 1 "ph (m/s) (a) A (cm) Fig 39.2. (a) 3X10 8 (b) A (cm) Fig 39.2. (b) 256 Solutions to H and R Physics II (cm) Fig 39.2. (c) 39.3. The group velocity v ff r is found from _d|stance _ 100 meter Vffr ~~ time 1.0X 10~ fl sec ~ vgr _ 10 8 meter/sec 1 c 3xl0 8 meter/sec 3 10 8 meter/sec Solving for A, A=2a Vl (v,r/c)=(2)(3 cm) V"l (1/3^=5.66 cm. _ (3 x 10 8 meter/sec) 2 Phase velocity, vph~ - = -.^ - 7- 10 8 meter/sec n IAg ^ , -- ~ ==: 9 X 10 8 meter/sec 39.4. Guide wavelength is given by A- A .' V"l-(A/2a> CJ *A \ ^ \ tjvi A^ = /A Use (2) in (1) and solve for A, A -(1) -.(2) whence, A= if 3 a. Electromagnetic, Wafts 257 39.5. The displacement current id by definition is d+s d(EA) A dE /t , "-'o-dr^* ~dT ~* 4 ~rfT - (l) where #E is the flux, E the electric field and A the area. The electric field for the plane wave is given by =J?m in (fcx co/) ...(2) ~*T-~--a>Em COS (fot w/) (3) Using (3) in (I), and setting A=\ /*= C 01 m COS (/CX oi/) 39.6. The density of energy stored in the electric field is i/=Je E* ...(1) The density of energy stored in magnetic field is -(2) The fields for the plane wave are JE=Em sin (fcx-o>r) ---(3) ^=5m sin (kx-vt) (4) Substituting (3) in (I) and (4) in (2) tt/1= =4 EO E? sin 2 (fcx-eof) --(?) oir) -(6) 2 ^o Dividing (5) by (6), ?-= /o^oj?.?. ...(7) M ^m But .^-i -(8) and Em^cBm (9) Using (8) and (9) in (7) we , - = 1 or UB == us UB 38.7. For the coaxial resonator of length L and closed at each end, the resonant frequencies are v r ,/iv/2L. 258 Soh fiows to H and R Physics II By Problem, L=3A/2, whence n=?3. .__. . V- T s,n _-_ . L cr L (See also Wave Guides by H.R.L. Lament, John Wiley & Sons 99.8. m j ;.;. . ^ conductor ^^ Inner conductor Fig. 39.8 39.9. The guide wavelength A? is eiven by 10cm ~- I 5 CTTI The cut-off wavelength A c for this guide in the fundamental mode fs given by m=12 cm. 39.10. Figure 39.10 shows the electric field lines for an oscillating dipole at various stages of the cycle. In the beginning the electric lines will be attached to the charges but as the charges are reversed the electric lines are detached, forming closed loops. The magnetic field lines are concentric circles around the dipole axis. Far from the dipole the wave is essentially spherical moving radially outward with velocity equal to that of light. Electromagnetic Waves 259 x - ^ ^*v s /'" N xx ^L \ v\ r? -;>\v, + *c-' *7~- ,^r Fig 39. 10. re) V-^ tj Fig. 39.10. (f ) Figure 39.10. shows four figures (e\ (/), (g) and (A) in sequence to Textbook Fig. 39.9 (a), (6), (c) and (</). The fifth figure. Fig. 39.10 (0 is identical to Fig. 39.9 (a). The lines of B form closed loop that move away from the dipole with speed c. 260 Solutions to H and R Physics 11 Fig. 39.10 (g) it,., Electromagnetic Waves 161 39.11. (a) E 1 f^ '\ * *s =ii ^ "VJ *r - S-J- z-direction Fig. 39.11 (6) Consider a length L of the conductor whose return path is remote. Let the voltage drop be V. Then .= ...(1) The Poynting vector is integrated over the surface of the cylinder. As 5 is directed radially inward, the caps at the ends do not make any contribution to f Su/A. The curved surface alones makes contribution to the integral. Thus. Observe that ^ anc * ^ arc mutually perpendicular to each L /S.<fA= J E z B,2nrdz L -T ' I *- w where use has been made of (1) and (2). Now the quantity Vi represents ohmic loss. We, therefore, conclude that the loss is supplied by energy entering through the surface of the wire. Within the conductors? the Poynting vector is directed radially inward so that the ohmic loss equal to i*R is supplied by the fields within the conductors. However, between the conductors the Poynting vector is directed towards the load so that the energy directed toward the load flows through the dielectric around the conductor and not through the conductor itself. 39.12. For the present problem, *=* sin (ityfb) exp [j (<o/ , .=0 B,=* ~ sin (xylb) exp cu 262 Solutions to tt and R Physics It where Jfc 2*A* a is the wave-number. The average value of the Poynting vector is where B* is the complex conjugate of B. i J k <Sa*=*Y-Re Em B,* A* i J_ R e (~.J?* J+0>*K) C Substituting the values of JE, U, and A in (2), we find that th first term in the parenthesis of (2) is imaginary whereas the secon term is real. The energy, therefore, flows only in the z-directioi and, The value of S is independent of x and is zero at the walli (y=0 and y=b) where E is zero and is maximum at y=bf2. 39.13. (a) The rate at which the energy passes through a face oi the cube is "2 As E and B are in the xy plane, the Poynting vector S points in the ^-direction, the direction of wave propagation. Thus, the energy passes only through faces parallel to xy plane. It is zero in the remaining faces paralled to yz and zx planes. For the xy plane, / ^ X ^r .^ C. ~~ ~- < 1 I B ^r | .^ ~~ y .>^ Ix^^U **/ *y Fig. 39.13 ExB "~ Mo Mo Mo where we have set A*=*cP. (6) As the energy flux within the cube is constant i.e. P is time independent, the net rate at which the energy in the cube changes is zero. Electromagnetic Waves 263 39.14. Radius of the wire, r=0.05x 2.54 XKT 1 meter = 1.27xl(T 8 meter. Magnetic induction at distance r is q ?~ 7 weber/amp-m)(25 amp) 2*r ~ T2wXT.27 X UT 1 meter) ==3.9 X 10~ 8 weber/meter*. The direction of B is tangent to the surface and perpendicular tb the current. Potential difference across 1000 ft or 305 meters is K=/ J R=(25 ampXO.l ohm)=25 volts The electric field is ~ c " =8.2 X 10-* volt/meier d 305 meters parallel to the current. The Poynting vector is given by Mo Mo since the angle between E and B is 90. (8.2X 10~* volt/meter)(3.9x 10~ weber/meter 2 ) (4n X 10" 7 weber/amp-ra) =255 watts/meter 2 . The vector S is perpendicular to the plane defined by E and B. It is directed radially inward, a result which is given from the rule of finding the resultant of the cross product of two vectors. 39.15. The average value of the Poynting vector is .-(I) Also, Em^cBm ...(2) 5- cB ^ (3) A - w; 2^o Intensity of radiation =5 =2.0 cal/cm 2 . ^DtL2^L_ MX 10 3 watts/meter'. (1.0 x 10 f meter ) f ( 60 sec) 264 Solutions tottandR Physics It From (3) we have .4X10* joule/mK4n X 1(T 7 webcr/amp-m)! 1 /* f 2&^ fteKl .4X10* joule/mK4n X 1(T 7 webcr/amp-m)! 1 / V~c L (3 X 10 s meter/sec) J =3.4 X Kr weber/meter'. Using this result in (2), -l =(3x 10 8 meter/sec)(3.4x NT* wcber/mctcr 1 ) =1020 volt/meter. 39.16. (a) *^= lO 8 meter/sec Qx 10* mcter/sec)(3.3x 1(T* webcr/meter*)* (2)(4itXlO- 7 wcbcr/arap-m) - = 1 .3 x 10~ u watts/meter 1 . 39.17. (a) E is parallel to the axis of the cylinder, in the direction of the current and B is tangential to the cylindrical surface. Since, S=(ExB),V, it is directed inward normal to the surface following the rule for finding the resultant of the cross-product of two vectors, i.e. S is perpendicular to the plane formed by the vector and B. (b) The rate at which the energy flows into {the resistor [through the cylindrical surface is P= J S.rfA ...(1) Since S is directed normal to the spherical surface, we need be concerned only with the surface area A of the cylinder. ...(2) ...0) O o where, we have used the fact that E and B are at right angles. V - iR At the surface, B=~- ...(5) Where we have assumed that a < /. Integrating (1) over the closed surface, P-S I dA~SA= ( - J Mo Electromagnetic Waves 26$ -JL i* Mo ' The quantity t*R is nothing but the rate of Joule !:e;tti;)g. 39.18. (a) E points in the direction of /, and B i> t uigcnti i! to the cylindrical surface. Hence, S whic-* is giver; by thv cro^-pioduct of the two vectors E and B points perp.-njicuUr to the plane containing E and B i.e. points radially inward. (b) Let the radius of the parallel plate capacitor be r, then 0=w* ...(i) The induction at the surface of a cylinder of radius r is Mo Mo where we have used the fact that E and B are perpendicular to each other. Now, P=Js.rfA=S f^W ...(4) where we have used the fact that d\ is normal to S, and that only the curved surface contributes to the integration. A=2nrd ...(5) Using (3) and (5) in (4), EB P =*Z(2Krd) ...(6) Mo Using (2) in (6) p~ Mo or r where we have used (1) and the identity ~3T "" "2 dT i ' 266 Solutions to Hand R Physics II SUPPLEMENTARY PROBLEMS S.39.1. (a) Let / be the length of the coaxial cylindrical capacitor. (Textbook Fig. 30.4). Applying Gauss's law c / E.</S=0 ...(1) we find e E(2*r/)=<7 -(2) the flux being entirely through the cylindrical surface and zero through the end caps. From (2) we have The potential difference between the plates is b b </L_ i_ , n b . 2nE l r ~2nE l a m I J _ I J Mr *f I a a Capacitance C-- ^= $- V In a Hence, capacitance per unit length of the cable is C a (b) The magnetic induction between the conductors is -w The energy density is 1 ^o Mo*" 2 where use has been made of (4). Consider a volume element rfv for the cylindrical shell of radii r andr+rfrand of length 6 (Textbook Fig. 36-7). The energy coo- tained in this volume element is Electromagnetic Waves 267 The total magnetic energy stored is given by integrating (6) jV<'i a But t/=tli* ...(8) where L is the inductance. Comparing (7) and (8), 2ic a Hence, inductance per unit length of the cable is _4Lln-* / -2 a S.39.2. ^=- -|, Textbook Eq. 39.10. / CX Ct I ~2 = Mo e M ( Textbook Eq. 39.13. vX ut Differentiating (39-10) with respect to x, Mo^o 3r where use has been made of (39-13). In order to obtain the equation satisfied by B, differentiate (39-13) with respect to x 3"JE_ J /3 \ exa* ~ M e er Ux / where use has been made of (39.10). We therefore obtain S.39 J. (a) |f OX 268 Solutions to Hand A Physics t Thus E and B satisfy (39-10) Mo'o Since c* - /*oo Thus and B satisfy (39-13) Thus, and satisfy (39-10). dE Ac 9 . a/ =-? ==x Thus and B satisfy (39.1 3). Thus (39.10) is satisfied. -fi Thus (39.13) is satisfied. ^,jl. to x+ct "' Thtts (39 ' 10) i8 Electromagnetic Waves 269 Thus (39.13) is satisfied. (*) To show that the given functions do satisfy equation (39.10) and (39.13) write for convenience =x ct !=> dy as, -c Thus (39.10) is satisfied. _ 8fl == _ / <i/ S J Sx Sj' 8jc dy 8r r "c 8 8> 8r ~ c* 5y ~ 8y Thus, (39.13) is satisfied. The corresponding situation for functions of (x-fcf) is S.39.4, (a\E~Em sin wf sin ^.v ...(1) BBm cos o/ cos A:x ...(2) --....(..< 1 $E ,,.. .13) Differentiate (1) with respect to x and (2) with respect to t and use the resulting expressions in (39.10), sin o>r cos D =-"aij? m sin wr cos 270 Solutions to H and R Physics II .'. kEm sin co/ cos kx=<*>Bm sin co/ cos kx or kEm~<x>Bm ...(3) Differentiate U) with respect to x and (1) with respect to / and use the resulting expressions in (39.13). SB = Bmk cos co/ sin . , y mco COS CO/ Sltt KX fc cos co/ sin fcx= = jEWo cos co/ sin c* or B m k= ...(4) Multiply (3) and (4), or (o=ck (5) Use (4) in (5) to find, /Tm ==: cBm () (6) The Poynting vector is 1 FR s= EXB=~ ...(?) ^0 ^0 Since E and B are at right angles. Use (1) and (2) in (7), \ c _ (Em sin co/ sin kx)(Bm cos co/ cos kx) i3 ' T ^ ~ Mo _ (Vgm)(2 sin wt cos o>/)(2 sin /ex cos kx) " EmBm ^ , _ = - Y sin 2 Ar# sin 2 I f o - (c) Sa*= I 5 sin <o/ dt -y j EmBm . S1D ^ w ' s ' n Electromagnetic Waves 271 r 1m sin 2 fex f . I si . . , sin 2 att cos <*>tdt J JumJBm Sin 2 f . , , . sin 2 ^/ rf (sin J =0 (d) The fact that Sai>=0 means that the net flow of energy across a given area is zero. This is reasonable since we are here concerned with a standing wave. S.39.5. (a) Sa*-~- ~ or E<n- * (2)(4 X : - ' ' ebc/ainm(3 xl 8 meter/sec)(10x =8.68 XlO' 2 volt/meter. The ciTective or root-mean-square electric field is ( r..)= -^ 6 - 14 x 10 ^ 2 volt/meter. (W ^= r^ 2-9x 1(r ,. Wcbcr/m2 ^ 3x i() 8 met cr/ sec J? rw .f,= w * -2.1 xl(r in wcber/m 2 . (c) Total power^radiatcd / ) -4 7 ir 2 5 =(4^)(10xl0 3 meter) 2 (10x 10" watts/meter 3 ) = 1. 26 XlO 4 watts. S.39.6. (a) By Supplementary Problem 39.1, the potential difference between the plates is T . q . b \ . b , n -^- In - == -- In (1) a 2*e a where A== is the charge density. Let V=E in (1) and using the result of Problem 28.17, __ _ In bla 272 Solutions to Hand R Physics ft For a < r < b, (6) S---EXB _ /* r In 6/a 2isrR ...(4) _ _ ~ 2rJ? In where use has been made of (2) and (3). (c) The total power flowing across the annular section a < r < b is given by b />= [ S(2rdr) -(5) a Using (4) in (5), b ^ [ J 2nr*R In b/a a - a f *i= 2 i A_!. "~R}tib/a J r ~~R\nb/a n a " /?' ^ Now, E 2 /R-=Pj, the rate of Joule heating. Therefore, the result obtained is reasonable. The Poynting theorem locates the flow of energy entirely in the field space between the conductors. (d) If the battery is reversed, then E-> E and because current is reversed, B- B, and consequently S->!(-E)X(-B)= EXB =S Mo Mo Hence S is unchanged in magnitude as well as in direction. S.39.7. (a) The charge on the surface is, <?=(2tt/?/) a co 2tc/?/aa) current /=^v=^ ^^ ~~2n~~ Electromagnetic Waves 273 Magnetic induction rn VU (6) Faraday's law of induction can be written as f where use has been made of (1) 2 ~ (c) S = ~ E X B = - - Mo Mo . ...(3) (</) Flux entering the interior volume of the cylinder =SA = ( i/io f R***L)(2nRl) = /i na/Z/a>a ... (4) Now, -=^^R~~=i^aR<*. ...(5) where use has been made of (1). Now (4) can be rewritten as dB d(R*IB*\ -~ =.^-( -^ ) "t at \ 2n J Mo / Mo where use has been made of (1) and (S). 40 NATURE AND PROPAGATION OF LIGHT 40.1. From Textbook Fig. 40-2 we find the relative sensitivity of 50% for (a) 510 mn i.e. 5100 A and 610 m/x i.e. 6100 A (b) Eye is most sensitive for A =5600 A. c 3xlO w cm/sec e .^ ,-,. . /. Frequency, v= T = 5600xlo - 8 - =5.4x 10" c/s. Time period, TC =~ =53^14-^ -1-85X lO' 1 ' sees. 40.2. Gravitational force between sun and the space ship is GMm r~ Mass of sun, M= 1 .97 X 1 kg. Mass of space ship, m= 100 slugs= 100 X 14.59 kg- 1459 kg. r=sun-earth distance = 1. 49 X 10 n meters. G=6.67xlO" u nt-m 2 /kg 2 . _ ^6.67^x IP" 11 nt-m z /kg a )(l .97 X 10 kg)( 1459 kg) F ~ (1.49x10" meter) 2 =8.64 nt. / Force/unit area== , c since it is assumed that the sail is perfect reflector. U= 1400 watts/meter 2 c=3xl0 8 meter/sec Total force = A, c where A is the area. Set A =8.64 nt/m C 8.64_c _ 8.64X3X10* or A=* 2U - - 2x1400 =9.2x10* meter*. Nature and Propagation of Light 275 40.3. Under the assumption that earth completely absorbs radiation, force on earth due to radiation pressure is t-S- A c where >4=K/? a , area of a flat disc whose radius is the radius of earth. With * 7=1400 watts/meter 2 c= 3 x 10 s meter/sec /*=6.4X 10* meter GMm ^2" O 6 m) 2 n=6.0x!0 8 nt. Gravitational attraction between earth and sun With M- 1.97x1 03 kg m-6.0xl0 24 kg r =: 1.49x10" meters G-6.67X10- 11 nt-m 2 /kg 2 (6.67 X 1Q" 11 nt-m 2 /kg 2 )(1.97 X 10 3 kg)(6.0x 10" kg) * ~" (1.49x10" m) 2 =-3.6 XlO 22 nt. 40.4. Let a plane electromagnetic wave fall perpendicularly on a perfectly absorbing surface. If the incident momentum per unit volume is p then the amount of momentum associated with the radiation falliag per unit time is given by the quantity pcA. By our assumption, the radiation is completely absorbed by the surface so that pcA is also the momentum absorbed per unit time by the sur- face of area A. Thus, the force on this area is Hence, the radiation pressure is given by A where u is the energy density. On the other hand, for a perfect reflector, the reflected light has momentum equal in magnitude but opposite in direction so that the change in momentum per unit volume is 2 p and the corresponding radiation pressure is 276 Solutions to H and R Physics II As the radiation pressure is doubled, the energy density is also doubled, half of the photons traversing towards the mirror and an equal number in the opposite direction after reflection. In general the above result holds irrespective of the incident energy that is reflected. 40.5. Let n bullets travelling with speed v strike a plane surface of area/meter 2 at right angles in unit time. Then, force Pressure -. ---- ^momentum delivered/sec/unit/area unit area ' ' =/?Xwv ..,(1) under the assumption that bullets are completely absorbed by the surface. Consider a box of length v meters and unit cross-section so that volume of the box is v meter 3 . Energy flowing/unit area/sec=A/X^ mv a . Energy density=wxi /wv a /v=$ nmv. ...(2) Compare (1) with (2) to conclude that 'prasuie' j s twice the energy density in the stream above the surface. 40.6. m=100slugs=14.59XlOOkg=1459kg. Fower^ 10 4 watts /=cnergy radiated in 1 day~(10 4 watts)(86400 sees) =-8, 64 X10 8 joules. Momentum delivered to the space ship is, c (3x 10* meter/sec) tr . . . p 2.88 nt/meter 2 ' Velocity increase- .- (M59 fcg) 0.00197 rneter/sec^2.0 m/sec. 10.7. Consider a sphere of radius r=1.0 meter. Let the bulb be located at the centre of the sphere. Consider a small area A at dis- A tance r. Energy falling/sec over area X = 500x -^ j- watts. Set A=--l.Q meter 1 and r = 1.0 meter. soo V r 39.8 watts/meter 1 . Momendum delivered P c Nature and Propagation of Light 277 V (Under the assumption that the surface is perfectly absorbing) Force - f-g - ^f^ -IJx UT' nt. Radiative pressure^ f ^=^>L^l = ,.3 X10 - nt/meter'- 40.8. The uncertainty in the distance to the moon. 8R=-0.5 mile =0.8 km =800 meters." Therefore, uncertainty in time, s , S/? 800 meter . .. a *"= "3 x16 =2 - 7 X 10 ~' sec = 2 ' 7 40.9. Roemer (1673) was the first person to establish that light travels with finite speed at about 186,000 miles per sec. The planet Jupiter which revolves around the sun once in, about 12 years, has 12 satellites of which four can be seen with the aid of a low power telescope. These satellites revolve around Jupiter in the same plane as Jupiter itself and each is eclipsed once in every revolution as it enters the shadow of Jupiter. The principle of the determina- tion of velocity of light consists of estimating the apparent time of revolution of one of the satellites obtained from successive eclipses when the earth is nearest to Jupiter and when it is farthest Roemer noticed that eclipses take place 16J minutes or 1090 sees late when earth is farthest from Jupiter compared to when the earth is closest. The delay is attributed to the extra time that light has to take in travelling across the diameter of earth's orbit. Now, the earth's orbit was known from the method of triangulation and is around 1.86X 10 8 miles. It follows that velocity of light, 1.86X10 8 ,- ., , c = - 100Q = 1 86,000 miles/sec. (a) Refering to Textbook Fig. 40.11 when earth moves from x to y, the earth-Jupiter distance has increased, Therefore, one expects the apparent time of revolution of Jupiter's setellite to increase. (6) Two observations are needed. (i) Timing of eclipses. (ii) Diameter of earth's orbit around the sun. 40.10. The uncertainty of measurement in c is less than 0.0001%' We should not have an error in the length more than 278 Solutions to H and R Physics 11 A* Ac 0.0001 _ 10 - R ~~ c ~~ 100 Set, #=1.0miles=63360in AjR=(63360 in.)XlO"=0.063 inches. 1 whence, v/c=Vo.02=014. 40.12. A=4340XlO- 8 cm A' =6562x10-* cm 6=180 > / v Tf A V /A J v A' y _ (4340X10^ cm) 2 __ "(6562X10- 8 cm)"" aw/ whence, v/c=0.39 v=0.39x3x 10 8 =1.2x 10 8 meter/sec. (6) It is receding. 40.13. (a) Let v be the frequency of the incident microwave beam. Let the car be approaching with spped v. Then the frequency seen by the car is v'= v (l+vAO ...(1) Upon reflection the microwave comes back as if it was emitted byja moving source travelling with speed v towards the observer. There- fore, the frequency observed ...(2) where we have used (1). Av^v'-v^vd+v/c) 1 v We can assume v/c < 1 Neglect the quadratic term Nature and Propagation of Light 279 For a receding car it can be shown, proceeding along similar lines, that Avc*2* v/c. -- c v=2450 mega-cycles/sec = 2450 X 10 6 c/s. Let v be expressed in miles/hour. c= 186,000 railes/sec=1.86x 1&X36QQ railes/hr. =6.7 x 10 8 mph Av 2v 2X2450X10* c/s , ., , . , . ~ "T'-SrTxKf mh = 7 -3 (cycles/sec). mph 40.14. v '=v(l-v/c) ...(1) f- where we have neglected higher order terms in the series expansion. 40.15. A=5500xlO" 8 cm Radius, r=7X 10 8 meter=7x 10 l cm Period of rotation, 7=24.7 days=24.7 X 86400 sec -2.13x10' sec By Problem 40. 14, AA=A =A C V C _(5500X IP" 8 cm)(2n)(7X 10 10 cm) (2. 13 X 10 6 sec)(3 X 10 l cm/sec) =3.8 X 10- 10 cm =0.038 A. 40.16. v=40xlOc/s v '=v (1 + cos 8) C y ' v^v^ cos 8. c 280 Solutions to H and It Physics 11 Now, AB*=vt, since on left side of B, time is minus. 250 miles. _ _ -vt "' COSU ~~^C ~V>f*+(250)* where t is negative and is expressed in hours. For the path AB, vv*/ ^ R7v 1O< f A ' 3 ' b7X 10 =JL (For the path AS) n~* c V>/ 2 +(250) V>+1~.9X 10 ^=18000 rnph 250 miles "t A'S) 4-5 mln. Fig 40.16 (a) Fig. 40.16 (b) For the path BQ, t is positive and consequently Av will be negative. Fig. 40.16 (b) shows the plot of Av against /. 40.17. v/c=0.2 AjH U ,=4750 X 10~ 8 cm _ ' v 1+0.2 A' v'=Av v , v 47SOxlO~ B cm -.., lrt , A=A V = - OJ816 -- 5821X10- cm. The colour would be yellow-orange (see Textbook Fig. 40,2). k ., (1+PCOS6) Nature and Propagation of Light 281 Set 6 = 180 Set 6=0 .87 x 10- 3 )*=0.824x 10" 8 = 2p a - 2(2.87 X 10~ 3 )= 1.65 X 10~ 8 V V See Textbook Table 40.2 for comparison. SUPPLEMENTARY PROBLEMS S.40.1. (a) v 4= 3X "f meter / sec =10 cycles/sec A 3 meter B is perpendicular k> both E and the direction of propagation. (c) EEm sin (fcx o>r) w=2v=(2w)(10 8 sec" 1 ) =6.28 X 10" radians/sec. (d) Time average rate of flow of energy is given by the average value of Poynting vector, , __ 1_ ., _(300 voUs/meter)(10" weber/m*) /1 (2)(4TXlO-'weber/*mp-m) 119.4 watts/meter*. 282 Solutions to H and R Physics- tt (e) For a perfectly absorbing surface of area A momentum delivered per sec, 5, t _ 119.4 watts/meter* 7 c 3 XIO 8 meter/sec ~ 4XU) A m ' _ ... force momentum delivered/sec Radiation pressure^ - ^ = : - '- * area area E x B V S.40.2. e ExB=e oA * - = A*o c 1 E x B Since c*= - and S= - is the Poynting vector 11 -[power/area] - -_ [LJ [volume] S.40.3. Gravitational force on the particle of mass m and density p at a distance r from the sun of mass M is Mm _GM =4 W (6.67X 10-" nt-meterVkg 2 )(1.98x 10 80 kg)f 1000 -^ }^ 3 \ m f r a =55.3 XlO 82 ^-nt ...(!) where /? is the particle radius. Solar radiation received per second by the projected area n/? at distance r from the sun is 7= (Mean total solar radiation) j^~- -(3.92 X 10 2 watts) ~- -9.8 X 10" ^ wa tts. Momentum delivered to the particle per second is, U P= Thus the radiation force - =( 9.8 X 10 M -* watts )f --J c \ r* /\3xl0 8 meter/ sec Nature and Propagation of Light 283 Condition that the particle is blown out of the solar system is Using (1) and (2) in (3), 55.3xlO a *^r< 3.3 XlO 17 - r 1 r* (a) or R< 5.9 X 10~ 7 meter ... (4) (b) Set R jR in (4) : The critical radius JV= 5.9X10"^ meter. j[c) Condition (4) is independent of r. Hence, R does not depend on r. 41 REFLECTION AND REFRACTION- PLANE WAVES AND PLANE SURFACES 41.1. A stationary disturbance produces spherical wave-fronts. However, in the present problem the wave-front of the disturbance can be found out by exciting in sequence a row of stationary sources of disturbance. We can determine ths disturbance at a subsequent time / 2 by drawing spheres of radii u (/2/i) around various points along the path of the object. The resulting wave-front is the envelope of spherical waves which originate at the nose of the fast moving object at successive instants in its flight (Fig. 41.1). The wave-front is a common tangent to the secondary wavelets given by the usual Huygen's construction. The envelope would be, a cone of semi-angle a given by sin a= = vt ^Should v be less then u than a would be imaginary and the spheri- cal waves have no envelope. In other words, no wave front is formed. Fig. 41.1 41.2. Mirror is rotated through angle a. The normal is also rotated through angle *, NBN'~x. The reflected ray BC becomes BD for the new position so that angle of rotation of the reflected ray is CBD. ABN'^N'BD = NBN' +ABN+ NBN' -NBC But MWV'= and ABN*=NBC Refaction and Refraction Plane Waves and Plane Surfaces 285 that is, the reflected ray is rotated through 2 if the plane mirror is rotated through a. 41.3. From Textbook Fig. 41.2, n=\M for A-5500 A v= = 3xl0^meter/sec =2os x 1Q9 meter/sec . A* A i \ c 3 XlO 10 cm/sec - AnwlAU -i 41.4. (a) v- T -__ nb ^_-5.09XlOM JC '. W A ^- 3875A- rt 1.52 (c) As the frequency io the medium remain unol'ianged v=vA / (5.09x 10 1 * sec^)(3875x 1(T 8 cm) ==L97xlO l cm/sec. -= . ^ 41 - S ' """ v 1.92X10 8 meter/sec ' 41.6. (a) 8/7-0.00001 c=3x!0 8 km/sec or , , = 7- a+ -- dv en dv Since n and v f are uncorrelated, v 2 +(*v) 2 n 2 ...(2) From (1) the extra uncertainty arising from uncertainty in refractive index is v8* = -S* =(3x 10* km/sec) ~~ = 3 km/sec n \ 1.0 (W Now 8v=( 1.0003X1 km/sec)=1.0 km/sec 286 Solutions to H and R Physics II If we want to reduce &c to 1.1 km Set S/i=0.0000015. or Thus, n should be measured with an accuracy better than a factor of 15. , 417. In Textbook Fig. 41.6 we find the angle of incidence 0=59 and angle of refraction a=35, so that the refractive index sin 6__ sin 59 sin a sin - ==1.49. The angle of deviation <|>=48 Fig, 41.7 (a) In Fig 41.7. (a), the ray is traced for the angle of incidence 6^45. The angle of refraction a t at the first refracting face is calculated from siu ^=^^==^^1. We find ct 1==2 8 20\ As the angle of prism #=70, we find ,==# a 1 ==702820' =4140'. The angle of emergence 6 2 is calulated from sin 6 2 ~n sin ^=1.49 sin 4I40' ; whence e i =82. The angle of deviation $0 l _ 1+ e a --a 2 =:45 28 20'+82 ~4140 / =57 , Fig. 41.7 (b) In Fig. [41.7 (6), the ray is traced for the angle of incidence 75. Proceeding as above we find l =40 9 25', ,= 29*35', ' and <!5222'. Reflection and Refraction Plane Waves ami Plane Surfaces 287 41.8. (a) =1.52 Critical angle of reflection, sin e,= =-- 1 vr n 1.52 \\'hcn <!* !* m.iximum, * will be least (Fig 41.8). But a is complimentary to p and so also ^ is complimantary to P, so that <*=--(V~tf a . But least value of 2 is 6 C for which total internal reflection takes place. /. Max. value of ^ is 90-a=90-~41=49 (6) Fig. 41.8. 1-52 1.33 w 1.14 Set 41.9. is small, so thai a is also small. Hence v--2ic x) is also small. Angle of prism <f> is also small. sn Simplify and rearrange to obtain or 288 Solutions to H and R Physics II Thus, the angle of deviation is independent of angle of incidence. 41.10. #=60 =1.6 (a) n=~ sin sin *<L~TT =0.625 l.o or sin 6 sin a .". sin 0=/i sir or 0=35 (6) <x, l "\-a l =^ Set oc l =ot <> 2l=1.6x0.358=0.5728. or a x =30 sin 6 . -- n sin a x .\ sin 0=/i sin or = 54. 30== 1.6x0. 5=0.8 Fif 41.lt Reflection and Refraction Plane Waves and Plane Surfaces 289 41.11. since, = sn sin 2 =n sin a 2 ; ^Oj 04+62 04 Light . 6 2 Blue 4750/1 1.463 2854' 3106' 4905' 3405' Yellow-green 5500 A 1.460 2858' 3l02' 4849' 33049. Red 6700,4 1.456 2905' 3057' 4822' 3329' 41.12. 10 20 30 40 50 60* 70 80 sin a Fig. 41.11 i e, sin a sin 6 8 " ___ n m 17365 745' 0.13437 1.292 +0.018 .34202 1530' 0.26724 1.280 +0.030 50000 2230' 0.38268 1.307 +0.003 ,64279 290' 0.48481 1.326 -0.016 .76604 350' 0.57358 1.336 -0.026 .86603 4030' 0.64945 1.333 -0.023 93969 4530' 0.71325 1.317 -0.007 .98481 500' 0.76604 1.286 +0.024 = 1.310 0.02 290 Solu ions to H and R Physics II 8= deviation from mean value =0.020 Observations are consistent with Snlll's law* 41.13. See Fig 41-13 or *= ; sin (90 i =tan n -0) -(2) ...(3) cos sil = or a= a -- /7 whence x= 41.14. For riB= 1.463 1.455 Fig. 41.14 For blue light 9cnticai=43.l. For Red light e c rica/=43.4 (a) At 6e*43, blue light will be internally reflected but red light transmitted, as the critical angle for the red light is slightly greater. Therefore, when white light travels fused quartz, at around 0^43' internal reflected light will contain blue component and red com- ponent will be transmitted. (6) As the angle of incidence is allowed to increase the internally reflected light will contain both blue as well as red components. Separation of blue colour is not possible. 41.15. sin0 c =l/rt The fraction of light energy that can escape is equal to the fraction of the solid angle through which the light can pass upward, towards the surface with- out total internal reflection. Let the light go through a cone whose half angle is equal to the critical angle 0<. such that sin 0c=l//j, where n is the refractive index (Fig. 41 15). The fraction of solid an^ ^fe 41.15 * *- r ^.-7 w n y Q-J V -~~ + / c ' \ h / \ / / r \ / / Reflection and Refraction Plane Waves and Plane Surfaces 291 2ip(l~ cos C ) , 1 cos 6c .~ f r 7 4* or =-< -sin 2 6, = i-J W^ For n=1.3 3, /== I -2XTT3 41.16. In Fig. 41.16 sin e t - J n sm a sin a==i or a=30 ; p-60 Y=180-(75+60)=45 30 Fig. 41.16 8=90 _ sin 6 8 ~ sin a, sin e t =/i/2==sin It id sufficient to show that the normals at the surface on which the ray is incident and the one from which r emerges are at right angles. But this is so, as these normals are parallel to the two faces which are inclined at 90. 41.17. Let fl0=refractive index of glsss in air, w=refractive index of liquid in air. 292 Solutions to H and R Physics II */, refractive index of liquid with respect to glass At small angle 8 incident ray will be able to penetrate liquid and escape into air. By increasing continuously, we can find when the total internal reflection occurs at air-liquid interface. Let the criti- cal angle be Qi. At the critical angle the ray suffers total internal reflection at the liquid-air surface and goes into the glass and can be observed. 1 sin 6/ If 6 is allowed to increase then at a higher angle internal reflection at the glass-liquid surface. = _l sin & g i -(2) tint the ray suffers -(3) The refractive index of liquid is found out from (2). With the use of (1) and (3) together with the knowledge of n t obtained from (2), the value of refractive index of glass (n ff ) can be found out. The method works provided n > /*/. 41.18. ^ 5 ^ sin a -.(I) Also, n- 1 sin p sin (90 ) cos a From (1), sina=~y~- v 2 n .". cosfa=l sin 2 a=l _ r= - 2 * 2n Eliminate cos a between (2) and (3) 2n /oc Fig. 41. IS = ~-(l + ^3) 41.19. sin e =- = 0.6667 Reflection and Refraction Plane Waves and Plane Surfaces 293 0*=41.8 , tane ' = ol r=0.5 tan 8c=0.5 tan 41.8 =0.5x0. 8941 =0.447 cm. Area of each circle=w*=n(0.447) a =0.627 cm 1 . Area of each side=1.0 cm*. The centre of each face must be covered with a circle of radius .45 cm. Fraction of area covered is 0.63. Fig. 41.19 41.20. Consider a ray to proceed from the point A and upon reflec- tion atin the plane P to arrive at the p int C. (Fig. 41.20). Construct the normal N through A perpendicular to the plane P. Extend the normal N as far behind the mirror as the point A is in front of the mirror i.e. A'D=AD. Join AB and draw the straight line A'BC. As the triangles ABD and A'BD are congruent, AB=*A'B. Hence, the path ABC A'BC. Now ABC is one possible path fron A to C via the reflecting surface. For any other path AB'C is equal to A'B'C which is greater than A'BC since the side of a triangle is shorter than the sum of the two other sides. In accor- dance with Fermat's principle light follows the path ABC which is the shortest. It is obvious that any path from A to C via a point B" on the plane P but outside the plane of paper will be greater than the straight line A'BC. Hence, the reflected ray lies in the plane of incidence, this being the plane containing the incident ray AB and the normal to the reflecting plane at B. Hg. 41,20 294 Solutions to Hand R Physics II 41.21. The total optical path is x)* ...(1) dl 1 2 2 -j 2 dx 2 2 . X M ^f V <j 2 +x 2 ~ V^ 2 +0^0* Solve for * to find x= 7 , "(2) a+o Use (2) in (1) to find ...(3) >/x 2 + (x-d)*+b* 2x(x-<Q _ X T X o (a+b)' _ ... -P sltlve where we have used (2). /. I given by (3) is a minima. Next consider the problem of refraction. /=i/i+"2/="i Vo+x*+n, 4+(d-x)* ...(4) di 2x dl or rfx" Reflection and Refraction Plane Waves and Plane Surfaces 295 ...(6) For any value of x, expression (6) is positive showing thereby t hat it is a minimum. 41.22. Optical path l=AC+CB.n Now, UC) 2 -(C7)) 2 +UD) a -2 UD)(CZ>) cos ^ where we have expanded cos $ upto two terms. Upon simplifying the right hand side we find, Similarly, l=<fa*+R(a+R) 2 . (i) 2 Setting 3//3^=0 as the condition for maximum or minimum, we find a 2 +7?(fl+ /?)^ V a 2 +/?(/? Differentiate (1) with respect to <f>, 3V _ a^V- 8 " ~ ~~(a 2 ~+R(a+R)<t>*} Use (2) to eliminate the dinomtnator of the second term in (3). Then (3) becomes R) P. (/f) 1 "l ;)#]/ L n(a-/?) J 296 Solutions to II and R Physics II As the expression outside the square bracket is always positive, the sign of ^ depends entirely on the sign of the expression in the square bracket. Thus, ^r a is +ve (condition for minimum) a+R ^ , or -,- - =?. < 1 n(a R) i.e. a+R < n (aR) or a(l-n) < -R(n+\) or n- Similarly, condition for / to be stationary (not changing) is and for / to be maxima, SUPPLEMENTARY PROBLEMS 5.41.1. The opening angle for the Cerenkov radiation is given by cos e- A- ...(i) with p=v/c and n the index of refraction. Minimum speed of electron is obtained by setting 6=0 in (1), in which case P(min)=^ 4 = 0.649 whence v (m < n) =0.649 c=(0,649)(3xl0 8 meter/sec) = 1.95x 10 8 meter/sec. 5.41.2. (a) Imagine that the atmosphere is divided into horizontal layers of increasing index of refraction in going from top to bottom as in Fig. S.41.2. Let the rays of light from a star be incident at an angle 6 1 with respect to the zenith. As the ray traverses various layers of atmosphere bending occurs in accordance with Snell's law. The curved path which the ray travels is approximated as a series of straight lines in various layers* The bending takes place towards the normal in each la>er as the index of refraction progressively increases in traversing down. Applying Snell's law to various layers, Reflection and Refraction Plane tfaves and Plane Surfaces Observer Fig. S 41.2 sin 6i=w 2 sin a 2 ...(1) sin 6 a =/j 3 sin <x a (2) m sin 6in f s j n a / ...(i) Add equations (1), (2)...(i). Note that 6,=^, 63=0,... we find, n i s * n QI^H/ sin a/ where / is the apparent angle of the star and) /i/ the index of refrac- tion at the point of observation. Furthermore, we can set 1^=1 as the uppermost layer of the atmosphere is so tenuous that it is as good as vacuum. We therefore find, sin 1 sin /= * nt showing thereby that the apparent angle of a star is independent of how n varies with altitude and depends only on its value at the earth's surface and on the incident direction. (b) Owing to the curvature of earth, the atmospheric layers of uniform refractive index are no longer horizontal slices but are now spherical shells. The analysis is complicated by the fact that the nprmals at various layers are no longer parallel but tend to coo- verge at the center of the earth. The angle of incidence at one layer 298 Solutions to H and R Physics tt of atmosphere is no longer equal to the angle of refraction at the previous layer. Thus, unlike the previous analysis, cancellation of various terms is no longer possible. In particular, the angle of the star with the zenith will be larger and will not be independent of the variation of refractive index with altitude. 5.41.3. The critical angle is given by A - -i 1 -i 1 c =sm l =sm * rr* n 1.33 =49. The radius of the circle is r==(80 cm) tan 6* =(80 cm)(tan 49)=92 cm. Therefore, diameter of the largest circle is 2r=184 cm. 5.41.4. From Fig S.4 1.4, sin6 sin r sin r= sin 6 sin 45 1.33 =0.5316 or =32*. Fig. S.41.3 D 0-5 1-5 C B 6 A Fig. S.41.4 Reflection and Refraction Plane Waves and Plane Surfaces 299 BG=DG tan r=(L5 meter) tan 32 -=0.94 meter GA=DE=EF=Q.5 meter AB=BG+GA=(Q.94+Q.5) meter= 1.44 meters. Thus, the length of the shadow at the bottom of the pool is 1.44 meters. S.41.5. (a) From Fig. S.41.5 (a), ^ sin r sin .'. sin 8=sin (90r)=cos r cos r=l//i where use has been made of (2). Fig. S.41.5 (a) From (1), we have sin 0! - sin r Squaring (4) and (S) and adding, Hence, n=Vl+sin6, N -.(I) .-(2) -0) -(4) (5) 300 Solutions to tt and R Physics // (6) The maximum value of 6 t can be 90. 1 <ma)=90 <> sin0 1( ma)=sin 90= 1 N, Fig. 8.41. 5 (b) /v A/ e>e Fig. S.41.5. (c) (c) For 0>0!, ray diagram is shown in Fig. S.41.5 (b). The ray emerges at the other side of the prism. For 8<0 lf the ray diagram is shown in Fig. S.41.5 (c). The ray suffers internal reflection. S.41.6. (a) For normal incidence the ray passes undeviated through the air-water interface. From the geometary of Fig. S.41.6 (a), the angle of incidence as well as the angle of reflection at both the mirrors will be 45. Consequently, the angle of total deviation will be 18Q. In other words the emerging ray will be antiparallel to the incident ray and will go out undeviated as it fails normal on the water-air interface. (b) Let the angle of incidence at the first and second mirror be 6 and * respectively. Reflection and Refraction Plane Waves and Plane Surfaces 301 Fig. S.41.6 (a) Fig. S.41.6. (b) From the geometry of Fig S.41.6 (b) it is obvious that 6+a=90. Total deviation, 8=* 26+* 2* Thus, in water the incident ray and the second reflected ray must be antiparallel. Hence the ray incident obliquely on air-water interface and the emergent ray must also be antiparaiel. (c) The three dimensional analog to the above problem is the arrangement of three plane mirrors hinged at right angle to earth other like two adjacent walls and the ceiling of a room. A light ray incident on any of the mirrors after single or multiple reflec- tions will emerge as parallel to the incident ray. The proof is an extension of that given for two mirror problem. 42 REFLECTION AND REFRACTION- SPHERICAL WAVES AND SPHERICAL SURFACES 42.1. (a) 6=45 2* 360 , _ "-- r- 1 45 1==7 (6) 6=60 360 , _ "-"a 11 " 5 (c) 0=120 " 120 42.2.* 6=90 90 42.3. Since the image will be 10 cm behind the mirror, the distance between the observer (who is standing 30 cm in front of the mirror) and the image of the object, will be 30+10=40 cm. Hence, the observer must focus his eyes at a distance of 40 cm. Mirror Image n- 10 cm Object Observer -^ 10 cm < 30cm Fig. 42.3. 42.4. O l and O fi are the images of object O respectively fjy mirrors A/! and A/ 2 . Rays are shown for this image formation 3 and O^ are images of images respectively of 0^ by mirror A/ 2 and O f by mirror MI (rays not shown for the simplicity of the figure). Obviously, O\f l ^O l M l , OM % =*0 % Mt, 0iAf,=B0,Af 8 2 A/ l =-0 4 ^/ 1> with 2 Reflection and Refraction Spherical Waves 303 Fig. 42.4 42.5. Paraxial rays, i.e. bundle of rays close to the axis upon striking the spherical mirror are brought to a sharp focus at F in the focal plane. However, rays which travel further from the axis do not form the image at a common point, give rise to spherical aberration. In Fig. 42.5 (a} are shown parallel incident rays at large distance h from the axis which upon reflection cross the axis closer to the mirror. Along the axis the size of the circular image is of least si?e, This is called circle of least confusion t 304 Solutions to H and R Physics -H Circle of least confuscln Axis Fig. 42.5 (a) Axis C F D Fig. 42.5 (b) That the parallel rays away from the axis upon reflection from the mirrors cross the axis inside the paraaxial focal point F can be easily proved with reference to Fig. 42.5 (6). According to the law of reflection the incident ray AB after reflection goes along BD such that angle of reflection <* is euqal to ^, the angle of incidence which in turn is equal to BCP. It follows that triangle BCD is isosceles so that CD=DB. Now in a triangle a side is smaller than the sum of the other two sides. In triangle BCD, BC<CD+DB Now BC is the radius of the mirror and is equal to CP. Therefore, CP < 2CD or \CP'i< CD But \ CP=CF, since focal length is half of the radius of curvature, thus, CF<CD Therefore, the point D lies within the paraaxial focus F. As the point B moves toward P, the point D approaches F. 42.6. Following formulas are useful for completing the information in the Table regarding the spherical mirrors. Reflectioa and Refraction Spherical Waves 305 1-+-L.1 I f m= / o r=2/ Type a b c d e / g h concave plane concave concave convex convex convex concave f r i o +20 +40 -20 + 10 00 00 -10 + 10 +20 + 40 + 60 + 30 + 20 +40 +3(1) +60 -20 -40 -10 +20 -20 -40 -18 -180 -20 -40 4 + 5 + 8 + 16 + 12 +24 m + 2 + 1 + 2 -0.5 +0.5 +0.1 +0.8 -0.5 Real Image No No Yes Yes No No No Yes Erect Image Yes Yes No No Yes Yes Yes No The results arc in agreement with the graph shown in Textbook Fig. 42.16. 42.7. JL +1=1. -d) o i f ' Longitudinal magnification (m* ) where m= = transverse or lateral magnification. o ' Multiply by (1) through by o and re-arrange to find ' -m- f m --- ;. o of m >-LJ( / -T m - i U-/J Ignoring the sign (6) Differentiate (t) holding /=constant. 306 S lutions to H and R Physics II -to = --# m '~~~ m * 9 ignoring the sign. (c) Since transverse magnification m=longitudinal magnifica- tion, m' But or m=l o-f Therefore, object must be placed at the centre of the curva- ture of the mirror. 42.8 (a) When viewed normally, __ actual ___ ~" apparant depth /. apparent depth = actual depth n - 8 - 0ft sin r ' whence sin r= ^ ' =0.376 or ' r=22 x =8.0xtan r = 8.0 x tan 22 =8.0x0.4=3.2 ft. Fig. 42.8, AB=AC tan 60-3.2x 1.73=5.54 ft. A<* A 1 42.9. . - ==/!. = 1.33 sin r tan BC Reflection and Refraction Spherical Waves 307 FH FH tanr,= ^77 = tan r, 1 DH 2.0 AG _ AG FG 4.0 s ' nr i "2 _ 1 46 , . ~ ~~ i ^T" 1 1 sin r a rtj 1.33 0,= G Fig. 42.9 Consider angles 6, r lt and r a to be small so that tan 6=6 etc., FH FH =2.0 r, _ a ~ 4.0 rj/r.,-1.1 CD _ FH+AG _ FH+AG e e 1.33 r, Solutions to H ai\d R Physics II 42.10. A useful formula for filling up the given table with informa- tion concerned with a spherical surface separatit^g two media with different indices of refraction is a b C ' d e / g h "l 1.0 1.0 1.0 1.0 1.5 1.5 1.5 1.5 "2 1.5 1.5 1.5 Indeter- 1.0 1.0 1.0 1.0 minate o + 10 + 10 +71 +20 + 10 + 10 + 70 + 100 i -12 -13 +600 -20 -6 -7.5 -105 +600 r +30 -32.5 + 30 -20 + 30 -30 + 30 -3 Real No No Yes No No No No Yes Image The rays are shown in Fig 42.10 for the situations (a), (b) (h). r=J (C) I 5f-- { OTT c V n A "2 (h) Fig. 42.10 42 *i. Following formulas are useful in completing the information in the given table regarding thin lenses. 1 , 1 _ 1 "7 7 m= Reflection and Refraction Spherical Waves 309 I o -S i ^> oo ai i V) ^ I S 1 - 1 . *5 "Sb f. ' n I + IS a i i ll rs O o T7T ON r ^ o S C3 >n V, 1 S? o o o ro f^ CO 1 1 + sf ^ ja *" d O < "3 So 1 + i co i / n g| -5? + M + 1 f ^ + a i / - 1 i 2, , en / 'v o S i 2 1 1 1 - j &p 8 *5b + T + ' A O r s> C3 <U >> i 2 I I o I c R? 4- ' T + + g g 'Si / <3 > ^o ** S! si , / 8 8; 1 - M O 5^ G O ^0 1 .^ - c E | 310 Solutions to H and R Physics 11 4112 Zi+^-= 42.12. Q f- . Set i-+ o i /*!= 1 for air 7I 1 =W n-l r If o==oo, then i=nrj(n l)=/ 2 . This value of i gives the distance of the second principal focus from the pole. On the other hand if 1=00 i.e. the beam becomes parallel after refraction, in that case /i/i=0 and o=rl(n !)=/! This value of o gives the distance of first principal focus from the pole and is called the first focal length of the surface. 42.13. (a) --H--^ /i /j=0 2 i=d From (3), 1^=0 Oj ...(1) ...(2) ...(3) (4) ...(5) ...(6) D Q Fig. 42.13 Use (5) in (1) and (6) in (2), J_. _JL ^j_ JL , 1 _ 1 2 J Subtract (8) from (7), 1 1 _ 1 . 1. _ - _ o l o, "*"/) GI Do t which simplifies to ...(7) (8) Reflection and Refraction Spherical Waves 311 ...(9) also, </=0j o t ...(4) Solve for 0j and o t , -CO) ...(ID Use (10) in (7) o obtain, A. 0! Use (10) in (12) and (11) in (13) to get m, fD-d\* m l \D+d I (a) l fa^u / "" i" Centre of curvature C' lying on the /{-side is -fve. So r' is posi- tive. / = +ve. Hence, it is a converging lens. (b) r'=oo r /r = +ve, since C'' lies on /?-side. Hence, /== ve, that is, it is a diverging lens. (c) Both r' and r* are +ve as C add C" lie on the /?-side. We note that r' < r'. Therefore, the quantity f -,- j is +vc. Thus, /=-fve. Hence, it is a converging lens. (rf) Both r' and r' are +ve as C and C* lie on the /{-side. But r' > r". 312 Solutions to H and R Physics II ( -,- ) is negative /= ve. Hence, it is diverging lens. 42.15. Let the object be placed at distance o from the first lens of focal length/!. Let this lens produce an image at a distance iV .--(I) 1 The image at i\ serves as a virtual object for the second lens of focal length / a . Let a real image be formed at / a , then, - - + - = -(2) Add (1) and (2) to obtain Fig. 42.15 1 L L__L , 1 o +/,-/, +/," ...(3) Now, consider the combination to be equivalent to one single lens of focal length F. Then, -- = o i a F Combine (3) and (4) to find -L L , -L_4./ 2 -~- 42.16. Let D=o+i or i=Do Use (1) in (2) to eliminate / JL+JL..! o + D-o & f .-(4) -(I) ...(2) ...(3) Reflection and Refraction Spherical Waves 313 re-arranging the equation, o*-oD+Df=0 Solving the quadratic equation In order that o be real, the expression (Z) 2 4 />/) under the radical must be +ve, thatjis, D a > 4 >/or D > 4/. 42. 17 . J- .(.-I) (-!--!.) H=:1.50 ;/=6.0cm r"=2 r' ; r' is +ve since it is on /J-side. r" is ve since C" is on F-side. Solve for r' ; we fiind r'=4.5 cm. r "==9.0 cm. 42<18 . 1+-L--1 ...(D Set x=o-f or o=x+f ...(2) x'=i-/ or /=*'+/ ...(3) Substitute o and < from (2) and (3) in (1) to obtain, x+x'+2f 1 (x+/)U'+/)~/ Cross multiply and simplify to find the desired result, / 2 =xx' 42.19. Let # be the angle of incidence and 6 the angle of refraction at A. Let r be the radius of the sphere and n the index of refraction of the material of the sphere. Obviously ABC=Q and EBF=<f>. Since DA is very close to GC, both 6 and <f> are small. As sin #=/i sin 6, we can write t=nB ...(1) Now BCF-M-BCA-+ 3l4 Solutions to H and R Physics tl Also BFC=EBF-BCF =2 (#-6) In A FBC, FC sin FBC _ sin ^ _ _^ C ~sin BFC ~" sin 2 (#-0) ~~ 2(#-l) Thus the equivalent focal length, {A M/* irr* nc y >_ r * r * ^ -l^Lx A / i r\\ /% / <\ FH=FC-HC=. nr Fig. 42.19 42.20. - + . 'i /i The first image is located at a distance 2/ t from the pole of concave mirror i.e. at the centre of curvature. ^ 2f} ~- ~*Jt___^ "'I _J t Fi x . 42.20 Reflection and Refraction Spherical Waves 315 The first image serves as an object for the concave mirror 1 . l l -f ~r-=T~ 01 h A or The image will be real, erect and at the same place as i t ; m=l. 42.21. (a) C x is on K-side and so r t is ve. C 2 is on fl-side and so r, is +ve. V-side > .. r Light ; 1 ' incident Fig. 42.21 (a) or Solve to find, / t 1 (c) The image is virtual and erect. (d) The ray diagram is shown below in Fig. 42.21 (b). Fig. 42.21 (b) 42.22. From the Textbook Fig. 42.32 we find y*i=7 mm /j25 mm <> 1 10 mm (object distance fron lens 1) rf=46 mm (distance of separation of lenses) .,_ + 'Y " A or > v i /i (10 mm 7mm) mm* 316 Solutions to H and R Physics // The first image'due to lens 1 is formed 23.33 mm behind the lens x i.e. 02= (46 mm 23.33 mm) or 22.67 mm in front of lens 2. The image is real, magnified and inverted. The magnification is given by m 1 = i 1 A> 1 =23.3 mm/10 mm=2.3. This image acts as an object for lens 2. The image distance for the second lens is calculated from , whence (25 mm)(22.67 mm) ____ _ ., . . , =-243 mm or ^24.3 cm. in front of lens 2. The negative sign shows that the image is virtual. As the image i a remains upright with respect to i l9 the final image i t is inverted as /\ is already inverted. The magnification due to lens ? alone is w 2 = /V0JJ =247 mm/22.7 mm=10.9 The overall magnification is m=w 1 w a =(2.3)(10.9) = 25. Aliter: The final image distance can be alternatively found from the formula, = 243 mm, from lens 2. ^10x46x25-7x25(46 + 10) (46-25)(lO-7)-10x7 The ray diagram is shown in Fig. 42.22. First image Reflection and Refraction Spherical Waves 317 42.23. (a) For L v 1 =20 cm /n=10 cm 1 + JL^-L <>i h A L_i.JL__ 1 20 "*" ij ~~ 10 whence, / x =20 cm. The image is formed 10 cm from LZ and is within the focal length / 2 . This image acts as an object for L t . 10 ^ /, 12.5 whence, / 2 =--50 cm. The final image is formed on the side of L 2 at distance 50 cm i.e. it coincides with the object. Overall magnification, m^niiXm^ x -^~ 20cm 10 cm (6) The lens system an ay-diagram is show below. 0, ; 2 Fig. 42.23 c) The final image is virtual, inverted and magnified, SUPPLEMENTARY PROBLEMS S.42.1. Number of images when two plane mirrors are inclined at an angle is given by 90 318 Solutions to H and R Physics II Due (o three combinations, (i) wall 1-fceiling (//) wall 2+ceiling (iiV) wall 1 +wall 2, total number of images is 3 X 3=9. Out of these 3 are to be subtracted as they are counted twice, due to three common boundaries. Therefore, total number of images is 6. S.42.2. (o) -~ + y =4 -(I) The image will be as far behind the plane mirror as the object is in front. Set o=(a +7.5) cm /=-(<! -7. 5) cm / = 30cm _J 1 1_ a+7.5 a-7.5 ~~ -30 Rearranging and solving for a yields, a=22.5 cm. Ray diagram is shown below in Fig. S.42.2. f "** 7 ^^-^ I -^ 1 D r m ^ ^ # < 22-5 cm -* V Fig. S.42.2 S.42.3. (a) Image A'B' is shown in Fig. S.42.3 at distance /=/), the distance of distinct vision. 9 Eye Fig. S.42.3 (b) Angular magnification of the simple magnifier lens is defined as Reflection and Refraction Spherical Waves 3 19 where p is the angle subtended at eye by image at near point and a is the angle that would have been subtended by object placed at near point. XT a A ' B> A>B ' Now, p- -__ ... (2 ) A C'B' AB .,,. and a== -- - 3) where D==25 cms is the distance of distinct vision. Now, triangles ABC and A'B'C are similar. A'B' CB' D where we have used (1), (2) and (3). As the image is virtual we have .. D D . t 25 . t or M= .+! = + 1 of f S.42.4. For the combination of two thin lenses of focal length /i and / a separated by distance d, the image distance i measured from the second lens (/ 2 ) is given by where the object distance is measured from the first lens (/j). Let the object be placed on the side of the lens with/! =12 cm (conver- ging lens). 0= =433-<//2=(43.5-3.5) cm-40 cm / 2 =--10 cm and d*=l cm. The image is virtual at a distance 710 cm from the second lens (/ 2 =- 10 cm) or 713.5 cm from the center of the system on the same side as the converging lens. Next, let the object be placed on the side of the lens with /i= 10 cm (diverging lens). Set/ 2 = + 12 cm and 0=40 cm. " OUcm 320 Solutions to Hand R Physics- II The image is formed at a distance 60 cm from the converging lens or 63.5 cm from the center of the system on the side of the converging lens. S.42.5. (a) For a combination of two thin lenses, where o and / are the object and image distances as measured in Fig. S.42.5. If we let 0->oo (incident parallel beam) then the Q . u. \ ^i , 2 F 1 L > Fig. S.42.5 express^n (1) is reduced to Setting /i=/,/ 2 = /and d=L, formula (2) becomes /=^= ...(3) Formula (3) shows that if / is to be positive L must be less than /. Also d should not be zero. Thus the condition that the parallel beam be brought to a focus beyond the second lens is o<L<f ...(4) (b) If the lenses are interchanged then /i= / ;/ a = / and d=L, in which case (2) becomes i^L^L^L ...(5) LJ Here, i is positive irrespective of the distance of separation and the condition (4) need not be satisfied. (c) When L=0, in both cases the emerging beam is parallel. S.42.6. When the two lenses are in contact, the beam of parallel rays remains parallel as it emerges from the other side, (Fig. S.42.6 (a). It is as if the beam has fallen on a glass slab. We can look at it in another way. As the beam falls on the concave lens, the rays diverge and appear to come from the focus F. The virtual image at F which i$ also the focus for the concave lens forms Reflection and Refraction Spherical Waves 321 the object. The rays arc therefore rendered parallel due to the con- vergent lens in contact wiih the divergent lens. If the lenses are moved apart the virtual focus Fof the concave lens will be beyond the focus of the convex lens and the rays upon falling on the convex lens will be brought to focur, the result being independent of the Fig. S.42.6 (a) Fig. S.42 6 (b) distance of separation. (b) When the two lenses are in contact, tbe rays of a parallel beam upon falling on the convex lens tend to converge at F, the focus of the convex lens. But Fis also the virtual focus for the concave lens. Thus, the rays are once again rendered parallel as they emerge from the other side as shown in Fig. S.42.6 (c). ] Fig. S.42.6 (c) If the lenses are moved apart a little then the focus F where the rays ought to have met will be within the virtual focus of the concave lens. Fig S.42.6 ( d) The rays will be focussed at a greater distance due to the exis- tence of the diverging lens. This result, however, depends on the 322 Solutions to H and R Physics II distance of separation of the two lenses. Should this distance be greater than the focal length then the rays fall on the concave lens as divergent rays and owing to the diverging action of the lens will diverge still further. Fig. S. 42,6 (e) S.42.7. (a) When the object is placed at distance o l in front of con- vex lens, the image is formed at distance ^ behind the lens given by 1 1 1 _ oJ_ _(\ meter)(0.5 meter) fl ~<V-/- (1-0.5) meter - 1 - 0meter f A F ! "'2 "^F Pin no rnirrnr \) * 1 mpfpr M- Fig. S.42.7 As the plane mirror is 2.0 meter behind the lens, it follows that the image is 1.0 meter infront of the plane mirror. Now, the image /! forms an object for the plane mirror. Therefore, the image in the plane mirror will be as far behind the mirror as the object is infront of it, i.e. 1.0 meter behind the mirror. Thus the image /, is formed 3.0 meters behind the convex lens. The image /, now acts as an object for the lens. Set f = 3.0 meter o 2 f (3 meter)(0.5 meter) A . '3- - -- ==0 ' 6 meter ' Thus the final image is formed 0.6 meter on the side of the lens away from the mirror. (b) The positive sign for / 3 shows that the final image is real. (c) /! is inverted and 80 also fe. However, i s in getting through the lens gets once again inverted, The two inversions cause the final image to be erect. Reflection and Refraction Spherical Waves 323 (d) The magnification for the first image i\ is /\ 1.0 _ f)i * = ~ 1.0 meter In the plane mirror the magnification m L is unaltered. The magnification due to the image / s .is image distance 0.6 meter ^ ~ m . 9 _ ---- _ ^ - =02 meter 3 object distance 3.0 meter /. Overall magnification /7j=m 1 /w 2 />7 3 =l X 1 x 0.20 2. 43 INTERFERENCE 43.1. The fringe-width for double-slit arrangement is given by where d is the slit spacing, D is the slit-sc'reen distance and A is the wavelength of light. A.y _ fl __ w_ _ B _A_ D 180 rf ,/= ^=(57.3)(5890 X 10- cm) =0.00337 cm "K =0.0337 mm. The slits must be 0.0337 mm apart. 43.2. r^-r^2a Transposing one radical, ./( ^-4- ff. y+x 2 =2a+. // Squaring and collecting terms, 2 Squaring and simlifying Divide through by a 2 f ~ 4 a 2 j to get </ </* Since y > a, -7 o* will be positive. Writing (</*/4) fl 1 ^^, we obtain tnterference which is in equation of a hyperbola with centre at the origin and the foci on the y-axis. In three dimensions, the locus of/? would be a hyperboloid, the figure of revolution of the hyperbola. 43.3. Angular fringe separation under water. fi , A6_0.20 ^^"""Tsr* - 15 - 43.4. Angular separation is given by where D is the slit-screen distance, d is the slit spacing and A is the wavelength of light. By Problem, A0=0.20 ; A-5890 A New angular separation A0'=0.22 , V=? 43.5. Shift in fringe system due to insertion of mica flake of thick- ness / is given by (n 1) f=wA m\ 7x5500x10"* cm 43.6. The approximate value for the location of the tenth bright fringe is obtained from y, ^= (10X5890 XiO-"c m )(4 cm) ==OOU78 cm a U.2 cm The exact value for y is obtained from the following equations. ...(1) = V (y' <//2) s -hD* ...(2) Transposing one radical Squaring, collecting terms and solving for y" 326 Solutions to H and R Physics 11 * ***** \ // ^r 4z>'~^> J/l 1 *- Since ~r'^ 1 y=i M> A/ 11 d * -/A/ i+J2ll!_=1.0003/ J A' 4Z) 2 ^ V (4)(4) a ' Fractional crror=(/-y')//=0.0003. .'. Percent error--=0.0003x 100-0.03. 43.7. Fringe width, &y=- Set D=/=1.0 meter =100 cm. Then, A,= -0.295 c m -3.0 43.8. (a) Condition for maxima at/> is r 2 1-!=^ ; m=l,2,3> or Jr^+d* -r^m^ Set A=-hO meter and J=-4.0 meter. c . . 16-m 2 Solving, ^1= 2m (/) w/ 3 : ^=1.16 meter. (//') m=2 ; rj=^3.0 meter. (Hi) m=\ ; rjL^V.5 meter. (6) The minimum in intensity along Ox is obtained by setting the path difference ...(1) where d=4 meters, X= 1.0 meter and m==0,l, 2,3 ...... solving for *, we find the minimum value of jc for m=3, viz., A'=0.53 meter in which case 8=^(4.030.53) meter = 3.5 meter. Since the path difference is a lot more than the wavelength of the radiation, the conditions approach that of partial coherence hither than complete coherence, and consequently the intensity will not stricity go down to zero i e. the minimum will not be zero. Interference 327 Fig 43.8 43.9. 1 = sin <>/ ^ a =2o sin (w/+^) = 1 +' i =o (sin cof +2 sin <u^ cos ^+2 cos o>/ sin = [sin <*>t (1+2 cos ^)+2 cos wf sin #] (1+2 cos #) 2 sin ^ ~,__ + =- cos cos ^ (sin a/ cos oe+sin a cos <ot) cos sin sin (cuf +*) With where we have set i:-. n 328 Solutions to tt and R Physics It / e=-^-(5+4 cos -[l+4(l+cos0)] . = sin 43.10. Sin 6ci Set ==-j. i+8cos> mA d l = /m COS 2 /./ ~2 \ where 6' is the angle at which intensity fall? to half of maximum value. J5 2 A Then cos = or 43.11. By construciing the suitable vector diagram, we get (a) y=yi+y2=nand^=14. y (b) where Fig 43.11 ^=10 sin <ot y,= 8sin (a/H-30) y = =yi+yt=W sin <ot+S sin (/+30) = 10 siiMs/+8 sin tat cos 30+8 cos or sin 30 =(10+4 V^3) sin oi/+4 cos / =v4 sin <at+B cos <r ^=10+4 cos Interference 329 (sin o>f cos sin ==Csin where 43.12. = V(10+ 4 V 3) f +4 1 = 1 7.39 = 17.39 sin 13.3) x Fig 43. 12 >==26.6 sin 43.13. The visible spectrum extends from 4000' A to 7000 A. Therefore, on the frequency scale, the visible spectrum extends from 3xl0 8 meter/sec 3x 10 8 meter/sec to -=. 4000 X 10~ 10 meter 7000 x 1Q- 10 meter i.e. from 7.5 x 10" c/? to 4.286 x 10 14 c/s. .'. Frequency range=(7.5-4.286)X 10 14 , or 3. 214 x 10" c/s Number of channels available = frequency width 3.214 ~ 8X1 ~ 4xlOc/s That is, 80 million t 43.14. (o) Radius of bright Newton ring, 1 2 330 Solutions to if and R Physics It (1.0 cm)' 1 - (500 cm)(5890 X 10' 8 cm) 2 " or m=33. (6) If the arrangement is immersed in water, number of rings that can be seen w '= w n=33.45X 1.33^=44.49 or m'=44. 43.15. ^=( In air, rm 2 =(ro+i) R* RX In liquid r'm a =(m+i) f^ 1 1-4* 43.16. For bright fringes formed by the air gap of a wedge due to light Incident normally, 2 rf=(ro+i) A where d is the diameter of wire (air gap) at the point the with fringe is formed _ 2 d 1 (2)(0.0048 cm) 1 m ~~T~- 2"* (6800xlO-cm)"~ 2 " 43.17. Condition for observing a bright fringe is 2nd 2(LS)(4 X 1Q- 5 cm) 12 X 1Q' S cm The integer m which gives the wavelength in the visible region (4000 A to 7000 A) is m= 2 in which case 4.8X10-* cm=4800 A> which corresponds to blue light. 43.18. For interference maxima, 2nd~(m^to*ma* . (!) For intereference minima, 2nd=m Xmi* (i) Combining (1) and (2), (m-i whence^ m2. Interference 331 Use (3) in (2) to get ,_ mAm.n (2)(4500 A) ,,,. . rf _ __ = ..__.___ _337; A 43.19. Here both the rays suffer a phase change of 180 and the condition for destructive intereference is 2m/=(m+i) A! ...(1) 2<f=(w+-~)A 1 ...(2) From (1) and (2) _ _ _ m-f 3/2 ~ A! ~~ 7000 A *" 7 whence /w=2. ...(3) Using (3) in (1) to get A- 0"i) *i _(2.5)(7000 A) d _ _ .__. ^ ..... _.._.. 43.20. Phase difference, $=i ~- \(2nd) By Problem, for A=5500 A, #=n ^w^=5500 A (,-) A 1 =4500 A Reflection intensity is diminished by 0.88 or 88%. A 2 -6500 A ( 2 * \ ( AT J f- -cos 2 -^- =0.057 /o ^ Reflection intensity is diminished by 0.94 or -94% 43.21. If n > 1.5, condition for minima is 2nd^mX l ...(1) 2nd=(m+\) A 2 ...(2) where m is an integer. 332 Solutions to H and R Physics H Combining (1) and (2) (m-f 1) A 2 =wA 1 (m+ 0(5000 A)=(m)(7000 A) whence m=2.5. This value is unacceptable since m is not an integer. If n < 1.5 condition for minima is 2/nMm-H) A! - (3) where m is an integer. Combining (3) and (4) (m+ \ )(7000 A)(ni+-|-)(5000 A) whence, /n=2, which is an acceptable value. Hence, we conclude that the refractive index of oil is less than 1.5. 43.22. Condition for maxima is 2mJ=(w+i) Amo (0 with Amo* 6000 A and m=0, 1,2,... Condition for minima is 2mMm+l) Ami. -(2) with Amtfi=4500 A corresponding to the violet end of the spectrum and m=0> 1, 2, ... Combining (1) and (2), (m+J)(6000 A)=(w+ 0(4500 A) whence, m=l. (3) Use (3) in (1) to find thickness A 0-5X6000 A) A d = (2X1-33) * 33 43 23. 2</=mA v A . MJ2XOJBM3 cm) .3^ 10 - cm==5880 A . FH 7"2 43.24. 2(-l) rf=mA 7x5890 A 43.25. (a) 6057.8021 A (b) The waveienRth of orange-red line of Krypton>86 has been Interference 333 adopted as a standard of length and the meter is defined as 1,650,763.73 wavelengths in vacuum of this line. Thus, this line is the basic standard of length from which meter is defined and is the reference wavelength in terms of which all other wavelengths of the electromagnetic spectrum are measured. The question does make since once the meter has been defined as above, it is meaningful to express the wavelengths in terms of meter by way of conversion of units. 43.26. (a) Let the wavelengths of sodium light be A t and A 2 . The fringe visibility will be high when the bright bands of A A nearly overlap with those of A 2 , and it will be poor when the bright bands of A x coincide with the dark bands of A 2 . The latter situation arises when the optical path difference is equal to the whole number of wavelengths of A, and an odd number of half-wavelengths of A r Thus, as one of the mirrors is moved, the fringes periodically dis- appear and then reappear, due to the reason given above and the variation in the visibility is explained. (b) Optical path -difference 2<?~m 1 A 1 ==/w 2 A 2 , is the condition for a maximum in brightness. Hence, Id 2d W 2 = -j A 2 Subtracting, we have 2 d where A The integer (m 2 '/HI) increases by 1 as d changes to d+ at the next occurrence of maximum brightness. We then have , A A m t - !!== - ^ -- Subtracting the last two equations A, A, A j- A * A (5890xlO~cmK5896xlO~ 8 cm) r A 2AA~" (2)(5896-5890)xlO-'cm =0.029 cm =0.29 mm, 334 Solutions to H and R Physics It SUPPLEMENTARY PROBLEMS S.43.1. The position of the fringe of order m on the screen' is given by AD y=m~ d - where d is the separation of the slits and D is the slit-screen dis- tance. The separation of the fringes due to wavelengths A t ^nd A 2 " mZ> ,, ,. (3H100cm)(6000-4bOO)Xl<r 8 cm =0.0072 cm = 0.072 mm S.43 2. The fringe width is given by AD , d$ _ (12cm)(18cm) " D ~ (200cm) = L08cm. Frequency of vibrations is =23.15 cycles/sec. S.43.3. Shift in the central bright band by n fringes is where j* ? and ^ are the refraction indices for the glass plate inserted in the slits and / is the thickness of the plate. , . .JUL. ^(5X4800 X 10- cm) ^ g x 1Q _ 4 cm=g microns . A*2--^i (1.7-1.4) S.43.4. (a] The given .arrangement is originally due to Lloyd (1834) and is called Lloyd's minor. The slit 5 acts as a source and its reflected rays from (he mirror MM' appear to diverge from the vir- tual image 5'. The interference occurs between the rays (SP) directly from S and those (S'P) from the virtual image at S'. Owing to the reflection from the mirror, the phase of the reflected ray is increased by TC. This leads to the shift in the band system ay-compared with that in Young's experiment. In particular at the center of the sys- tem lies the dark band (minimum) rather than the bright band (maximum). Interference 335 Drop a perpendicular SQ on S'P. Let angle S'SQ be called 0. Then the path difference Condition for maxima is *. 2 '' =2A sin The additional A/2 corresponds to the extra phase difference w which arises due to reflection. Condition for minima is 1 8=2 h sin 6=wA where m is an integer. (6) As the reflected rays cannot conte below the plane of the surface of mirror, the fringes can appear only in region A. Region A M' Fig. S.43.4 S.43.5. M jr \ fi(t)f,(t)dt A - 1 -*- J sin sn 336 Solutions to H and R Physics H T \ A A* f 2 T J 12 where we have used the identity Sin A sin fi=i [cos (A -B) -cos (A+B)] i A A r The second term in the parenthesis of the right side vanishes when the upper and lower limits are inserted and use is made of a>T=2'n " (fif^ at>:== i ^1^2 cos (^i $2) (!) Let the phasors be Pi e =A l sin ' and Pt=A 2 sin ( The dot product of the phasors is P^.P^A^i cos (<f>i ^ 2 ) --(2) since the angle between the two pjjasors is (^^2). Comparing (1) with (2), we find the desired result S.43.6. (a) Let S^/^ sin (ai t+jJ+Az sin ( +. . ,+^iii sin (a>t+<f>n) At (sin to/ cos ^4-cos a>/ sin t ) 4j (sin co/ cos ^ 3 +cos cor sin -h +/4n (sin o>r cos ^n+cos a>r sin 0n /i n =sin a>r S ^7 cos rfi+cos cor S y4 il i-l ==J? sin a>r+C cos wr n where 5= S y4 cos ^ and C= 2 1 Ai sin /=l /-I (6) fl*+C a = L x i icos t #i+ S A' sin 8 1 /l /-I +22 S Ai4i (cos ^i cos ^#+sin fa sin y Interference 337 -O) But (S ^,) = S X'-f 2 S L^ .-(2) Now, the right hand side of (1) is equal or less than the right nd side of (2) be .'. 5 2 +C 2 < (S hand side of (2) because | cos (0. <; ) j < 1. or (c,> The equality of sign in (3) holds when ^=^2 ..... . =i4 rt that is, all phase angles ^/ must be same. S.43.7. As reflection is from a medium or greater refractive index, we have the conditions, 2rf/? = (w + J) A, w=0, 1 , 2, ...(minimum) ...(1) 2*7.7 =wA, w-~0, 1, 2, ...(maximum) ---(2) where rf is the thickness of the film and n its refractive index. (2X1.25) ,/=(6QOO A)(m + J) -(3) (2)(l.25)rf=f7000 A) m ...(4) Combining (3) and (4), (6000 A)f m+ i)-(7000 A) m or m = 3 - (5) Using (5) in (3) or (4), </=8400 A. S.43.8. Condition for minima is 2rfrt = (/H-hJ) A, ;w=0, 1, 2, ... Solving for J, with /;i=(), 1, 2,-. S.43.9. Condition for maxima for the film thickness t l A where in is the order of the fringe. At a greater thickness / 2 of the film, let the order of the frincc be tn+p. Then A ...(2) 338 Solutions to H and R Physics II Subtracting ( 1 ) from (2), or t -t- pX (9)(6300A) - or '' /i - = "- As the dark bands are flanked by bright bands, on either side, the information on the number of dark bands is irrelevent to the problem. S.43.10. (a) Here, a phase change of 180 is associated with the rays undergoing reflection both at the upper and lower surface of the oil drop since at both the surfaces the reflection is from a medium of greater refractive index. The thinnest region of the drop must therefore correspond to a bright region. This is in contrast with the dark central spot observed in the system of Newton's rings under reflection where the ray from the bottom of the air film alone under goes a phase change of 180. (6) Condition for brightness for interference under the given conditions is Setting A=^4750 A for the blue colour, /i= 1.2 and m=3, we get A) 2n - (2X1.2) (c) Unless the film is reasonably thin i.e. a few wavelengths of light, interference fringes would show up localized on the film. When the oil thickness gets considerably larger, the path difference between the two ra\s, one getting reflected on the top surface and the other one at the bottom of the film, would amount to several wavelengths, leading to a rapid change in phase difference at a gwen point as one moves even a small distance away. The inter- ference fringes are blurred out and the pattern disappears. 44 DIFFRACTION 44.1. Condition for getting minima in the diffraction pattern from single slit is a sin 0^wA, m = l, 2, 3, where a is the slit-width. If xis the distance of the minima from the central maxima and D is the slit-screen distance then sin 0^0--^ ^ , provided x < D. ^ mX m\D SI r~*t . a ~ e - " x But x= J (Distance between first minima on either side) -4 (0.52 cm) =0.26 cm Set m\ for first minima. Then ' a = (5460x 10~ 8 cm)(80 cm)/(0.26 cro)=0.017 cm. 44.2. Z)=/=70cm ^~^ (a) Linear distance on the screen from the center of the pattern to the first minima is given by setting ml. V _n ; nfl ^n fl_ DA (70cm)(5900x10" 9 cm) AI LJ Sin v U v = /f\f\A \ a (0.04 cm) -=0.103 cm = 1.03 mm (b) Linear distance on the screen from the center of the pattern to the second minima is given by setting m 2. jr z =206= -(2)(1.03 mm) = 2.06 mm a 44.3. (a) sin G=m\ a sin Oa^A, a sin 6',=2A?> (a) But 6a-e,> A a = 2^. (6) a sin Oa^AWaA a~2^< ;A/> a sin 0^=:w&A& Set 0a=0b. Then minima in the two patterns coincide when JH& = 2fMa. 44.4. (6) / 8 ---/.., i^- \Tcxtbook Eq. 44.8b) 340 Solutions to H and R Physics II Differentiate with respect to a and set 3a - f 2a 2 sin a cos *" a] __ or a sin a (a cos a -sin a)=0 One solution is a^=0. Other solutions are obtained from a cos a sin a^O or tan a a (6) y -.at 44.4 (c) We find the first root is zero, then we have a series of roots less than but gradually closer to (w + J)K= 1, 2, 3 ...... The first three values of m are 0.93, 1 959, 2.971. 44.5. Figure 44.5 shows the graph of y~ ( - - ) . The ordinate y has the value 0.5 for cr-=7944, biffraction 341 A 1-0 0-5 30 60 90" sin oc Fig. 44 5 44.6. (') For the first maximum beyond the central maximum since rather than 3^, the vector E is not vertical. (b) The angle which En makes witli the vertical is given by (3ft 2.86*) 44.7. sin O a (a) a/A: 3* 1.4A (540)-25.2. sin Ba;-*- 1 - K 6 X - 26.45 ' Half width, A0-2e a -52.9. (a) a/A -5 sin 6 9 =^~ 5* 342 Solutions to H and R Physics II (c) <7/A^iO sin 0*=^ AB=(2)(2.55) = 5.r. 44.8. (a) Condition for obtaining the first minimum from diffrac- tion at a circular aperture is given by sin 6=1.22 ~ (7 where d is the diameter of the aperture, A the wavelength and 6 the angle between the normal to the diaphragm and the direction of the iirsi minimum. r 1450 meter /sec n v 25000 cycle/see Cr D . .. 1.22A . -. (1.22);0.058 meter) sin l -- ---sin x -~ 1 . 6.8 d (0.6 meter) , , ._1450 moter/sec , 1000 cycle/sec , .. 1-45 meter . (1.22H 1.45 meter) n/lo sin 8~ _ - - -- - =2.948 (0.6) meter) This is impossible as sin cannot be greater than 1. Hence, the condition for minimum cannot be obtained. 44.9. Raylcigh's, criterion for resolving two objects is , . . 1.22A ^ 1.22A a e*=sm' -y- ~d-D where Jis the diameter of the pupil, A the wavelength, a the dis- tance between the two headlights and D the maximum distance of the automobile at which the eye can resolve the lights. n _ ad (1.22 mcter)(5.0x KT 3 meter) __ onol f "TO^J meter) ~ 9091 meters " 44.10. As in Problem 44.9. 44.11. (a) Angular separation of two stars is given by .. adiui 1.22A (1.22)(5000XI(T 8 cm) ON/lft . T .. ---- -8X10 'radiui =0.165 seconds of arc. \ (6) Set 6jta/D where a is the distance between the stars and > is the distance of the stars from the earth. Diffraction 343 />=10 light years-(10)(9.46xl0 13 km^9.46X 10 18 km. a=-D8R=(9A6x\Q l * km)(8XlO~ 7 radians) -7.57 x 10 7 km. (c) Diameter of the first dark ring is </'=(2)(1.22/A)A/=2/e* where /is the focal length of the lens and d' is the diameter of the lens. cm)(8X 1(T 7 radians)-2.24x 10~ 3 cm. 44.12. Separation of two points on the moon's surface that can be just resolved is given by = where D is the distance of moon from the earth, d is the diameter of the lens and A the wavelength ot light that is used. D -240,000 miles 3.86 X 10 8 meters J-200 in -5.08 meters A 5500 X iO~ 8 cm 5.5 X 10" 7 meter (1.22)(3.86X10 8 meter)(5.5XlO~ 7 meter) , t a~- ----- -' ---------- T - c A - - = 5 1 mete (5.08 meter) 44.13. Diagrams (a) to (h) in Fig. 44.13 correspond to the points labelled on the intensity curve. The contributions from the two slits Let -, Then, o b c d e f g h i TT I an srv i in l 5TV IT (a) (c) A=A,-A 2 = / \ (e) (y) 344 Solutions to Hand A Physics It \ to the amplitudes in magnitude and phase are indicated by A^ and A t . The intensity is found as the square of the resultant amplitude A where A is given by the vector addition of AX and A2. 44.14. For the double slit, the combined effect due to interference and diffraction is given by A, =/ m (cosfi) 2 / S -^) Z ...(1) Condition for minima in the interference pattern is p=(w+i)*, m-0, 1,2,. . . --(2) where p= y sin 0. Condition for minima in the diffraction is OC=TC, 2rc. . . (3) In Example 8, the Cth minimum of the interference pattern coin- cides with the first minimum of diffraction pattern. So - So ' T~T-2 For the second minimum in diffraction pattern, set a=w. Then, 0=- a- 11* From (2), w=10 gives p-^(lO^) TT which is the maximum value less than 11^. Thus with w=0, I, 2, ........ 10, eleven fringes in all are present in the envelope of the central peak. Hence, the number of fringes that lie between the first and the second minima of the envelope is (11 6)=5. 44.15. Since d/a^2 is an integer, the second order will be missing. There will be one fringe on either side of the central fringe. Hence, in all there will be 3 fringes in the central envelope of the diffrac- tion pattern. 44.16. For the double slit, the combined effect due to interference and diffraction is given by / e =/m (cos p) . ...(44.16) wtth ^ T sin . n sm 6 Diffraction 345 Set d=a. Then p=a, __ / (2 sin a _^L5) 2 - sin 2a \ 2 . / sin " / " V : The last equation is appropriate for the diffraction pattern of a single slit -of width 2a. 44.17. The missing orders occur where the condition for a maxi- mum of the interference and for a minimum of the diffraction are both fulfilled for the same value of 6. Thus, d sin 6 /M.t, /fi--0, 1,2,... a sin ~///i, p=l, 2, ... , dm so that =s a /> Since both m and p are integers, rf/ must be in the ratio of two integers if the missing orders are going to occur. If order 4 is mis- sing then we must have d/a-=4. (b) Other fringes which are missing are 8, 12, 16- SUPPLEMENTARY PROBLEMS S.44.1. (a) The water droplets in the air act as diffracting centres giving rise to the observed rings. The diffraction pattern due to a large number of irregularly arranged circular obstacles is similar to that for a single obstacle. (fr) Airy's formula for the intensity distribution for diffraction from a disk of radius a is given by -< . 2n floe where jc= -, ...(2) and /xOt) is the Bessel-function of the first order. The position of the first secondary maximum is given by the condition jc=5.136 Now, the radius of the ring is r=(1.5)(1740) km-2610 km. 346 Solutions to H and R Physics 11 If d is the distance of the moon from earth, the angle subtend- ed by the racius of the ring is given by Using (3) and (4) in (2) and A=5.6x 10" 6 cm for the center of the visible spectrum, the diameter of the water droplets is found from n , (5.136)(5.6xlO- 5 cm) nm . . tl D=2a= -- ------ =0.013 cm-0.13 mm. S.44.2. Two objects like A and B are said to be complementary screens. Let VA and U# denote respectively the values of the vector implitude of the light wave when screen A or B alone is placed and let U be the value when no screen is used. A generalisation known as Babinst's principle relates the diffrac- tion patterns produced by two complementary screens. It states that the resultant of the two amplitudes, produced at a point by the two screens, gives the amplitude due to unscreened wave. Thus we may write Setting U=0, we get U* = U/?, showing thereby that at P where 7=^0 the phases of U^ and UB differ by and the intensities IA= \UA\ 2 and 7n= } UB \ 2 are equal. 5.44.3. Rayleigh's criterion gives e =i ='- 22 7 where x is the distance between the two point sources, D is the dis tance of the sources from the observer, d is the pupil's diameter. *=1.2 2 A Z)=(1.22)( 5500 X 5 10 c ^ cm )(100x5280ft) =71 ft. 5.44.4. (a) Linear separation of two objects on Mars is given by x= 1.224- D a =( 1 .22)( 5500 X 5 l ^' <:m )(80.45xl0.tm) = 1.08x10* km. (b) x=1.22 4 D a ~ (80 - 45x 10 km)==10 - 6 km - Diffraction 34? S.44.5. Assume for simplicity that the sources have equal intensity /. Then the intensity pattern is given by, /(e)=4/ (cos p) a ( sin- a -) -(1) TCfl with a= -y- sin 8 ...(2) , . 2m d . a and r- sin 6 The term (sin a/a) 2 is due to diffraction and the term (cos p) 2 due to interference. The quantity 2[i is the phase <f> due to path differen- ce plus the phase difference of the sources, (E t 2 ). If si""-^ is allowed to vary from to 2rc, the intensity pattern at a given point goes alternately through maximum and minimum twice. 45 GRATINGS AND SPECTRA 45.1- rfsin 0=/wA, where w=0, 1, 2,. . . Set 9-90 rf=^-crn-2.5xlO-cm For A-4000A on 8/A = (2.5xlO- 4 cm)(sin90) (4000X10"* cm) For A-7000 A, (2.5xl(T 4 cm)Xl -=3.57 (7000xl(T 8 cm) Therefore three complete orders of entire visible spectrum can be produced. 45.2. Let E c and co be the amplitude and the angular frequency of the wave incident on the three-slit grating, Let <f> be the phase dif- ference between the diffracted waves emerging from 6\ and S t and a further <f> between those from S 2 and Sg. The waves from Si, S 2 and S 3 are given respectively by Fig. 45.2 Grating and Spectra 349 ^EQ sin cot ...(1) ...(2) ...(3) *u / sin 6 ... with 0= - ^ ..(4) Here 6 is the angle of diffraction anJ d is the distance between the centers of two neighbouring slits. The resultant amplitude at a point P on the screen is given by E=E l +E 2 +E^E 9 fsin a>f+sin (w/ + 0) + sin (wf + 20)] Expanding the second and third terms in the parenthesis and re- arranging, = E [sin cuf (l+cos0 + cos 20) + cos <t (sin 0-fsm 2*6)1 .-.(5) Let the resultant wave at P be given by -5 sin (CU/+Y) .-(6) or /?==/? (sin o^ cos y + cos wt sin y) (7) Since (5) and (7) represent the same wave, the coefficients of sin o>/ and cos cot must be separately equal. =E {} (1 + cos B sin y~/f (sin + Squaring both sides, adding and simplifying, ! 4cos 2 0) 45.3. (a) Set -?- = 1 = J- (1+4 cos f -4 cos 2 /w z 9 or cos 2 I -cos 0-7/8-0 _. 2V2 0-r=56 f =-0.977 radians. Width at half maximum is given by Ae - 20 - A * - 977 A - Ae " 2t) ~ ^^ 3.14 J "" (t) For two-slit interference frings, (c) Yes. 350 Solutions to H and R Physics II 45.4. 7 e ^ m - (1+4 cos #+4 a; 0) -(1) Differentiating with respect to 0, 8 4 77 = -x-- 7m (sin + 2 cos sin 0) ..(2) u0 9 Set-~=0 sin (H-2 cos 0)=0 The solutions with < < 2r, are sin 9 = or = 0, rc, 27? jO y The correspond to maxima as ~~n^~ ~~ ve The other solutions are given by 1+2 cos 0=0 2rc 4rc * = r , y < These correspond to minima as ,. 2 =+ve. Upon using the values = 0, TC and 2rc in (1) we find the intensity 7 1 as 70=7w -~ and 7 respectively. Of course for the minima 0=-r- and are expected as 70 ^0. Tlvus, the secoedary maximum is located at 0=n and lies half way between the principal maxima, and has intensity In,/9 or 0.1 17m. 45.5. As the grating is blazed to concentrate all its intensity in the first order for A-- 80,000 A, we have With the use of visible light (4000 A < A < 7000 A) the order numbers are given by d sin 6 8 x 10~ 4 cm , , c , .. . A ,^ ~*f\f\ A \ A T\Qm = ' ' for red " ght (Ar= /00 A) and m ^ X [JJ_' cm = 20, tor blue ligl.t (A=4000 A). Q x, i \t c m Grating and Spectra 351 Thus, the intensity is concentrated near the llth order for red and 20th order for blue light, the orders overlapping to such an ex- tent as to give the appearance of white light for the diffraction beams. Af na . ,. 45.6. a= ~ =-r- sin Z A where ^ is the phase difference between rays from the top and bot- tom of the slit and the path difference for these rays is a sin 0. For first minimum, o; rc. Also, set a Nil, where N is the number of slits and d is the grating element. Replace by A# , then, sin where the angle A0 corresponds to zero intensity that lies on either side of the central principal maximum. For actual giantings, sin A0 will be quite small so that we can replace sin APo by A0 to a good approximation and obtain the desired result 45.7. The gratine element, d^ > : = 3.175 x 10~< cm oOOU Condition for maxima is d sin 0-^/;/A, w^O, 1, 2.. . . For the maximum wavelength that can be observed in the fifth order set = 90 and m-5. Then, ^.nsx^cm ^ 6 ?50X |0 .. cm or 63SO A . Thus, for all wavelengths sliortcr than 6350 A the diffraction in the fifth order can be observed. ? S4 45.8. (a) The grating element d-=~- cm = 5.0S x 10" 4 cm Diffracted light at angle 0=^=30 is observed such that d sin 6=^mA or mA = (0.5)(5.08xlO~ 4 cm) = 2.54x 10' 4 ,m. In order that A mny be in the visible region (4000 t< < ^ < 7000 A), the corresponding diffraction orders and the associated wavelengths are given as follows: 352 Solutions to H and R Physics II m =4 . A-6350 A (red) m = S ; A=5080 A (green) w =6 ; A=4233 A (blue) (b) The wavelengths can be identified by observing their colour 45.9. The grating element, </= ~^-cm-3.33x I(T 4 cm d sin =mA sin 6=m-7- d /5890xl(T 8 cm\ A ,_- -=m -=TT^ r n - 1=0. 1767m V 3.?3x 10 4 cm/ As sin cannot exceed unity, the' order of dilTraction in is restrii ted to 0, 1, 2, 3, 4 and 5. The corresponding ?ngles of diffraction are then, 0o=sin~ 1 e^sin" 1 (0.1 767x1)- 10' 6,-sin- 1 (0.1767X2)=21 ej^sin" 1 (0.176' 7 x3) = 32 e^sin- 1 (0.1767 x4)=45 8,= sin' 1 (0.1 767X5) = 62 45.10. rfsin 6=mA Set m=l For ^=4300 A rfsin 6 1 =(1)(4300X IO* 8 cm)-4.3xiO- 5 cm ...(1) For A =6800 A rfsin e 2 =(l)(6800xiO~ 8 cm) ==6.8 x 10" 8 cm ...(2) By ProMem, e 2 -0,-20 ..(3) Adding (i) and (2), 2 d si n 2 -t- fl -L cos ^~ - 1 1 . 1 X 1 0' 5 cm . . . (4) Subtracting (2) from (I), 2Jcos' i 'Y t sin - 6 ^== 2.5X10-' cm -.(5) Dividing (4) by (,5) tan J - cot ^4.44 .-(6) Grating and Spectra 353 Use (3) in (6) to find fl I fl A _ n tan - z ~* =4.44, and tan-^r- 1 -4.44 tan 10=0.7828. / L ^ -\ ~t where use has been made of (3). Hence, 6,4-0^76 ...(7) Solve (3) and (7) to find e i= =28 ; 6 2 =48. -.(8) Use (8) in (I) or (2) to get , 4.3xl(T 5 cm cm. = 10900- - ^^ sin 28 /. Number of lines per cm, N= -i- Number of lines per inch = 10900 x2.54= 27700. 2 54 cm 45.11. The grating element, d= - =3.17x 10~ cm. The angular range through which the spectrum can pass is defined by 6 a and 6, (Fig. 45.11). 6^9 28' sin 9 l =0.1 648 d sin Oj-^/wAj ^O.nxlO' 4 cm)(0.!648) =5220x10-" cm = 5220 A Screen 4 30cm .*. sin 6 a =0.196 d sin 6,=mAj A,=(3. 17 X 1(T 4 coi)U96) 5cm i / Grating Fig. 45.11 354 Sol <tions io H and R Physics II Thus, t ic range of wavelengths that pass through the hole is 5220A to 6210 A. 45.12. The path difference between ^i adjacent rays like 1 and 2 is :d sin ty+d sin 6 =(/(sin ^+sin 6) Condition for obtaining diffraction maxima is 8=mA, w=0, 1, 2, ... i.e, d (sin 4*+ sin 0)~mA, m 0, 1,2, ... Fig. 45.12 45.13. Maxima in the interference pattern occur when d sin 0~mA where w=0, 1, 2, ... Minima in the diffraction pattern result if a sin G~w'A where now w'=l, 2, ... When both the conditions are satisfied for the same angle 0, jrf m _,- a ~~ m' If alternately transparent and opaque strips of equal width are used in the grating then d=2a and consequently M=2. There- fore, all the even orders w=2, 4, 6, ...(except w=0) will be missing. 45.14. For various values of angle of incidence ^ ranging from to 90, we calculate the angle of diffraction 8 using the grating equa- tion, d (sin ^-J-sin l )=mA with w=l for the first order, A=6x 10~ 5 cm and <f=1.5 X IO" 4 cm. The angle of deviation 8=^8> the 4-ve sign is taken when the incident and diffracted light is on the same side of the normal and ve when on opposite side. Fig. 45.14 shows the plot of 8 versus ^. Grating and Spectra 355 6 60 20 20 <;0 u<J 90 j + Fig 45.14 45.15. The principal maxima occur when d sin 0=mA. The path difference between the light which gives rise to princi- pal maximum and the immediate minimum is A/W, where Wis the number of slits. Let a change of angle A0 produce this path diffe- rence. Then dsin (6-f A0)-rfsin 8= A/TV rf(sin 6 cos Afl+cos 6 sin A0-sin 0)=A/AT For small angles, sin A0J-* A0 and cos A0 - 1. Hence, or Thus for order m, 45.16. The half-width of the fringes for a three-slit diffraction pattern is given by A <* A/3.2 d. (1) The half-width of the double-slit interference fringes is given by ..(2) If the middle slit in the three-slit grating is closed then it reduces to a double-slit and the grating element beomes Id. Replacing d by 356 Solutions to H and R Physics II Comparing (3) with (1), we find that the half width of the inten- sity maxima has become narrower by 20%. 45.17. (a) The grating element, d= ~~ = 1.9 X 1(T< cm = 1.9X10 4 A d sin 8=wA For 1st order e^sin" 1 ^ =sin~ 1 f -- -^-- ^ Wsin"" 1 0.31 = 1 rx- r* k Dispersion />=;; --- A r d cos 6 = 11.9X10*^)008 1-8^0.553x10-' radians/A =3.17xlO~ 8 deg/A. For 2nd order, 8,-siiT* ^-==si n - 1 f^^- =38,2. (1.9 X 10* )00838.2 =7.67XlO" 8 deg/A. radians / A ^ x 5800 A For the 3rd order, e^sin" 1 r^ T^ ^sin" 1 0.93=68.5 1V X 1U A 3 (1.9X10* A) cos 68.5 =4.31 X 10"* radians/A =24.7 X 10~ 3 deg/A. (6) The resolving power is given by RNm, where N is the number of rulings and m is the order. For the 1st order, /?=40,000x 1=40,000 For the 2nd order, K=40,000 X 2= 80,000 For the 3rd order, ^=40,000x3 = 120,000 45.18. A/l = 5895.9 A=5890 A5.9 A 6=80" Mean wavelength X =$(5890+5895.9) A=5893 A d sin 6=w/\ y,^ .. . . (a) Gratmg spacmg, mA (3)( 5893X10-" cm) --- = 1 7952 X 10~ 8 cm = 1 8000 A, Grating and Spectra 35? (b) Resolving power, R= ~^= .'. Total width of the rulings, M/=(333)(I8000 A)=6x 10* A =0.6 mm A* in \T A 6563 A ~ f Af .. 4519 ^- - =3646 lines ' 45.20. d sin 9=wA d& m d\ ~~d cos Equation of grating is d sin 6==/wA 4 m sin -j = : : a A n__d^ _^!5^ tan 9 <M ~~ Acos6~ A 45.21. Number of rulings, tf= -(6 crn)=36000 cm A1 A 5000A AA , rA A A= - == 7 r 77 =0.046 A . (36000)(3) (&) Resolution is generally improved by going over to higher order m for a given grating (N fixed) and given A. However, in the present case m=3 is the highest order that can be employed for normal incidence. 45.22. In order that dispersion be as high as possible (condition 2) maximum value for the diffraction angle must be chosen. Since the first and second maxima are restricted to angles upto 30, (Condi- tion 1), set m=2 and 6=30 in the grating equation, dsinfl=mA (2X6000 A) " where we have used the value of the higher wavelength. 358 Solutions to ft and R Physics tt (b) Since the third order is missing we have d i =3 a where a is the slit width. a- -y =-1^(24000 A)=8000 A (c) The maximum order that can actually be seen is given by setting 6=90 and with the choise of A=6000 A. d sin 0=/wA -^A in J. - (24000 A)(sm90) , m ~~ A - 6000 A The orders that actually appear on the screen are w=land2. since w=3 is missing and w=4 occurs at 6=90. 45.23. (a) Grating equation is, d sin 8=wA. For order m and angle 6^ we have d sin 1 =mA ...(I) whilst for order (m+1) and angle 2 , rfsine 2 =fm+l) A ...(2) Dividing (2) by (1) m+\ sin 6 a _ 0.3 m "" sin e^"" 0".2" whence m=2. Using (3) in (1), we find the separation between adjacent slits, (fr) Since the fourth order is missing, we have -i-4 a where a is the slit-width. /. Smallest possible slit-width, a==4= 6 -l^=1.5Xl0 4 A. 4 4 (c) The maximum possible order is obtained by setting 8=90 in the grating equation d sin 6=mA </sin6 (6xlQ*A)(sin90) 1A m - ..-^ - " Grating and Spectra 359 Also, since m=4 is missing, it follows that m=8 will also be missing. The orders actually appearing on the screen are m=0, 1, 3, 5, 6, 7, 9. The tenth order occurs at 0=90. 45.24. For oblique incidence the grating equation is rf(sin '/'-sin 8)=wA .. (1) where <J> is the angle of incidence, 6 the angle of diffraction and m the order of diffraction. We have assumed that the diffracted beam and the incident beams are on the opposite sides of the normal. ^=90-Y ...(2) 6-90-2p ...(3) Hi = l ...(4) Use (2), (3) and (4) in (1) to find d[sin 90-Y)~sin (90-2?)]= A or cos Y~~COS 2p= /. 2 sin (Y + 2p) sin or s As Y is small one expects (5 also to be small. Replacing since function by the argument which upon simplification and re-arrangement yields the result Bv Problem ___ --- __ '-- uy Problem, - A - 3000 ' Therefore p. 45.25. Bragg 's law is 2d SID fl^iftA, 01= l f 2, 3, . . </=2.8l A .-.--. . A i.: R. 360 Solutions to H and R Physics II Hence, the angle through which the crystal must be rotated is (45- 12.3) or 32.7 in the clockwise direction. For w=2, sin 6= 0-25.3 The crystal must be rotated through an angle (45 25.3) or 19.7 in the clockwise direction. For m3 sin 6- g)(1 ' 2 A) -Q 64 6=39.8 The crystal must be rotated through an angle (45 39.8) or 5.2 P in the clockwise direction. For 6-58.6 .*. The crystal must be rotated through an angle (58.6 45) or 13.6 in the counter clockwise direction. Higher order reflection is not possible as sin 6 would exceed 1. 45.26. The sodium chloride crystal has face-centered cubic lattice. The basis consists of one Na-atom and one Cl-atom. There are four units of NaCl in each unit cube, the positions of the atoms being: Na 000 ; iJO ; OJ ; Oii Cl *H ; OOi ; 0*0 ; 00 Each atom has in its neighbourhood six atoms of the other kind. The reflection from an atomic plane through the top of a layer of unit cells is canceled by a reflection from a plane through middle of this layer of cells because a phase difference of K rather than 2rc is introduced. 45.27. Bragg's law is 2d sin 0=mA A== (2X2.75 A)(sin 45)_3.889 A m m For m=3, A=1.296 A. For m=4 f A=0.972 A. Thus, diffracted beams of wavelength 1.29 A and 0.97 A occur. Grating and Spectra 361 45.28. Bragg's law is 2d sin 0=mA For Irce X, ' _(30*gA) .in 30 m 1 SUPPLEMENTARY PROBLEMS S.451. I II Ml IVV VI 5=0 _7T 36 47T ,-ZL *5 A 6 (iv) s / = - Fig. 44. 13 S.45.1. In Fig S.4S.1 (a) the diagrams corresponding to the points (o)to(/)of the intensity plot for six slits are shown. For the central maximum light is in phase from all slits as well as that from 361 Solutions to H and R Physics It a single slit as in (0 of the figure and gives resultant amplitude A which is N times as large as from a single slit. In (ii) is shown the condition half way to the first minimum. This point corresponds to a=rc/12 so that the phase difference from corresponding points in adjacent slits is 8=75/6, this being also the angle between successive vectors A x to A 6 . The resultant amplitude A is given by compound- ing these vectorially and the intensity is given by A 2 . For the derivation of th^ general intensity formula, consider Fig. S.45.1 (b) wherein are shown the six amplitude vectors with phase difference slightly less than in (//) of Fig. S.45.1 (a) The magnitude of each of these is identical and is given by Fig. S.45.1 (b) (sin p)_ A An - r - A (,!/ P The amplitude An is represented by the chord of-an arc of length A subtending an angle 2(1 at the centre (single slit diffraction pattern). Successive vectors are inclined to each other by an angle 8=2. Also each of the vectors subtends a constant angle 2<x at the center, indicated by the broken lines in the figure. The six vectors form a part of a polygon with center at O, the total angle subtended at O being ..... (2) From the triangle OBC, we find that the resultant amplitude is where r==O5 is the radius of the inscribed circle. Similarly, from the triangle OBD we find that the individual amplitude An is given by A l ^An ss 2r sin -'-(4) Dividing (3) by (4), we get A 2r sin (+12) sin NCL An~~ 2r sin a ~~ sin a '"^ where use has been made of (2). Eliminating An between (1) and (5), Grating and Spectra 363 A A sin p sin y4=^4 - sin a * j A* A 2 sin 2 (3 sin 2 Ma sin 2 MX ,_ Intensity. /-^-^y? -^^ - / -^^ ...(7) where / is the intensity for a single slit. The principal maxima occur in directions for which the waves are all in phase. with m=0, 1, 2, ... ...(8) We note that when the condition (8) is used in (7), the quotient (sin Mx/sin a) becomes indeterminate. We, therefore, resort to L'HospitaFs rule i.e., differentiate both the numerator and the denominator of (7) and use (8), Lim sin MX Lim N cos MX = it M <x~*mft sin a a-*wrc cos a Then (7) becomes These secondary maxima bear a strong resemblance to those of the secondary maxima in the single slit pattern. Therefore, repla- cing / with h 9 we have This establishes the result required in part (6) of the Problem. Secondary maxima are positioned for the approximate value 2N In this case | sin Ma | =1. Also, since M is very large a will be small and sin 2 a c* a 2 . Also, sin p/(J^l, as p will also be small. From (7) / = lk ~ A 7m a This is the result for part (a) of the Problem (c) As the number of slits becomes large the polygon of vectors approaches the arc of a circle and the analogy with the pattc rn due to a single slit of width equal to that of grating is complete. The diagrams for the grating become identical with those for a single slit if Ma is replaced by p. 364 Solutions to tf and R Physics // S.45.2. e sin6 m sin e/m 17.6 0.30237 1 0.30237 17.6 0.30237 1 0.30237 37.3 0.60599 2 0.30300 37.1 0.60321 2 0.30161 65.2 0.90778 3 0.30259 -65.2 0.90778 3 0.30259 (sin e//w) =0.30242 \=*d sin e/m=(1.732 X 10~ cm)(0.30242) =5238XlO- 8 cm=5.238A S.45.3. (a) e sinO m sin e/m 640' 0.11609 1 0.11609 1330' 0.23345 2 0.11672 2020' 0.34749 3 0.11583 3540' 0.58306 5 0.11661 (sine/m)o.=0.1163) A=<f sin e/w=(5.04xlO~ 4 cm)(0.1163)=5861 A. (6) The missing fourth order must lie between 6=2020' and 6=3540'. With e=20'20' With _ 5861 X IP" 8 cm a ~ 0.34749. 6=3540' 5861X10-' cm = 1 .69 x 1 0~ cm = 1 .69 micron. >1.0x 10~ 4 cm 1.0 micron. 0.58306 Thus, the slit width must lie between 1.0 and 1.69 microns. S.45.4. Grating equation is ...(1) where d is the distance between the rulings and m, which is an integer, is the order of diffraction. Diffrentiating (1) with respect to A, Grating and Spectra 365 d^ m _/> _ m OT ~ = l-sin'O ~ N/d a ^ sin 1 6 77) 1 V d*-(mA) v/(d a /J*) - A If the wavelengths are crowded then d\ can be replaced by AA and </6 by AO. __ Thus, Ae=AA/V(d/w)-"A. S.45.5. (a) ^ sin 6=mA ...(1) Set e=90 j/i 900 A i Then, w^M-^" 1 ' 5 ' Only m=l is allowed. Thus, there can be only one line on each side of the central maximum. (b) d sin e=/wA Set m=l . A 6000 A The angular width is A tanB cos Af =^-~^~-=9xl<r 4 radians where use has been made of (2). (c) Resolving power, /?=A/AA. ...(3) Multiply (3) and (4) to get where use has been made of (I). =tan 6/R 366 Solutions to H and R Physics II 8.45.6. (a) /?=A/AA=Mw Now, v =c/A A i c AA Av|=c -T. - where use has been made of (1). (b) Path difference between the extreme rays is S=M/ sin 6. Therefore time of flight difference, 8 Nd sin 6 -(I) ...(2) ...(3) ...(4). ...(5) (c) Multiply (3) and (5) / A % / A , N c Nd sin (Av) (A0= dsm 6 mA where use has been made of the grating equation, d sin 6=mA. S.45.7. Refer to Textbook Fig. 45.14 (a) and 45.14(6). Next five smaller interplaner spacings are shown in the sketch of Fig. S.45.7. Fig. S.45.7 d 1 =a 9 o . *o j _?9_ A 9 Ir 6== " ' ;/> 46 POLARIZATION 46.1. Let 7 be the intensity of the incident unpolarized light, /, the intensity of beam through the first sheet and 7 2 the intensity trans- mitted through the second sheet. By Problem, / 1 =|/ /,= /! cos 2 6 ( fl ) ~-=cos 6= -i .'. cos6=l/V3=0.577, or6=55. r F (6) / 2 - 3 -M cos0= /^ = 0.8165 or 0= : 46.2. Let 7 be the incident intensity, /j, / / 3 and / 4 the intensity from successive sheets. Then the transmitted intensity of light 7 4 -/ s cos 3 ~ Also, / 8 =/j,cos a .6 /j-/! COS 8 9 A-t/o / 4 =/ 8 cos 8 e=/ t cos^ e^/i cos 6 e=(V2) cos 8 e. /. Fraction of incident intensity that is transmitted is given by 46.3. (a) fx-fsin (kz-a>t) ...(I) Ey^Ecos (/T2 cor) ...(2) Square (1) and (2) and add J =E* ...(3) This is an equation of a circle. It is circularly polarized. The wave has the form =ia+ j,=[i sin (**-/)+ j cos (jfcz-r)] We can examine its behaviour at some fixed point in space, say r=0. At r=0, r/4, 7/2, 37/4 and T, E (0, r) has values of j, i, j, i and j, respectively. These values are indicated jn Fig 46.3 (a) over one period, using a right-handed system, has 368 Solutions to H and R Physics II constant magnitude but rotates counter-clockwise (looking toward tfce source). The field is therefore left circularly polarised. (6) En=E cos (kzo>t) E,~E cos (kz o>/+/4) Equations (4) and (5) can be combined to obtain -(4) ...(5) -(6) This is an equation of an ellipse whose major axis is tilted at an angle 45 to the E* axis. We may write (4) and (5) in the vector form E (z, /)=i cos (kz o>/)+ JE cos (kz a)t+n/4). We can examine its behaviour at z=0 and for various values of /. E(0,0)=iJM-J J* Fig. 46.3 (a) Pig. 46.3 (b) E(0, Polarisation 369 It is seen from Fig. 46.3 (6), that the -field rotates counter- clockwise and is therefore left handed. (c) Ex~E sin (A-z a>t) This is an equation of a straight line. The light is plane polarized with the major axis inclined at an angle of 135 with the E* axis. 46.4. (a) For water tan 0*>=fl=1.33 (b) Yes, the angle does depend upon the wavelength as n is a function of A. 46.5. From Textbook Fig. 41.2, we note that the refractive indices range from 1 470 to 1.455 for the white light. The corresponding polarizing angles are 8,,= tan- 1 1.470=55*46' and tan- 1 1.455=5530' 46.6. (a) For the ordinary ray, sin i sin 45 sin ro= "0 r n =2514' 1.658 -0.4265 For the extraordinary ray sin / sin 45 ~ .. ro sin r= = 1 AQ , =0.4758 n 1.486 tan r =tan25 14 / =0.471=FC/BF 370 Solutions to H and R Physics II Since J5F= 1 cm, FC=0.471 cm tan r,=tan 2825'=0.541=FE/J3F FE-0.541 cm C=F-FC=0.541 cm-0.471 cm=0.07 cm Z)=C'/^2=0.05 cm=0.5 mm (&) Ray x is extraordinary whilst ray y is ordinary. (c) E in ray y lies in the plane of figure whilst E in ray x lies at right angles to the plane of the figure. ((/) If a polarizer is placed in the incident beam and rotated, for every rotation of 90 one or the other ray will be alternately extinguished. 46.7. Set the prism in the minimum deviation position for the ordinary ray n nd then for extra ordinary ray and if the angles of minimum deviation > an d D*> respectively for the two rays be determined then the refractive indices can be found out using the following formulae. n =sin \ C4 + D )/sin A rt=sin | (A+D*)/sin i A with ^=60. 46.8. Thickness of the quarter-wave plate, A S890xl(T 8 cm ) ~4 (1.61171.6049) =0.0021" cm = 0.0217 mm. 46.9. Let the two linearly polarized plane waves be given by E l (r, r)=E 0l cos (k 1 .r~ E 2 (r, /)=E 02 cos (k t .r--of Then the resultant field is E-'Ej+E, The intensity is where J?*=E.E U. f =( Taking the average we find where 7 l -< l >, /,= <,*> and 7, 2 -2 Polarization 371 the last being the inteference term. But since E t and E a are perpen- dicular to each other, the dot product E^E 2 vanishes and cons- equently /=/!+/, cos 2 Similarly < 2 2 >=-y rr^l r **0l , *-^02 Thus /== 2 ~ + 2 ~ which is independent of the phase difference. Hence, interference effects cannot be produced since the intensity remains constant as one moves from point to point in space. 46.10. Consider a parallel beam of circularly polarized light of cross-section A to fill up a box of length rf. If U is the energy and 01 the angular frequency then the angular momentum is by Textbook Eq. 46.5 given by j_j^z. CO But L*^^ =z~ d \ where we have written VdA Also, the distance travelled d=ct so that L ^ _y_^p_ <*>ctA <*>c since P = U/tA . 46.11. Rate of transfer of angular momentum, P P PA (100 watts)(5x 10~ 7 meter) ^ c IA . u . - 0/ . /^-w-r v fn - r~/ x - =2.65 x 10 u kg-m 2 /scc 2 - 2*)(3 X 1 0* meter/sec) If the angular momentum is transferred in time t sees, then where L' is the angular momentum acquired by the flat disk. /'^J MJR? is the rotational inertia about its axis and <' is^lhc angular fr equency of rotation. 372 Solutions to H and R Physics II j(1.0~ g kg)(2.5 X 1(T 8 metcr) 2 (2n X 1.0 rev/sec) /= , sec 7400 sec= 2.06 hrs. SUPPLEMENTARY PROBLEMS S.46.1. Let Ip be the original intensity of plane polarized light and 7 that of randomly polarised light. In the light transmitted through the polaroid sheet for the randomly polarized component /*' = */ -U) and for plane polarized component, /p'=/p.cos 2 8. ...(2) where 6 is the angle between the plane of polarization and the characteristic direction of the polaroid. The maximum transmission intensity is given by adding (1) and (2) and setting 6=0. /' C m., = /,'+// ==/ + $ I ...(3) and the minimum transmission intensity is given by setting 6=90. /'(m in) =//+/;==0+i I ...(4) Dividing (3) by (4) I whence 7j>=2/ Thus, the relative intensity of the polarized component in the incident light is -A- 2/Q __ 2 Ip+I ~ 2I +I - 3 and that of randomly polarized light is _JL_ = _A_ ___ L Ip+I. 2/ +/ ~ 3 S 46.2. (a) Let the light pass through a stack of polaroid sheets such that each sheet causes the plane of polarization to rotate through an angle less than 90 but the total angle of rotation addes up to 90. (b) Transmitted intensity through the first polaroid sheet /i/. cos 1 Palarization 373 where 1 is the original intensity. Similarly, the transmitted inten- sity through the second sheet will be /a^/i cos 2 0=7, cos 4 6 If N sheets of polaroid are used, then where we have assumed that angle of rotation through each sheet is the same. By Problem, T, .u a 90 Furthermore, 0~ ^-yrr radians As N is expected to be large, the cosine function can be approxi- mated by the power series retaining only the first two terms and then use the binomial expansion as follows: Solving for N 9 we find 46.3. The function of a polaroid is to convert unpolarised light into plane polarised light. The quarter wave plate introduces a path difference of A/4 or a phase difference of rc/2 between the two emergent waves, where A is the wavelength of the incident mono- chromatic light. Its function is to convert in general the incident plane-polarised light into elliptically-polarised light though the special cases of emergent plane-polarised light occur for 0=0 or 90 and circularly* polarised light for 0=45, where 900 is the angle between the electric vector in the incident plane polarised light and the optic axis. Upon reversing the direction of the light the role of the quarter-wave plate is to change the elliptically-pola- rised light to plane-polarised light. Let the side A have the polaroid and let the polaroid axis be oriented at 45 to the principal axes of the quarter-wave plate. The unpolarised light first enters the polaroid and the emergent plane- polarised light upon traversing through the quarter wave plate is changed to circularly-polarised light which after reflection from the coin returns and on traversing the quarter-wave plate is changed back to the plane-polarised light but now the polaroid is "crossed" and so the extinction of light occurs. Thus, the same combination of polaroid and the quarter-wave plate acts both as polariser and 374 Solutions to H and R Physics If analyser. On the other hand if this combination is placed with the quarter-wave plate away from the coin (with the face A against .the coin) then unpolarised light would first enter the quarter-wave plate and will be unaffected, and the resultant light upon traversing the Polaroid would not be extinguished. S.46.4. For the right circularly-polarised light the eletric vector E r6tates clockwise on a circle around the direction of propagation as we look toward the source. (a) Upon reflection the direction of propagation is reversed and the reflected beam becomes left circularly-polarized. (6) As the direction of light beam is reversed, the direction of linear momentum of light is also changed. (c) Angular momentum is a pseudo vector (axial vector) i.e. a vector given by the vector product of two polar vectors. Under mirror reflection, the direction of angular momentum of light does not change. As the incident light is 'right circu- larly-polarized, its spin points in the direction of propagation. Upon reflection the light has become left circularly*polarized and its direction of propagation has been reversed so that its spin still points in the original direction. In other words the direction of spin has not changed. (d) The impact of light beam on the mirror causes radiation pressure. 47 LIGHT AND QUANTUM PHYSICS 47.1. Planck's formula is ( with C l =2iec 2 /i and C 2 hclk The peak in the intensity distribution at the given temperature T is obtained from the condition C 2 1 This leads to Am = 7 <+ KT\ Jl I + -r-C2M ^ This is a transcendental equation in A m . Set C 2 /rA m = ^ then> f^h^ 5 The equation is satisfied to a good approximation with x=5. C , , Ac or * , (6.63 X 1CT 84 joule-secJO X 10 8 meter/sec) .(5X1.38 X 10- joule/K)(6000 K) 4.8 XlO" 7 meter==4800 A. 47.2. Area under the lower curve (Tungsten) is 2.35 cm 1 . Area under the upper curve (Cavity radiator) is about 9 cm 2 /. Emissivity of tungsten=2.35/9==0.26 47.3. r=6000K Area of the hole= /4 (0.01) 1 cm*=7.85 X 1(T 8 cm 1 376 Solutions to ti and R Physics tt Mean A=" w ^" -=5505 A=5.505x lQ- cm <A=iOA=10-cm. Let N photons/sec energy through the hole. Power radiated through the hole^Mzv or N= N hc _ 6.63 X 10" M joules-sec)(3 X 10 9 meter/sec) W ' AJtr (5.505 X 10~ 7 m)(1.38x lO" 38 joule/K)(6000K) =4.36 (2n)(3 x 10 10 cm/sec)(10- 7 cra)(7.85x 10~ 6 cm) N (5.505 x =2.1 X 10". 47.4. r=4000K Area, ^=n(0.0025)-1.94xlO- 8 meter 2 Mean A=' 4+0 ' 7 ^m=0.55 Mm=5.5xlO- 7 meter JA=0.7-0.4=0.3 x 10~ cm=3 x 10" 5 cm=3 x 10~ 7 meter (a) Energy/sec escaping from the hole in the visible region No hc _ (6 ' 63 X 10 ~ M Jo^es-secXS X 10 meter/sec) INOW ' (5.5 X 10-' meter)(1.38 X 10' 2 ' joule/KH4000K) 6.55 (2)(3 X 10* meter/sec) 8 (6.63 X 1 Q- j-s)(3 x 10~ 7 m)(l .94 x 10' m 1 ) " (5.5 X 10"' rae1er)(e-w-l) 63 joules/sec =63 watts Light and Quantum Physics 377 (6) Total cavity radiation, /? c =ar 6 where a is Stefen-Boltzmann constant, a=5.67 X10~ 8 watt/(meter 2 )(K 4 ) Total cavitation radiation escaping through the hole 1(T 8 ) X (4000)* x (1.94 X 10~ 6 )=285 watts. .". Fraction of eneigy in the visible region = R ^- =0.22 285 1 47.5. Planck's law is, /?* = -;* If A is small or T is small then the exponential term in the paran- thesis is much larger compared to unity. A i* ^^(Wien's law > 47.6. Solar radiation falls on earth at the rate of 2.0 cal/cm 2 -min, or 2.0 X 2.613 X80 1 ' ev/cm a -min. Energy of each photon of A=5500A is i7 L he =Av = -r- A =(4.14 x 10' 18 ev-sec)(3 x 10 10 cm/sec)/(5.5 X 10~ 6 cm) =2.26 ev Therefore, number of photons incident/cm 2 -min 2.0X2.613X10 1 " ev/cm 2 -min _. mlt = * ir\ r\f \ - ft.Jl X 1U (2.26 ev) #i n i?_j. _ hc ^ 4 -l 4 x 10~ u ev-sec)(3 X 10 to cm/sec 47.7. 7 -Av-= - -- _(4.14xlQ- ev-sec)(3 x 10 cm/sec) . . v , A _, _ -(2icm) =5.9X10 ev. 1.826 ev Sodium cannot show photoelectric effect as the photon energy is less than 2.3 ev, the energy required to remove an electron from sodium. 378 Solutions to H and R Physics II 47.9. (a) Energy of photon =- - A (4.14X10-" ev-sec)(3xl0 10 cm/sec) _ r (2000X10-* cm) ~" 2l Energy of fast photoelectrons=Av W=6.2 4.2^2.0 ev. (6) Energy of slowest photo electrons is zero since originally electrons in the metals have Fermi-distribution and for these elec- trons which are sitting down the well, greater energy is required. Therefore, the photo-electrons have an energy spectrum, the mini- mum energy being zero. (c) Stopping potential is 2 volts. (d) Cut-off wavelength corresponding to threshold energy ' of 4.2 ev is /^ (4.14 X 1(T 16 ev-sec)(3 X IQ^cni/scc) E 4.2 ev si3Xl(r 5 cm=3000A 47.10. Photo-electric effect equation is faE^V Q with 0=23 ev for Lithium. For a given value of K , the frequency of the incident radiation is calculated E +K if* f\ tTl ^ t ~ """ " V *T* L~ n nnlrknln irom v . ine calcula- ted values of v are tabulated. Fig. 47.10 shows the plot of stopping potential versus the frequency of the incident radiation. y. Av v (Cps) 2.3 ev 5.5X10" 0.2 2.5 ev 6.0X10" 0.4 2.7 ev 6.5X10" 0.7 3.0 ev 7.2X10" 1.0 3.3 ev 8.0X10" 1.5 3.8 ev 9.2X10" Light and Quantum Physics 379 1-0 0-5 10 *(X10 U CPS) Fig. 47.10 47.11. Intensity of light 8ource=10~ 6 watts. Binding ene/gy of electron=20 ev=20 X 1.6 X 10~ w joules 3.2 X10~ 18 joules Target area=n (10" 10 meter)=10- n meter* The area of a 5-meter sphere centered on the light source is 4n (5 meter)=10Qjj meter* If the light source radiates uniformly in all directions, the rate P at which energy falls on the target i given by watte/ meter*) meter*) 40~* 7 joules/sec. 380 Solutions to H and R Physics It Assuming that all this power is absorbed, the time required for the ejection of electron is 47.12. A=5890xl(T w m. (a) Let N photons be emitted/sec. hcN 100 watts=/iv#= ~- A xr 100 A 100 X 5890 XlO" 10 tn . a or " -^ 2 - 96 * 10 per sec At distance r cm, number of photons going through 1cm 8 is N N At distance r cm, density is - =10 =8863 cm=88.6 meter. 40nc V 40 X 3 X (Z>) As the density is inversely proportional to r z , the density at (ftft f\ \2 ~- ] =1.96 XlO 4 /cm 8 . 47.13. Suppose a photon of energy hv is completely absorbed by a free electron. Then the photo-electron must be ejected in the for- ward direction in order to conserve momentum. Conservation of energy gives _ "o 2 ^ 4 (1) Conservation of momentum gives Av ,,_ - c =P -(2) Using (2) in (1), Av + w c 2 = v J +w V ...(3) Squaring both sides and simplifying 2Avm c 2 =0 > which is absurd since and / for 6=90 Light and Quantum Physics 381 E (a) for A=3.0cm, hv A =<4.l4x ICT* ev = 4 -' X KT' ev. ' 3 cm E ~ l Percentage change in energy=(A/) 100=8x.lO~ ll xlOO =8xlO~ 9 % which is practically zero. ft>\ c i cnr^AO , he (4.14xlO- 15 ev-sec)(3xl0 10 cm/sec) (o) For A=5000A ; hv= T = ^nn ~~ =2.48 ev. Percentage change in energy =5 X 10~ 8 X 100=5 X 1 1~ 4 % (c) For A=,.OA ; k..^* = 12420ev. AE,, //. , 0.51X10* cv\ 1 _ o ~F " ' A ! + 12420 ev r~4L2 ~ Q Percentage change in energy =0.0242 X 100=2.4 % (<0 ForAv=l Mev ,0.51 /. Percentage change in energy=66%. For small photon energy (small compared to rest-mass of elec- tron i.e. 51 Mev) there is practically no loss of energy in the scattering process. But for high energies such as those associated with X-rays, energy loss may be significant. 4715 _.+,*, -, Textbook M7-.3) cos 8= A _ C os *, Textbook (47.15) V* A ~ 382 Solution? to H and R Physics- II *L!JLL * sin ^.Textbook (47.16) V 1vVt:* A Eliminating 6 by the squaring (47.15) and (47.16) and adding, m V _ / 1 I 2 cos * \ l-v/c 2 ~ V A" + A'~" AA' ) Cancelling the common term c in (47.13) and re- writing Square the last equation and from the resultant equation substract the previous equation 2/^cosj^ " AA' or W 2 c 2 =m c 2 - f(l-cos <f>)+2m hc (1/A-l/A') AA Cancelling m c* on both sides Multiply throughout by AA', and simplify, AA=A'-A= A. (i-cos #) f Textbook (47.17). First three longest wavelengths in the Balmer series are obtained by putting / =2 and fc=3, 4 and 5. 2tt a me 4 _, (9.1 1 X 1(T 2S gm)(4.8xlO~ 10 esu) 4 ^ i o ~" \fcT5 I ~/~ ~*^T V^THa* "^"^ =1 "J^fcO X l\l A 8 (6.625x10 * T erg-sec) 8 (o) v t =3.28x10" {^Jr-4iV4.556 XlO 1 *- \ 2* 3*/ A 1= = ~- = - 4 - 5 3 5 g-^QM =0.6585 X 10~ cm=6585 /I J \ v,=3.28 x iO 1 * I - 2 m , - ^ J-0.6150X 10". *. v 10 Light and Quantum Physics 383 ---- W.6888X10 15 3x 10 10 (6) ^=3.28X10^ -^-^ W0.82X10 1 ' v 3X10 10 __ 00 ~ 0.82 x 10 1 * -3.458XIO *cm= Balmer series lies between A x and A^ i.e. 6585A and 3658A 47.17. (a) n=l (K ~ /|2 ^^ r== 4 2 W p2 w here e is in e..v.u., w in gms and /; in ergs-sec ._ (6.63xlO-") 3 Q 1 v io~2 v r4 v in~ii ^.->3x ju cm (c) L= 2~ =6.63 X lO-*/2is= 1.1 X 10'" erg-sec n = mv - * e * 0.511X10* 1 V /|r *~' . ^ " |J ** s^: " --- - _- ----- - - __ h c c tc 137 c -3730ev/r. (e) = v __^ r h 3 ' " (6. 63xiO~" erg-sec) 8 =4.1xl0 18 radians/sec. (2)(4.8 x lO-'- - (67 , 3x 1Q - 27 erg . secl =2.18 XW cm/sec. (9.llxlO- 28 gm) s (4.8xlO- in " r z /* "" (6. 63 X 10"" erg-sec) 4 = 8.17xlO- a dynes. 16n 4 me 8. 17 X 10'' dyne . .... _, |= - = - - - =Q v 1ftl rm *r^ /i 9.lxlO- g m c 05llXlO/ 1 \ ev=13.6ev. , ^ ,, 2n t me * , , , *') = ----- - =-i3.6cv. 384 Solutions to H and R Physics II Aliterf <&) r- - nme* - (6.63 X IP-*" joule-sec) 2 (8.K5 X 10~* 2 couP/nt-m 2 ) n(9.1 X 10" 81 kg)(1.6x ICr 1 ' coul)* = 5.3xlO~ u meter. ,^ T h 6.63xlO~ 84 joule-sec f A . IA - M . , (c) L==r = ~ =1.06x10 * 4 joule-sec. _ /-^ V 4 *e i 2 mr _ f(l .6 XJO' 1 ' coul) 2 (9 X 10 9 nt-_m a /cpui a ) V ~'~(9J x 10~ 31 "kg)(5.3"x 10" 11 meter)" =2.19 X 10 6 meter/sec. (d) /?=mv=(9.1xlO"" 31 kg)(2 19X10 6 meter/sec) =2 x 10~ 24 kg-meter/sec. f . v 2.1 9 XlO e meter/sec A t , .^e ,. (e) =-= ^ 3 _ I ^ i _L._ =4.1 X 10>- rad,ans/sec. (5.3 x ID'" meter)* coul) 2 . , n _ 8 =8-2x10 nt. (i) 8 =J (9x 10' nt-m 2 /coul*)(1.6x 10~ M coul)/(5.3 x lO" 1 * m) =2.17 X10~ 18 joules =(2.17 x 10~ 18 joule)/(l .6 x 10' 19 joule/ev)= 13.6 ev. --2K=-272ev. 47.18. (*)ran s (c)Locn(<f)/oc /I Light and Quantum Physics 385 (A) c*~f (0 47.19. As the energy required to remove the electron is oc (l//t f ) and since ionization potential (i.e. energy required to remove the electron from the ground state of hydrogen atom with = i) is 13.6 ev, energy required to remove the electron from the /i=8 state is 13.6 " 8* or 0.21 ev. JIM / 47.20. /b- =(3.28 X 10" sec^)(4.14X 10~ 15 ev sec) f ^-JJT (a) For 7=1 and k =4 .!__ _ 1 = 17 7S ev |S "" Jp ) l * ti:> ev * (b) For transition from /* =--4 to fl3, =0.66 ev. For transition from /i=4 to n=2, =2.55 ev. _ 08S v . For transition from n=4 to/i=l, = 12.75 ev. -t.5i - For transition from n=3 to For transition from /i=3 /o n=i f =12.09ev. For transition from /i==2 to n=l f =10.2ev. -13-6 Fig. 47.20 (c) Transition energy, = 12.75 ev. This energy must be shared between the emitted photon of energy Av an<i recoiling hydrogen atom, a+/,v= 12.75 ev -U) Momentum conservation gives i.e. </2~E*MH Av c 386 So/i tfo/w toHandR Physics II or kv*=*<f2E R M H <? Using the value of Av from (2) in ( 1) M<* =12.75 ...(2) 1 2.75 or 25.5 25.5 25.5 ev <* T c ~~M H <* 940X10' ev PH-2P-2.7X 10-=0 where p=v ff /c Solution of the quadratic equation gives, p=1.35xHr 8 VH=pc=(1.35x 10-")(3x 10 10 cm/sec)=405 cm/s. 47.21. (a) Av= 0.85 ( 3.4)=2.55 ev 0=00 n=A -U-6eV Fig. 47.21 With Et - ,= 10.2=13.6 ( ^5-71 or fc=./iM=2 3.4 With ,= -0.85 -(-13.6)= 13.6 1 -p,-^r we find /=4. 47.22. A= 1216 A he (4. 1 4 X 1 Q-" ev)(3 X 10 10 cm/sec) ._ , *-7- (12l6xlO~"cm> ),21 C v Light and Quantum Physics 387 10 .21 = 13.6 -Ja-fcif -3-4 ev Assigning different integral values for j and k 9 we find that the above equation is satisfied n =^ fory==l and fc 2. The energy level diagram is shown in Fig. Fig 47.23 47.23. 47.23. Since the excitation energy for the state /i~2 is 10.2 ev, the kinetic energy of neutron (6.0 ev) falls short of it. Therefore, the collision can be only elastic one. The initial energy of 6.0 ev will be shared between the scattered neutron and the recoiling hydrogen atom. 47.24. For singly ionized helium, the nuclear charge will be +2e. The expression for photon energy will be multiplied by a factor of 2 2 i.e. 4. The corresponding spectrum (apart from correcting for reduced mass) will be pushed up on the frequency scale by a factor of 4 compared to hydrogen spectrum (or A shortened by a factor of 4). 47.25. /iv~ (4)(13.6 ev) = 54.4 ev. 47.26. In Bohr's formulae m should be replaced by the reduced mass ft. Ill _ m " = + ' r >*~ (a) Compared to hydrogen spectrum, it is seen that the frequencies in positronium spectrum are shrunk by a factor of 2, (or A increased by a factor of 2). '" 2 " - where rj/= ground state radius of hydrogen atom=0.53 A. /. r<p*>2x 0.5 3=1.06 A. 47.27. (a) where n is the reduced mass of p and proton system. . _ _ /u 1836m~~267ra where m is the mass of electron. 388 Solutions to H and R Physics II h* __ra 0.53x10-' ***** = 186x4**me* " F86 = - 186~ cm =0.285 X10~ W cm. (b) As the reduced mass is 186 times greater, the ionization energy is enhanced by the same factor; i.e. ionization energy for muonic atom with proton as nucleus is (13.6 ev)(186) or 2530 ev, i.e. 2.53 kev. (c) Maximum transition energy is provided for transition between n 2 and n 1. For muonic atom, ~ _=1897 ev 1897 4.14X10~ 16 (3 x 10 10 cm/sec)(4.14x 10~ 16 _ (1897 ev) =6.5xtO- 8 cm. a wavelength which falls in the X-ray region. 48.28. Equating the electrical force to centripetal force, ~=mo> 2 r .. (1) Quantization of angular momentum gives Squaring (1) and re-arranging, m*<*>*r*=e* ...(3) Raising both the sides of (2) to power 3, Divide (3) by (4) to get The frequency of radiation is given by _ it V/ f * for transitions between the states k and/ Setting y=-r;j and * w-H Light and Quantum Physics 389 i If n is large then in the parenthesis of both numerator and deno- minator of equation (6), 1 can be neglected so that 2n 2 Next lettingy=/i and k=n+2 ri_ ' 1 "1 L a (+2j 8 J~ = v For large values of n, (H+l) We may repeat the calculation for transition between states with quantum numbers /t+3 and n to find v==3v . Thus, in the limit of large quantum numbers Bohr's theory predicts v=v , 2v , 3v etc. 47.29. For fc=, j~n 1, frequency of emitted radiation is given by _2n^m 4 f 2/1-1 "j Vi 3 L (-D a /i J J The orbital frequency v is given by '. Percentage difference is 100 I - J. 2 ^^ ___ -2-1 -l)*' n 8 J 100 (3/1-2) 2/1-1 For large n, 3/i 2 > 3/i and 2/il > 2 390 Solutions to H and R Physics 11 .'. Percentage difference=100X - n (2n) n 47.30. L=mvr = Mass of earth, m=6 X 10 14 kg. Mean orbital speed of earth, v=29770 meters/sec. Planck's constant h =6.63 X 10" 84 joules-sec. Orbital radius, r 1. 5 X 10 U meters. _2nX6xlOX2.977xl0 4 Xl.SXlO u 10=2.5X10 74 Such a quantization can not be detected. 48 WAVES AND PARTICLES 48.1. (a) deBroglie wavelength, A= == p m\ A=(6.6x 10~ 34 joules-sec)/(0.04 kg X 1000 meter/sec) -1.65X10~' 5 meter (b) Diffraction effects are noticeable for obstacles which have dimensions of the order of wavelength. But, here X is too small. Thus, for obstacles of the size of atoms (/?^10~ 10 meters) the diffraction angle ^ = Tjpier" 6 ! rS = 10" 25 radians an angle which is beyond detection. 48.2. deBroglie wavelength, A= = T"^^ P V 2rrinK (6.6 X10~ 84 joule-sec) *= lL x (1 .67 X 10~ 27 kg)(0.025ev x 1 .6 X 10~' joule/ev)" = 1.84xlO" l metcr=1.84 A. Mt% ^ , . * t r i * h 6.6 X10~ 84 joule-sec 48.3. (a) Momentum of electron, p= -y= 2 X 10^ 10 meter =-3.3X 10" 24 kg-meter/sec Momentum of photon, /?= =y=3.3x 10" 14 kg-meter/sec. n2 (b) For electron, kinetic energy, K- Ln\e ^=(3.3 x 10'" kg-meter/secW(2x9.1 x 10' 81 kg) =0.6 xKT 17 joules = (0.6X 10~ 17 joules)/(1.6 X 10"" joules/cv) =37.5 ev. For proton, energy, E=k*cp =(3x 10* meter/sec)(3.3 x 10"" kg-meter/sec) =9.9 XUT" joules =-(9.9 x 10'" jou!es)/(1.6x 10~ w j =6188 ev 3 02 Solutions to ff and R Physics 11 48.4. Kinetic energy of electron, JT=50 Bev = 5X 10 l0 ev ' So K > m,c f ; the rest mass energy of electron which is o 0.51 x 10 ev. Now, total energy, E^K+m^ is given 'by relativistic formula or K*+2Km<)<*=c*p* If K >> m c s 9 the second term on the left-hand side will be much smaller than the first term, ^ K E or p=* = r c c E Momentum, />= = (5x 10 10 evx 1.6 X 10~ w joule/ev)/(3.0 X 10 s meter/sec) =2.7 xl(T 17 kg-metcr/sec. deBroglie wavelength, A= =6.6X 10" 84 joules)/(2.7 X 1(T 17 kg-meter/sec) -2.45X10" 17 meter, On the basis of constant density nuclear model, nuclear radius, /?=roX l "=1.2x Hr 15 ^ 1 /* meter, where 4= mass number. For a typical medium size nucleus with X = 125, wh get =6.0 xHT 1 * meter. Thus A is comparable with R and diffraction effects will be prominent. 48.5. (a) deBroglie wavelength of electrons of kinetic energy 7==54cv==(54ev)x(1.6xl<T lf joules/ev)=86.4xl<r u joules is h 6.6 X10"" M joule-sec .1 X 10-" kg)(86.4x 10~ l * joules) =1.65 XHT 10 meter Now, condition for Bragg-rcflectiqn ia Set m=2 (second order diffraction) Waves and Particles 393 2A l^SxlO" 16 meter tt,^ then, which is impossible since the value of sin can not exceed unity. Hence, second order diffraction cannot occur. Similarly, third order diffraction also cannot occur with the given accelerating voltage (which defines A) and the set of planes (which define d). (b) Set F=60 volts Then, kinetic energy of electrons, AT=60 ev=(60 ev)(1.6x 1(T joule/ev)=9.6x KT 18 joule. Momentun,/>= Vim^VUHP.l xiO'* 1 kg)(9.6x 10~ 18 joules) 4.18X 10~** kf meter/ice. A= =(6.6 X 10~ joule-ec)/(4. 1 8 X lO'" kg-meter/sec) - 1. 59 XlO' 10 meter For first order diffraction, \=>2d sinO or 8in= =(1 .59 X 10~ meter)/(2 x 0.91 x 10~ 10 la =0.874 /. Bragg angle, 8=61. But, 6=90-tf whence ^=180-2e=180-2x61=58 . 394 Solutions to Hand R Physics H 48.6. (a) 10A versus K for Electrons 12-3 A x where K is in eV 10 15 K *.(eV) Fig 48.6 (a) 20 25 eV (XiO 13 cm) 4 30 20 10 A versus AT for Protons . 267XlQ"' 13 cm ~~ where K is in MeV 20 30 K i ig. 48.6 (b) Waves and Particles $9$ Mean kinetic energy of hydrogen atom at temperature T vin) is 3 = ~- kT, where /c=Boltzmann's constant JL constant = 1 .38 X J 0~ 3 joule/K r=273+20=293K =(- Vl.38 x lO-2 joule/K)(293K)=6.0x 10-" joules H = 1.67X10-" kg , = _ h _ __ _____ 6.63 X IP"'* joule-sec w V(2K6 x lfr~ joule)(l .67 x 10~ 27 kg) = 1.48 X10- 10 meter =1.48 A. 48.8. (a) Av- 3 -: 2 =3.4-1.5=1.9ev. (6) Energy of photon, =1.9 ev=(1.9 ev)(1.6X 10~" joules/ev) =3 xlO" 1 ' joules - ~ E (6.63 X 10~ 4 joalc-sec)(3x 10* meter/sec) =6.63xl(T 7 meter=6630 A. 48.9. The normalized, time independent wave function for the parti- cle trapped in an infinitely deep potential well of width L is , 2 . nnx = /T sin -r- Probability of finding the panicle between x and x+dx in state /t, is P(x)dx= | # dfc- ~ sin* 55? dx LI LI Probability of finding the particle between x=L/3 and x<=Q, is LI3 L/3 L/3 I 2 f ri-(cog2nx/L)-K. o 396 Solutions to If ana R Physics It LI3 L/3 1.1 r * -- sin L 2 /m L/3 J or (o) (c) 1 1 . 2/m T ~2n* Sm "T " = T-=- sn - 15 ^ - 3 2Jt J )=-.--*" sin.^- 3 4n 3 )= -r --- sin 2n=0. J OJt - X0.866=0.4 (J) Classically, probability for the particle between x a ? is dx/L. L/3 p 4810. Time independent normalized wave function for 1 1 atom in the ground state is ^0= J"~~ 3 e~ r l a - r r gen _i^ r ~a* J r*e dr- Set 2r Integrate by parts, 2 1 f xe~*dx\ --+ }