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 noncalcului
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 hoe 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*ntm/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 couiVDtm a )(0 ! 01_kg)(98m/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 ntm 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 xcomponent of force is given by
n* nt
(H) =<
4ne. V2 a 2
O' 7 coul) 2 (9xlO ntm 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' ntm/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 ntm 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,
^"91.6x10" 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* ntrnVcoui'i
^" 4K Gc J a ~~ """" (2.0 meter) 1 " l^nt
Simplifying,
9x iO f ^ 45 X 10 4 <?+40
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 xcom $ 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 xaxis is T: ,
=F cos 45=
The xcomponent 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 +16X 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 ?*r3.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 tfivei by
n^&N^CM) (J^xlO*)^. 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 aparticle and
Thorium nucleus and r is the distance of separation.
F (2) (9Q)(1.6xlO" 19 coul)(9x 10 ntm a /coul*_)
" "(9x i'J" ia meter) 3
= 512 nt
(ft) Mass of aparticle 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. (1x?
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 ntmVcoul 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 workenergy 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 10ntm 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,
..na
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 ntm 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 ntmVcoul 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 <?,= 48.5 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 ntm 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 Wfx 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 xcomponcnt 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 jaxis 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 xcomponent 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 xcomponent 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  (aHx 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 xcomponent (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. (af 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 zaxis. The transverse
component will lie in the direction OA. Hence, its projection on
the zaxis averaged over the azimuth angle .gives the zcomponent. 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 xcomponent 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)
=8.92xlOnt
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 aparticle F=mg=(6.68x 10""*' kg) (9.8 meter/sec 1 )
=6.55xiOnt '
Charge of aparticle, 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 iQn 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 cmuan ^ ) 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 nonuniform 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 nim 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 fr/ 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 06
\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 semicircle. 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 yaxes 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 xcomponents + 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
rffAAf ...(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 ydirections.
The resultant xcomponent 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 ycomponent,
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,(/Ma)
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*ntm 2 /coul
8.9xl(T 12 coul 2 /ntm 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* ntm 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*/ntm 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 ntmVcoul 2 ) (2 x 10~ 7 coul)
E
(c) E
2 (0.25 meter) 2
=2.9xi0 4 nt/coul*
q ^ (9 X 10 9 ntm 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 /ntm 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 ntm'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 jj
(//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 inHcouP'ntmsd > 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 fnin sin eriitir 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 tir .: 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
aTc/^K(6.9x 1(T 13 cm) 2  1.5x KT'cnr
S (1.76x 10 18 ) U5 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 .tcomponent
of the field as E y  f;.= 0.
Influx, 6,n ^ fxj7) <J a
Outflux, >.,< . ab v/2a  S2hu*
Net outward flux. 6 ^u/ 6 aM '
\ 2
Gauss's Law 53
l) (800 nt/coulm*) (0.1 meter)*' 2
1.05 ntmeter a /coul
(8:9 X 10" ls coul 2 /ntm a ) (1.05 ntm 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 /ntm 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 + Q2KAa*
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*
=56xlOnt/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.JT8ne.*
(6) Yes. Actually, F=0 at oo.
29.3. The potential is given by
q (l(r coul) (9 X 10 DtmVcoul 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)
JC3x
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
2xl=0
lf3
= 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 coulmeter.
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 Vfl/
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* ntmVcoul 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
= 13.6ev27.1 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 ntm 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* couiVntm 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 coulVntm 1 )
== 1.8x 10"* ntmVcoul.
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 ntm 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 rewritten 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 690 or 270i
29.21. Field on the surface of the sphere is
rJL _ (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 :35xlOT7ne^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 iOntm 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 ntmVcoul 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'ntmVcoul*) ( ** , + VjyLJS. )
\ 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 ofcharge 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 aparticle just touches the surface of gold
nucleus, the original kinetic energy is completely transformed into
electric potential .energy. ]
4716. r
Ct 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 aparticles 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 ocparticles 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 ntm 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 ntm 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 aparticle 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 aparticle.
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 couWntm 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* ntmVcoul 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 fJJ + J\
47ie 6 4 ice a 4 T5 \ ^ a )
,n tnft . </ n\ /" 5xlO"" coul 2xlO"" 6 coul \
=(9xlO* ntm 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 midpoint 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.
Rewriting (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
=JM 2
<2)(9 X 10 9 ntmVcoul^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 headron 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
* yOHD
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/)(8xlOf).
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!if10 ^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=(2xlOf)(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 ntm*/coul 2
=711XlOf=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]!
~~ joulmeter "" ntmeter^
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 parallelplate
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 parallelplate 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
AKH100T21)=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 IMMhi
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 fiff 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 ^f2.08 /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 . ,
3x0  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 parallelplate 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 parallelplate 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 /ntm 2 ) (100 X 10"'' meter')
(7) (l.Ox 10~ a meter)(0.5x 10' 2 meter)(7l)
= 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 rr A 2 *o A
Cl c *~R^r7=&~
The combination of these two capacitors in series has the capaci
tance
30.21. For a parallelplate 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"" coulVntm)(lQ a meter 8 )
____
=8.9 M"
Thee capacitance with the slab in place is given by
_ v , ._
8.9xlO
Kdb(K\)
__ (7)(8.9XlO"coulVnt.m 8 KlQ meter')
(7)(10> raeter)(0.5x 10~ a meter)(7l)
 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 /ntm 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 /ntm 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
=(5xlOcoul) 15
=4.lXlOcoul
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 coulVntm*)(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 parallelplate 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"c6uiVntm 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'/ntm^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 parallelplate 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 airgap 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/ntm)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/=CFK100X10 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(2xlOf ) (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'/ntm*) (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
CC (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 Whr, 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 ^^,=(10079) 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 =00 9 =(7.3x 104x KT) coul
= 3.3 XlO~ coul
lh \ v i _.3.3xlQcoul_ 33 u
(6) K I IoxTof  33volt
v _ <? 3.3 X1Q* coul
Vz ~C^ 5xlOf 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= iC.
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. S30.4
/ / /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 crosssection.
.'. r ( iron ) / _P (iron)~_ /I
r (copper) V p (copper) V 1 .
x m> M 2 4
.7 x I0t~ ohmm
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" ohmm)( 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~ ohmm
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.Wii ... (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
jibei 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 ohmm)(10 meter)
^J Ml^xTo^meterT 2 '" ~ 5 ' 4X 10 hm '
Resistance of iron wire is
ff _Pi / _(.0x 10^ 7 ohromHlO meter) . A _,
^"  ~ ~~" s3I  8x0 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 855 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 ohmmeter
The materials is nickel (Textbook Table 31.1).
(a) Resistance, *=4 ,= (6  8x 1(r> ohmmeterKl.Ox 10 meter)
wr a *( 1 .0 X 10"" 1 meter)'
= 2.17 Xl(T 7 ohm
31.10. Crosssectional 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 ohmin 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 ohmm  rt1 ^ f _
 SS5  91 x 10
For carbon at 20C
p '=3.5xlO ohmm and <x' = ~5x 1(T 4 .
The resistivity of carbon at 0C is
9' 3.5X10' ohmm _ 354x  1(ri ohm ~
p < ~ ''^ 120X5X1CT*  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 ohmm)(5 X KT 8 ) __ ^ 9 fi , a
(3.54 xlO'ohmmX 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 jouleheating 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' ohmm)
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
Crosssection area, A=ttr z 'n (5.1 x 10~* meter)*
= 8.17 X10~ 7 meter 3
Resistivity for copper, p=1.7x 10~ 8 ohmm.
n ?i (l.7xlO~ 8 ohmm)(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 jouleheating, P=~r = ^'^ ' ^t 6 watts *
/( 0.63 ohm
31.19. Radius of wire r^O.05 in1.27x 10~ 8 meter.
Crosssection 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 ohmm)(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) (100C20 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, RwR* (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 vxiAift ~=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 aparticles 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 aparticles,
v= f^ = f (2) (20X 10" eV)(1.6x
A/ m V
6.68X10"* 7 kg
=^3.1X10 7 meter/sec.
Number of aparticles crossing a given plane area per second,
___J_^ 0.25 XlO" amp
q (2)(l.6xlOcoul)
=7.8 XlO 11 aparticls/sec.
Number of aparticles per meter length of the beam
Hence number of aparticles per 0.2 meter length of the beam
=(2.5 X 10 4 per meter)(0.2 meter)=5tXX) aparticles.
(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
apatricle z=>2, hence, the potential required to accelerate
clcs to 20 MeV is
20xlOeV 1nf .
V= ~ = jo volts,
Current and Resistance { 1 5
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1 16 Solutions to H and R Physics 11
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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 volumeelement is
...(1)
where use has been made of (1).
1
P/Vx
* [la+(ba)x]*
o
Set, y=la+(b a)x
dy(ba)dx
So that,
bt
at
=__ f d y = L(\ 
it(fttf) J y* x(ba)\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 ohmmeter.
/ (40 meter)
(20 ohm KO.l meter 2 ) Anc ,
=  7177   :  =0.05 ohmmeter.
(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 07
. . 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 kwhr energy produced is
1.25 kwhr=(1.25 kwhr) lOTS kcal
(1 kwhr)
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 =(273f 27)=300 K
that of surroundings, 2itr/=surface area of wire (neglecting the
area of crosssection 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 '' ohmm and =5x 10~/C
and solving for T, we find 7=670K
=(670273)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 jouleheating 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
20 1X10 
10
"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+'tftO ...(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 tt5 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 H4) volt ...(1)
Applying loop theorem to the entire circuit starting from c and
going in the counter clockwise direction
(42) 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 rearranging
. = _ 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 1010/
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^=30i30/ 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 14) 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=19amp.
' f
Fin 32.22
(a) The Joule heat produced in JRj is
P 1 =/i a /^==(jamp 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 jamp 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^'gf f^i,
= (1.263160.15789) 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*_(0.45450.4S04) 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 + 4020^ 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 + rR (40 + 20 r47,6) 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__Kr__(2.272.21)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,
EitiRii 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, counterclockwise in the left loop
/ 2 =0.33 amp, up in the center branch
/ 3 =0.33 amp, counterclockwise 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 _ 1001 . 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 (ltT' //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 firstbattery 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
502 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
' Rjr/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 (7V1) CRr)=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. oioo
Taking logarithms on both sides,
^ in 1004.6
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 r7T^==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 (iocoui) 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^KooOe'/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 ccparticle, 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
= 00 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* gaussj524 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=9030=60 .
/. T=(20)(0.1 ampXO.l X0.05 meter 2 )(0.5 weber/meter 2 ) sin 60
=4,3xl(T 3 ntmeter
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/meterj
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,
51.5 webers/meter 1
For copper, n=8.4x lO^/meter*
p= 1.7x10"* ohmmeter
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 ohmmeter)
=0.0066.
33.17. (a) Kinetic energy of proton, Kv=eV
Kinetic energy of deutron, K* =eV
Kinetic energy of aparticle, 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 ydirection perpendicular to the plates. The
positive plate is along the ;*>axis and magnetic field is applied
along the zaxis 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 zaxis, 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 xaxis. The maximum
displacement of electron along the yaxis 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 breakup is
p* (4.8 X KT 1 * kgm/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 kgmeter/sec
Radius of the new orbit is
, p'_ _ 2 A X IP"" 19 kgmcter/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 nonzero component of velocity albng B viz. v 11 makes the
plane of the circular orbit of the pdsitroq path advance along the
J3axis. 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 X10M joule) _ 25 , X10 .
(1.6xlOj^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 counterclockwise 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 rewritten as
...(2)
__ Ar v='y
Writing v=<o r, (2) upon reorganization 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 ) iti"* _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 lefthand 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
_ (041 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 *~xxtA4 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 1.0xlOvolt ,_ .
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, /'= 
: 7v=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)(iOvoJt7meter)
=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/ampm)(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/ampm)( 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/ampm)(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)
dgW'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. (aLByAmjjere'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/ampmctcr) ( 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/ampmeter)(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 HTH1 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. 3425 (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 xcomponents 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 weberamp/meter(20 amp)(l//2)
(2n)(0.2 meter)/ V^2
=8 x 10"" 5 weber/meter 1 , along yaxis.
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 weberamp/m)(20
(2)(0.2 meter) * 45 V /
=4xl(r 4 nt. /
The current in wire 2 and 3 being / ^4
antiparallel, the force of 3 on 2 will '
be repulsive and acts towards left.
Fig. 34.12
weberamp/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 weberamp/m)(20 amp) 2
(2Tc)(\/2)(0.2 meter)
The xcomponent 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 antiparallel 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 weberamp/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
2cos6 = 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 ffl 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 BiotSav'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 semicircle 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
semicircle 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 tan5
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 midway 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
025
B
50
+ 50 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(zx)(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/ampm)(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/ampm)(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/ampm)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) iv^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
^%Hr  (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/ampm)(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 ampmeter 1
T=(1.57x 10" 1 ampmeter > )(9.4x 10" 1 wcbcr/mctcr 1 )
1.48x 10~ e Dtmcter.
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
counterclockwise 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 semicircles respectively. Now, the current i divides itself
equally in the two semicircular paths. By Problem 34.18, the con
tribution to the induction from the upper semicircle 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 semicircle
viz.,
But since the current is running in the counterclockwise 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 BiotSavart 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 semicircle of radius R l is, ff 1 ==/o741Z 1 . (current being
clockwise) and that due to outer semicircle of radius /? a is
^i^fj?' (current being counterclockwise)
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 crosssection.
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,
Crosssection area of the solenoid,
 (0.03 meter) 8
Field B in the solenoid .
=:(4Tr x 10~ 7 weber/ampm)(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 crosssection 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 crosssection 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 ohmmeter)(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 crosssection 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 flP'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 crosssection of wire, a=nr 2 =n(5.08x !0~ 4 meter)*
=81xl0~ 8 meter*.
B e u D P 7 (1.7X10 8 ohmm)(0.5 meter)
Resistance of the wire, *o=  =  CSlX lO's meter 2 ) 
= 1.05x102 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 counterclockwise, 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 turnmeter 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 (ba)
' 2nr
b
<* \
Faraday's Law 197
_(4rcX 10~ 7 weber/ampmXIOO 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
r2
=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 counterclockwise.
(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, yJL ... (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 nonuniform 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) gausscm
=6.84 X 10 5 gausscm
 A 6. 84 XlO 5 gausscm Ol _
B= = 5r 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. [emfjfclectric 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~vot.
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/xn * 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 semicircular 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 selfinductances 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 = selfinduced emf

~ l dt
emf in coil 2 = L ,
~ (it
Net emf = (^ +L 2 )
From its definition, the equivalent selfinductance is
L^L,+L 2 (I)
(ft) Separation should be large so that mutualinductance 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
da
.
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 crosssection.
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
crosssection.
n= _ 10 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/ampm)(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==
hK5XlO3
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 crosssection 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/ampm)(618> 3 (0.02 meter) . J^2
L   __ , n 1Q
=0.28X10 'Mi
=0.28 mh
(6) Perimeter of each turn for the square crosssection 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= ' = ^Plil^^.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=lnlOOO69
210 Solutions to H and R Physics //
36.13. (a. For an LR 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 50xlQh
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 ohmXie" 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/ampmeterKlOO 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/ampm)(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/ampm)
%
=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)
rMo/
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/ampm)(10 amp)*
2.5 X 10~ joules/meter.
(b) For the coaxial 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/ampm)(10 amp) 1 . 4.0
UB =  4jj in JQ
= 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 coaxial 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(i6Vc)J
10~ T weber/ampm)(10 amp)'
4
/ la (5/4) 3(16/2S) \
V (l16/25) 4(116/25) /
=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/ainpm)(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 couWntm 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/ampm)
= 1. 5 XlO volt/ meter.
SUPPLEMENTARY PROBLEMS
S.36.1. (a) By Problem 36.2, the equivalent selfinductance 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 samedhrcction 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 voltzero=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.
() FtJ^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,
Ei^itR^O ...(2)
Using (1) in (2),
. . E 100 volt
(6) At f=oo, / 3 will have a nonzero 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) iiO
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/ampm . b
2 V 8.83 x 10 coul/ntm 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) [coulombohmmetcr/wcber]
(W
(c) [c >ulombampere/farad]
,.. [kilogranvvoltmctcr 2 ]
[henryampcrej
(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
MM (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 arnpm 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* ntm 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/ampm)(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/ampm)(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 crosssection area of
the toroid.
.1 = *<f/4 ...(6)
With the diameter of the crosssectional 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
= (016 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 ampm 2 )=7.6 ampm 2
(b) The magnitude of the torque is given by
T=^J8 sin 0=(7.6 ampm')(1.5 weber)(sin 90)
= 11.4 ntmeter.
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 zdirection. 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 righthandscrew 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 ampm 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'ii 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/.vfl 1 s 1 + L
LT 0 ti^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 crosssection.
1 T 1.0 meter . M ^1
" (4KXlO*weber/ampmML 5000 rv "' "~""J
The flux,
8121 , . ,
= j amp/ weber ..(I)
)/l = 1.8^ weber ...(2)
Magnetomotive force (mtnf)Ni ampereturn ...(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)(2xlOf)
= 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/(secdegree) 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 rearranging (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
_.__<//
VLL 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 timeconstant
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 108 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/
Rearranging 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
le
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
Rearranging 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,'Ll/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 KL1KCP + /P 2
Squaring and simplifying
or > V LC V3 w' CR1=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&AIiC, (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 parallelplate 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 crosssection 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 parallelplate 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 TT //=
JLJUk *u CA , AI^^*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/metersec,
Electromagnetic Oscillations 24?
~
By Problem, ^ = ^ f ^ L = x 1.4x 10 M volt/metersec
=7 X 10 U volt/metersec.
Set ^=0=2.5 X 10"" 1 meter
Also
B( 1=: dE  ( 2>Sx 10 ~ g meter X 7x 1Q1S volt/metersec)
^ 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.Dp
with ,ff==/x //, D=e
We shall verify that Maxwell's equations satisfy (r)and (r)
as given in the above table.
(a) VXBO; (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 2x ^ 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 selfconsistency 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 selfconsistency 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
vprc 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
_dstance _ 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 ,
"'odr^* ~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 (fcxo>r) (3)
^=5m sin (kxvt) (4)
Substituting (3) in (I) and (4) in (2)
tt/1= =4 EO E? sin 2 (fcxeof) (?)
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 cutoff 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  SJ
zdirection
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 wavenumber.
The average value of the Poynting vector is
where B* is the complex conjugate of B.
i J k
<Sa*=*YRe 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 zdirectioi
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/ampm)(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/ampra)
=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/ampm)! 1 /*
f 2&^ fteKl .4X10* joule/mK4n X 1(T 7 webcr/ampm)! 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/arapm)
 = 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 crossproduct 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,
PS 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. 367). 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 (3910) with respect to x,
Mo^o
3r
where use has been made of (3913).
In order to obtain the equation satisfied by B, differentiate
(3913) 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 (3910)
Mo'o
Since c* 
/*oo
Thus and B satisfy (3913)
Thus, and satisfy (3910).
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 (xfcf) 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 rootmeansquare electric field is
( r..)= ^ 6  14 x 10 ^ 2 volt/meter.
(W ^= r^ 29x 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) SEXB
_
/* 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. 402 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^ 185X 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=sunearth distance = 1. 49 X 10 n meters.
G=6.67xlO" u ntm 2 /kg 2 .
_ ^6.67^x IP" 11 ntm 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
tS 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
m6.0xl0 24 kg
r =: 1.49x10" meters
G6.67X10 11 ntm 2 /kg 2
(6.67 X 1Q" 11 ntm 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 crosssection 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  fg  ^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 earthJupiter 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 megacycles/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(lv/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)
45 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 yelloworange (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/*mpm)
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" ntmeterVkg 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 wavefronts.
However, in the present problem the wavefront 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 wavefront 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 wavefront is a common tangent to the secondary wavelets
given by the usual Huygen's construction. The envelope would be, a
cone of semiangle 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 A5500 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) 8/70.00001
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 90a=90~41=49
(6)
Fig. 41.8.
152
1.33
w 1.14
Set
41.9. is small, so thai a is also small. Hence v2ic 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'
Yellowgreen
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 4113
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
QJ 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, = iJ 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 ; p60
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 airliquid interface. Let the criti
cal angle be Qi. At the critical angle the ray suffers total internal
reflection at the liquidair 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 glassliquid 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 +
(xd)*+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+(dx)* ...(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(ln) < 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
05
15
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 airwater 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 waterair 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 airwater
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 rearrange 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
of
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 603.2x 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
nl
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, fDd\*
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 + Do & f
.(4)
(I)
...(2)
...(3)
Reflection and Refraction Spherical Waves 313
rearranging 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 /Jside.
r" is ve since C" is on Fside.
Solve for r' ; we fiind r'=4.5 cm.
r "==9.0 cm.
42<18 . 1+L1 ...(D
Set x=of 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
BCFMBCA+
3l4 Solutions to H and R Physics tl
Also BFC=EBFBCF
=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=FCHC=.
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 Kside and so r t is ve.
C 2 is on flside and so r, is +ve.
Vside
> .. 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.
^10x46x257x25(46 + 10)
(4625)(lO7)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 aydiagram 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 1fceiling (//) 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 a7.5 ~~ 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 ^ ^
#
< 225 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=(43.53.5) cm40 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/ (10.5) 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 fringewidth for doubleslit arrangement is given by
where d is the slit spacing, D is the slitsc'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 yaxis.
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 slitscreen distance, d is the slit spacing and A is the
wavelength of light.
By Problem, A0=0.20 ; A5890 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 1000.03.
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 . . 16m 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/H30)
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=(7.54.286)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 14*
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"* (6800xlOcm)"~ 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),
(mi
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)
_ _ _
mf 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)
(mf 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/nMmH) 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 05X6000 A) A
d = (2X133) * 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 orangered 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 halfwavelengths 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)(58965890)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 slitscreen dis
tance. The separation of the fringes due to wavelengths A t ^nd A 2
"
mZ> ,, ,. (3H100cm)(60004bOO)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 (1.71.4)
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 aycompared 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 ^4cos 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 il
==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 = (/HhJ) 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 slitwidth.
If xis the distance of the minima from the central maxima and
D is the slitscreen 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 6ae,>
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
10
05
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, A02e 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
, ..
145 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 ~dD
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
J200 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)*, m0, 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~T2
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, /fi0, 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 Besselfunction 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) km2610 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 cm0.13 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 990
rf=^crn2.5xlOcm
For A4000A
on
8/A =
(2.5xlO 4 cm)(sin90)
(4000X10"* cm)
For A7000 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 threeslit 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 0fsm 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 07/80
_.
2V2
0r=56 f =0.977 radians.
Width at half maximum is given by
Ae  20  A *  977 A 
Ae " 2t) ~ ^^ 3.14 J ""
(t) For twoslit 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 (H2 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 m5. 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 . A6350 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, </= ~^cm3.33x I(T 4 cm
d sin =mA
sin 6=m7
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, 6,40^76 ...(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 ^Jsin 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 4ve 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 (6f A0)rfsin 8= A/TV
rf(sin 6 cos Afl+cos 6 sin A0sin 0)=A/AT
For small angles, sin A0J* A0 and cos A0  1.
Hence,
or
Thus for order m,
45.16. The halfwidth of the fringes for a threeslit diffraction
pattern is given by
A <* A/3.2 d. (1)
The halfwidth of the doubleslit interference fringes is given by
..(2)
If the middle slit in the threeslit grating is closed then it reduces
to a doubleslit 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 18^0.553x10' 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
i4
a
where a is the slitwidth.
/. Smallest possible slitwidth, 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.
^=90Y ...(2)
6902p ...(3)
Hi = l ...(4)
Use (2), (3) and (4) in (1) to find
d[sin 90Y)~sin (902?)]= 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 rearrangement 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=
025.3
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
658.6
.*. 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 facecentered cubic lattice.
The basis consists of one Naatom and one Clatom. 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 ofan 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 ~ =
lsin'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
At/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) fxfsin (kza>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 (jfczr)]
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 righthanded system, has
368 Solutions to H and R Physics II
constant magnitude but rotates counterclockwise (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)=iJMJ 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 (Az 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
FE0.541 cm
C=FFC=0.541 cm0.471 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 quarterwave 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 .rof
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
crosssection 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/ .
/^wr v fn  r~/ x  =2.65 x 10 u kgm 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
planepolarised light into ellipticallypolarised light though the
special cases of emergent planepolarised 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 quarterwave plate is to change the ellipticallypola
rised light to planepolarised light.
Let the side A have the polaroid and let the polaroid axis be
oriented at 45 to the principal axes of the quarterwave plate. The
unpolarised light first enters the polaroid and the emergent plane
polarised light upon traversing through the quarter wave plate is
changed to circularlypolarised light which after reflection from the
coin returns and on traversing the quarterwave plate is changed
back to the planepolarised light but now the polaroid is "crossed"
and so the extinction of light occurs. Thus, the same combination
of polaroid and the quarterwave 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
quarterwave plate away from the coin (with the face A against .the
coin) then unpolarised light would first enter the quarterwave plate
and will be unaffected, and the resultant light upon traversing the
Polaroid would not be extinguished.
S.46.4. For the right circularlypolarised 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 circularlypolarized.
(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
larlypolarized, 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 + rC2M ^
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 joulesecJO 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=10cm.
Let N photons/sec energy through the hole. Power radiated
through the hole^Mzv
or N=
N hc _ 6.63 X 10" M joulessec)(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=0.70.4=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^essecXS 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 js)(3 x 10~ 7 m)(l .94 x 10' m 1 )
" (5.5 X 10"' rae1er)(ewl)
63 joules/sec =63 watts
Light and Quantum Physics 377
(6) Total cavity radiation, /? c =ar 6
where a is StefenBoltzmann 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 evsec)(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 evsec)(3 X 10 to cm/sec
47.7. 7 Av=  
_(4.14xlQ evsec)(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" evsec)(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 Fermidistribution and for these elec
trons which are sitting down the well, greater energy is required.
Therefore, the photoelectrons have an energy spectrum, the mini
mum energy being zero.
(c) Stopping potential is 2 volts.
(d) Cutoff wavelength corresponding to threshold energy ' of
4.2 ev is
/^ (4.14 X 1(T 16 evsec)(3 X IQ^cni/scc)
E 4.2 ev
si3Xl(r 5 cm=3000A
47.10. Photoelectric 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
10
05
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 5meter 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 photoelectron 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 evsec)(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 restmass 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 Xrays, 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 * \
lv/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(lcos <f>)+2m hc (1/Al/A')
AA
Cancelling m c* on both sides
Multiply throughout by AA', and simplify,
AA=A'A= A. (icos #) 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 ergsec) 8
(o) v t =3.28x10" {^Jr4iV4.556 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 ,  ^ J0.6150X 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 ergssec
._ (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'" ergsec
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~" ergsec) 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"" ergsec) 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*" joulesec) 2 (8.K5 X 10~* 2 couP/ntm 2 )
n(9.1 X 10" 81 kg)(1.6x ICr 1 ' coul)*
= 5.3xlO~ u meter.
,^ T h 6.63xlO~ 84 joulesec f A . IA  M . ,
(c) L==r = ~ =1.06x10 * 4 joulesec.
_ /^
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 kgmeter/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
=82x10 nt.
(i)
8
=J (9x 10' ntm 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.
136
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
PH2P2.7X 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
U6eV
Fig. 47.21
With Et  ,= 10.2=13.6 ( ^571
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 Jafcif
34 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 Xray region.
48.28. Equating the electrical force to centripetal force,
~=mo> 2 r .. (1)
Quantization of angular momentum gives
Squaring (1) and rearranging,
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 * wH
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 2/11 "j
Vi 3 L (D a /i J J
The orbital frequency v is given by
'. Percentage difference is 100 I  J.
2 ^^ ___ 21
l)*' n 8 J 100 (3/12)
2/11
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 joulessec.
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 joulessec)/(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 joulesec)
*= 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 joulesec
48.3. (a) Momentum of electron, p= y= 2 X 10^ 10 meter
=3.3X 10" 24 kgmeter/sec
Momentum of photon, /?= =y=3.3x 10" 14 kgmeter/sec.
n2
(b) For electron, kinetic energy, K
Ln\e
^=(3.3 x 10'" kgmeter/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"" kgmeter/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 lefthand 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 kgmetcr/sec.
deBroglie wavelength, A=
=6.6X 10" 84 joules)/(2.7 X 1(T 17 kgmeter/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 joulesec
.1 X 10" kg)(86.4x 10~ l * joules)
=1.65 XHT 10 meter
Now, condition for Braggrcflectiqn 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~ jouleec)/(4. 1 8 X lO'" kgmeter/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=90tf
whence ^=1802e=1802x61=58 .
394 Solutions to Hand R Physics H
48.6. (a)
10A
versus K for Electrons
123 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 lO2 joule/K)(293K)=6.0x 10" joules
H = 1.67X10" kg
, = _ h _ __ _____ 6.63 X IP"'* joulesec
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 =3.41.5=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 joalcsec)(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\
+ }