3. Electromagnetic Theory NET-JRF June 2011 - June 2014

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ELECTROMAGNETIC THEORY

JRF/NET(JUNE-2011) Q1. The electrostatic potential V(x, y) in free space in a region where the charge density is

zero is given by ( ) ( ) 22 34, yxfeyxV x += . Given that the x-component of the electric field Ex, and V are zero at the origin, f(x) is

(a) xex x 843 22 + (b) xex x 1643 22 + (c) 84 2 xe (d) xex 22 43

Ans: (d)

( ) 22 34 yxfeV x += . Since ( ) 061600 22 =+== xfeV x . Since 0=xE at origin ( )[ ]xfeEVE xx +== 28GG

( ) ( )[ ] ( ) 800080,0 ==+= ffEx . Since ( ) ( ) ( ) 4000400,0 ==+= ffV Solve equation ( ) 0616 2 =+ xfe x ( ) ( ) 122 86166 cexxfexf xx +== , since

( ) 080 1 == cf . Again Integrate ( ) xexxf 286 = ( ) 222 43 cexxf x += since ( ) 040 2 == cf . Thus ( ) xexxf 22 43 =

Q2. For constant uniform electric and magnetic field 0EEGG = and 0BB

GG = , it is possible to choose a gauge such that the scalar potential and vector potential AG are given by (a) ( )rBAand GGG == 0210 (b) ( )rBAandrE GGGGG == 00 21 (c) 00 == AandrE

GGGG (d) tEAand 00GG ==

Ans: (a)

Let ( )zyxEE 0 ++=G and ( )zyxBB 0 ++=G since they are constant vector. Lorentz Gauge condition is

tVA = 00

GG





( ) ( ) ( ){ }zxyxzxyz ByBBrBsince 000 += GG (a) 0=

t and 0= AGG (b) ,0

t and 0= AGG

(c) 0

t and 0= AGG (d) 0

t and 0 AGG

Q3. A plane electromagnetic wave is propagating in a lossless dielectric. The electric field is

given by

( ) ( ) ( ){ }[ ],3exp,,, 00 zxctikzAxEtzyxE +++=G where c is the speed of light in vacuum, E0, A and k0 are constant and x and z are unit

vectors along the x- and z-axes. The relative dielectric constant of the medium r and the constant A are

(a) 4=r and 31=A (b) 4=r and 3

1+=A

(c) 4=r and 3=A (d) 4=r and 3=A Ans: (a)

( ) ( ) ( ){ }[ ].3exp,,, 00 zxctikzAxEtzyxE +++=G Comparing with term ( )trkie GG ( )zxkk 30 += and ck0= . Since

23 2020

0 ckk

ckk

v =+

== Refractive index .42 === rrn

Since ( ) ( ) 030 0 =++= zAxzxknk ( ) 310310 ==+ AAk Q4. A static, spherically symmetric charge distribution is given by ( ) Kre

rAr = where A and

K are positive constants. The electrostatic potential corresponding to this charge

distribution varies with r as

(a) Krre (b) Krrer

1 (c) Krrer

2

1 (d) ( )Krer

11





Ans: (b)

since 02 /= V

V2 must be proportional to krerA where .1 22

2

=

rVr

rrV

Q5. The magnetic field of the TE11 mode of a rectangular waveguide of dimensions a b as shown in the figure is given by ( ) ( )yxHH z 4.0cos3.0cos1= , where x and y are in cm.

A. The dimensions of the waveguide are

(a) a = 3.33 cm, b = 2.50 cm (b) a = 0.40 cm, b = 0.30 cm

(c) a = 0.80 cm, b = 0.60 cm (d) a = 1.66 cm, b = 1.25 cm

Ans: (a)

Since ( ) ( )yxHH z 4.0cos3.0cos0= 1nwhere4.0

bn and1mwhere3.0 ====

am

cmbcma 50.2,33.3 == B. The entire range of frequencies f for which the TE11 mode will propagate is

(a) 6.0 GHz < f < 7.5 GHz (b) 7.5 GHz < f < 9.0 GHz

(c) 7.5 GHz < f < 12.0 GHz (d) 7.5 GHz < f

Ans: (d)

+

=22

, 2 bn

amcf nm zGHba

cf 5.7112 221,1

=+= .

z

b

a

y

x





For propagation, frequency of incident wave must be greater than cutoff frequency.

JRF/NET-(DEC-2011)

Q6. Consider three polarizers P1, P2 and P3 placed along an axis as shown in the figure.

The pass axis of P1 and P3 are at right angles to each other while the pass axis of P2

makes an angle with that of P1. A beam of unpolarized light of intensity I0 is incident on P1 as shown. The intensity of light emerging from P3 is

(a) 0 (b) 20I (c) 2sin

820I (d) 2sin

420I

Ans: (c)

20 cosII = (Malus Law)

( ) .2sin8

90coscos2

,cos2

,2

202203

202

01 IIIIIII ====

Q7. Four equal point charges are kept fixed at the four vertices of a square. How many neutral

points (i.e. points where the electric field vanishes) will be found inside the square?

(a) 1 (b) 4 (c) 5 (d) 7

Ans: (a)

Inside the square, there is only one point where field vanishes.

(unpolarized)

0I

1P 2P 3P





Q8. A static charge distribution gives rise to an electric field of the form ( ) 2/ 1 rreE Rr= , where and R are positive constants. The charge contained within a sphere of radius R,

centred at the origin is

(a) 20 Re (b) 2

2

0 Re (c)

eR

04 (d) eR 2

0

Ans: ( ) ( ) == rddrrreadEQ Rrenc sin1 22/00 GG ( ) =

0

2

0

/0 sin1 dde

Rr

at Rr = ,

=e

Qenc114 0 . So none of the options given are correct.

Q9. In a Youngs double slit interference experiment, the slits are at a distance 2L from each

other and the screen is at a distance D from the slits. If a glass slab of refractive index

and thickness d is placed in the path of one of the beams, the minimum value of d for the

central fringe to be dark is

(a) ( ) 221 LDD

+ (b) ( )L

D1

(c) ( )1 (d) ( )12

Ans: (d)

For central fringe to be dark, ( ) ( )1221 == dnd

Q10. Consider a solenoid of radius R with n turns per unit length, in which a time dependent

current I = I0 sint (where R/c





trnIE cos21

00=G

Q11. A constant electric current I in an infinitely long straight wire is suddenly switched on at t

= 0. The vector potential at a perpendicular distance r from the wire is given

by ( ) += 2220 1ln2 rtcctrIkA . The electric field at a distance r (< ct) is (a) 0 (b) ( )ji

tI

21

20

(c) ( )ji

rtc

Ic 2

12 222

0 +

(d) k

rtc

Ic 2 222

0

Ans: (d)

( ) .2 212 2222

222

0

+

+=

==

rtctcc

rrtcctrIE

tA

tAE

GGGG

krtc

IcE 2 222

0

=

G

JRF/NET-(JUNE-2012)

Q12. The magnetic field corresponding to the vector potential rr

rFA GGGG 310

21 += where FG is

a constant vector, is

(a) FG

(b) FG (c) r

rF GG 4

30+ (d) rr

F GG 430

Ans: (a)

( )[ ] +== 31021 rrrFABGGGGGGGG . Since FG is a constant vector, let

( )zyxFF 0 ++=G , ( ) ( ) ( ) 000000

FxyzFxzyFyzxzyx

FFFzyx

rF +== GG





( )( ) ( ) ( )

[ ] [ ] [ ] ( )zyxFFFzFFyFFxFxyFzxFyz

zyx

zyx

rF 2

0000000

000

++=+++=

= GGG

( )[ ] ( ) FzyxFrF GGGG =++= 21

0 , 03 = rrGG . Thus FB =

Q13. An electromagnetic wave is incident on a water-air interface. The phase of the

perpendicular component of the electric field, E , of the reflected wave into the water is

found to remain the same for all angles of incidence. The phase of the magnetic field H

(a) does not change (b) changes by 3/2

(c) changes by /2 (d) changes by

Ans: (d)

Q14. The magnetic field at a distance R from a long straight wire carrying a steady current I is

proportional to

(a) IR (b) I /R2 (c) I2/R2 (d) I / R

Ans: (d)

Q15. Which of the following questions is Lorentz invariant?

(a) 2

BE (b) 22 BE (c) 22 BE + (d) 22 BE Ans: (b)

Q16. Charges Q, Q and -2Q are placed on the vertices of an equilateral triangle ABC of sides

of length a, as shown in the figure. The dipole moment of this configuration of charges,

irrespective of the chice of origin, is

(a) iaQ 2+ (b) jaQ 3+ (c) jaQ 3 (d) 0

Ans: (c)

a a

aA

C

BQQ

2Q-j

i





Let coordinates of A is (l, m), then

[ ] ( )[ ]

++

+++++== jamialQjmialQjmilQrqp ii 2

32

2GG

[ ] ( )[ ] ( ) ( )[ ]jamialQjmialQjmilQp 322 +++++++=G jaQp 3= G Q17. The vector potential A due to a magnetic moment m at a point r is given by 3r

rmA = .

If m is directed along the positive z-axis, the x-component of the magnetic field, at the

point r , is

(a) 53

rmyz (b) 5

3rmxy (c) 53r

mxz (d) ( )523 r xyzm Ans: (c)

zmm =G and ( ) ( )[ ]mrrmr

rrmAB GGGGG =+== 31sincos2 33

++= zmrr

rzzyyxxzm

rB

313GG

5

3rmxzBx =

JRF/NET-(DEC-2012)

Q18. Three charges are located on the circumference of a circle of radius R as shown in the

figure below. The two charges Q subtend an angle 90 at the centre

of the circle. The charge q is symmetrically placed with respect to

the charges Q. If the electric field at the centre of the circle is zero,

what is the magnitude of Q?

(a) 2/q (b) q2 (c) 2q (d) 4q

Ans: (a)

20

21 41

RQEE == and 203 4

1RqE =

Resultant of 21 EandE is 122

21 2EEEE =+= , Thus 23

qQEE ==

Q Q

q





Q19. Consider a hollow charged shell of inner radius a and outer radius b. The volume charge

density is ( ) 2rkr = (k is constant) in the region a < r < b. The magnitude of the electric

field produced at distance r > a is

(a) ( )20r

abk for all r > a ,

(b) ( )20r

abk for a < r < b and 2

0rkb for r > b

(c) ( )20r

ark for a < r < b and ( )2

0rabk

for r > b

(d) ( )20a

ark for a < r < b and ( )2

0rabk

for r > b

Ans: (c)

For r < a

2 2enc 20 0 0

1 1 1 kE.da E(4 r ) Q dV r sin drd dr

= = = = G Gv

2 ra

0 0

4 k 4 kE(4 r ) dr (r a) = = 20k r a E r

r =

G

For r > a

2 ba

0 0

4 k 4 kE4 r dr (b a) = = 20k b a E r

r =

G

Q20. Consider the interference of two coherent electromagnetic waves whose electric field

vectors are given by tEiE cos 01 =G

and ( ) += tEjE cos 02G where is the phase difference. The intensity of the resulting wave is given by 20

2E

, where 2E is the

time average of E2. The total intensity is

(a) 0 (b) 200 E (c) 2200 sinE (d) 2200 cosE Ans: (a)

Since waves are polarized in perpendicular direction hence there will be no interference.





Q21. Four charges (two + q and two q) are kept fixed at the four vertices of a square of side a

as shown. At the point P which is at a distance R

from the centre (R >> a), the potential is

proportional to

(a) 1/R (b) 1/R2

(c) 1/R3 (d) 1/R4

Ans: (c)

Given configuration is quadrupole.

Q22. A point charges q of mass m is kept at a distance d below a grounded infinite conducting

sheet which lies in the xy - plane. For what value of d will the charge remains stationary?

(a) 04/ mgq (b) 0/ mgq (c) There is no finite value of d

(d) qmg /0 Ans: (a)

There is attractive force between point charge q and grounded conducting sheet that can

be calculate from method of images i.e. ( ) 022

0 4241

mgqdmg

dq ==

Q23. An infinite solenoid with its axis of symmetry along the z-direction carries a steady current

I.

The vector potential AG

at a distance R from the axis

(a) is constant inside and varies as R outside the solenoid

(b) varies as R inside and is constant outside the solenoid

(c) varies asR1 inside and as R outside the solenoid

(d) varies as R inside and asR1 outside the solenoid

Ans: (d)

q q

P

qq

a R

R

z





Q24. Consider an infinite conducting sheet in the xy-plane with a time dependent current

density iKt , where K is a constant. The vector potential at ( )zyx ,, is given by ( ) izct

cK

A 4

20 = . The magnetic field BG is

(a) jKt

20 (b) j

cKz

20 (c) ( )izct

cK

20 (d) ( ) jzct

cK

20

Ans: (d)

yz

AAB x

== = ( ) jzctcK

20

Q25. When a charged particle emits electromagnetic radiation, the electric field EG

and the

Poynting vector BESGGG =

0

1 at a larger distance r from emitter vary as nr

1 and

mr1 respectively. Which of the following choices for n and m are correct?

(a) n = 1 and m = 1 (b) n = 2 and m = 2

(c) n = 1 and m = 2 (d) n = 2 and m = 4

Ans: (c)

JRF/NET-(JUNE-2013)

Q26. A particle of charge e and mass m is located at the midpoint of the line joining two fixed

collinear dipoles with unit charges as shown in the figure. (The particle is constrained to

move only along the line joining the dipoles). Assuming that the length of the dipoles is

much shorter than their separation, the natural frequency of oscillation of the particle is

(a) 2

50

6eRmd (b) 40

6eRmd (c)

2

50

6edmR (d) 40

6edmR

RR

me,

d2 d2

++





Ans: (d)

Let us displace the charge particle by small amount x at A. Then the resultant electric field at point A is given by

( ) ( ) xRd

xRxRpE 4

033

0

61142

=

+= where 1.2 2p d d= = .

40

6edF eE xR= =

Then 40

6k edm mR

= = Q27. A current I is created by a narrow beam of protons moving in vacuum with constant

velocityuG . The direction and magnitude, respectively, of the Poynting vector SG outside the beam at a radial distance r (much larger than the width of the beam) from the axis,

are

(a) uS GG and 20

2

2

4 ruIS G

G= (b) ( )uS

GG || and 40

2

2

4 ruIS G

G=

(c) uS GG || and 20

2

2

4 ruIS G

G= (d) uS

GG || and 40

2

2

4 ruIS G

G=

Ans: (c)

Let charge per unit length be , hence uI = in z-direction. The magnetic field at a distance r is

2

0

rI

B = .

The electric field at a distance r is rur

Irr

E 2

2 00 == .

RR

me,

d2 d2

++x

A





Hence Poynting vector zur

IBES 4 20

2

2

0 ==

Q28. If the electric and magnetic fields are unchanged when the potential AG

changes (in

suitable units) according to ,rAA + GG where ( ) ,rtrr =G then the scalar potentialmust simultaneously change to

(a) r (b) r+ (c) t /r (d) t+ /r Ans: (c)

rAAA +=+= 1/ = r Cr += trVtVV == //

Q29. Consider an axially symmetric static charge distribution of the form,

2/2

00 cos0

rrerr

=

The radial component of the dipole moment due to this charge distribution is

(a) 4002 r (b) 400r (c) 400r (d) 2/400r Ans: (a)

( ) ddrdrerr

rdrrp rrV

== sincos 22/

20

00

400

0

2

0

2

0

/200 2cossin

0

0 rddrderrpr

r

rr

== =

Q30. The components of a vector potential ( )3210 ,,, AAAAA are given by ( )xytzxtyztxyzkA ,,,= where k is a constant. The three components of the electric field are

(a) ( )xyzxyzk ,, (b) ( )zyxk ,, (c) ( )0,0,0 (d) ( )ztytxtk ,, Ans: (c)

( )zyx AAAA ,,, = kxytAkzxtAkyztAkxyz zyx ==== ,,, Since ( )zxyyxzxyzk = and ( )zxyyxzxyzk

tA ++=G





( ) ( ) 0 =++++== zxyyxzxyzkzxyyxzxyzk

tAEGG ( )0,0,0= E

Q31. An oscillating current ( ) ( )tiItI = exp0 flows in the direction of the y-axis through a thin metal sheet of area 1.0 cm2 kept in the xy -plane. The rate of total energy radiated per unit

area from the surfaces of the metal sheet at a distance of 100 m is

(a) ( )300 12/ cI (b) ( )30220 12/ cI (c) ( )3020 12/ cI (d) ( )3020 24/ cI

Ans: (b)

JRF/NET-(DEC-2013)

Q32. A horizontal metal disc rotates about the vertical axis in a uniform magnetic field pointing

up as shown in the figure. A circuit is made by connecting one end A of a resistor to the

centre of the disc and the other end B to its edge through a sliding contact. The circuit

that flows through the resistor is

(1) zero (2) DC from A to B

(3) DC from B to A (4) AC

Ans: (2)

Q33. The force between two long and parallel wires carrying currents 1I and 2I and separated

by a distance D is proportional to

(1) DII /21 (2) ( ) DII /21 + (3) ( )221 / DII (4) 221 / DII Ans: (1)

B

A B

S





Q34. The electric field of an electromagnetic wave is given by

( )[ ]ktyxEE 10004.03.0cos0 += G . The associated magnetic field B

G is

(1) ( )[ ]ktyxE 10004.03.0cos10 03 + (2) ( )[ ]( )jityxE 3410004.03.0cos10 04 + (3) ( )[ ]( )jityxE 4.03.010004.03.0cos0 ++ (4) ( )[ ]( )jityxE 4310004.03.0cos10 02 ++

Ans: (2) ( ) tyxk 1000,4.03.0 =+= ( ) ( )[ ]ktyxEyxEkB 10004.03.0cos4.03.01 0 ++==

( )[ ]( )jityxEB 3410004.03.0cos10 04 += Q35. A point charge q is placed symmetrically at a distance d from two perpendicularly

placed grounded conducting infinite plates as shown in the figure. The net force on the

charge (in units of 04/1 ) is

(1) ( )1228 2

2

dq away from the corner

(2) ( )1228 2

2

dq towards the corner

(3) 2

2

22 dq towards the corner

(4) 22

83dq away from the corner

Ans: (2)

2

2

214dqkFF == and 2

2

38dqkF =

Resultant of 21 , FF is 22

22

2112 8

22dqkFFF =+= .

d

d

q

3Fq

d

d

qq+

q d

d2

d

1F2F





Net force ( )1228 2

2

=dqkF (towards the corner)

Q36. If the electrostatic potential ( ) ,,rV in a charge free region has the form ( ) ( ) cosrfrV = , then the functional form of ( )rf (in the following a and b are

constants) is:

(1) rbar +2 (2) 2r

bar + (3) rbar + (4)

bra ln

Ans: (2) 2

2 22 2 2 2 2

1 1 1sin 0sin sin

V V VV rr r r r r

= + + =

( )22 21 1cos sin sin 0sinfr f

r r r r

+ =

( )222 2 2cos 2 2sin cos 0sinf f fr r

r r r r + =

( )22 2 2 2 0f fr r f rr r + =

( ) 2rbarrf += satisfy the above equation.

Q37. Let four point charges qqq ,2/, and 2/q be placed at the vertices of a square of side a . Let another point charge q be placed at the cnetre of the square (see the figure).

Let ( )rV be the electrostatic potential at a point P at a distance ar >> from the centre of the square. Then ( ) ( )rVrV /2 is

2/q

q

2/qq

q





(1) 1 (2) 21 (3)

41 (4)

81

Ans: (4)

According to multipole expansion 022

=++= qqqqqQmono

( ) ( ) ( ) ( ) ( ) 002

2

=+++++= yaxaqyaxaqyaxaqyaxaqyaxaqp

Thus ( )( ) 8121

3 = rVrV

rV .

Q38. Let ( )AV , and ( )AV , denote two sets of scalar and vector potentials, and a scalar function. Which of the following transformations leave the electric and magnetic fields

(and hence Maxwells equations) unchanged?

(1) += AA and t

VV = (2) = AA and

tVV

+= 2

(3) += AA and t

VV += (4) = AA and

tVV

= Ans: (1)

JRF/NET(JUNE-2014)

Q39. A time-dependent current ( ) zKttI =G (where K is a constant) is switched on at 0=t in an infinite current-carrying wire. The magnetic vector potential at a perpendicular

distance a from the wire is given (for time cat /> ) by

(a) ( )

++222

2222/122

220

4

atc

atc zazactdz

cKz

(b) ( ) +

ct

ct zatdzKz 2/122

0

4

(c) ( ) ++ct

ct zazactdz

cKz 2/122

220

4

(d) ( )

+222

2222/122

0

4

atc

atc zatdzKz





Ans: (a)

( ) ( )0 0 / 4 4

rI t K t R cA z dz z dzR R

= = JG

( )2 2 2

2 2 2

2 20

1/22 2

4

c t a

c t a

K ct a zA z dzc a z

+ =+

JG

Q40. A current pi flows through the primary coil of a transformer. The graph of ( )ti p as a function of time t is shown in the figure below.

Which of the following graphs represents the current Si in the secondary coil?

(a) (b)

(c) (d)

( )tip

1 2 3 t

1 2 3 t

( )si t

1 2 3 t

( )si t

( )si t

1 2 3 t

1 2 3 t

( )si t

Idz

Pa

Rz





Ans: (c) psdi

idt

Q41. If the electrostatic potential in spherical polar coordinates is

( ) 0/0 rrer = where 0 and 0r are constants, then the charge density at a distance 0rr = will be

(a) 20

00

er

(b) 20

00

2re

(c) 20

00

er (d) 2

0

002r

e

Ans: (a) ( )2 200

= =

( )0 0 0 0/ / / /2 2 2 2 20 0 02 2 2 20 0 0 0

1 1 1 1 1 2r r r r r r r rr r e r e r e rer r r r r r r r r r r r

= = = = + 0 0/ /2 0

0 0

1 2r r r re er r r = +

At a distance 0rr = , 2 1 10 020 0 0 0

1 2e er r r r e = + =

0 0 00 2 2

0 0r e r e = =

Q42. If kxyjzxiyzA ++=G and C is the circle of unit radius in the plane defined by 1=z , with the centre on the z - axis, then the value of the integral

CA d GG Av is

(a) 2 (b) (c)

4 (d) 0

Ans: (d) ( ) ( ) ( )

0

i j k

A i x x j y y k z zx y z

yz zx xy

= = + = JG JG

Since ( ) 0CS

A d A da = = G JG JG GG Av Q43. Consider an electromagnetic wave at the interface between two homogenous dielectric

media of dielectric constants 1 and 2 . Assuming 12 > and no charges on the





surface, the electric field vector EG

and the displacement vector DG

in the two media

satisfy the following inequalities

(a) 12 EEGG > and 12 DD

GG > (b) 12 EEGG < and 12 DD

GG <

(c) 12 EEGG < and 12 DD

GG > (d) 12 EEGG > and 12 DD

GG < Q44. A charge ( )e is placed in vacuum at the point ( )0,0,d , where 0>d . The region 0x

is filled uniformly with a metal. The electric field at the point

0,0,2d is

(a) ( )0,0,19

102

0de

(b) ( )0,0,1910

20de

(c) ( )0,0,120de

(d) ( )0,0,120de

Ans: (b)

( )2 20 01 1 4

4 4 93 / 2e eE

dd + = = and ( )2 20 01 1 4

4 4/ 2e eE

dd = =

Thus resultant electric field at point P is

2 2 2 20 0 0 0

1 4 1 4 1 40 1 104 9 4 4 9 9

e e e eE E Ed d d d + = + = + = = 20

1 10 9

eE xd =

JG

Q45. A beam of light of frequency is reflected from a dielectric-metal interface at normal incidence. The refractive index of the dielectric medium is n and that of the metal is

( )inn += 12 . If the beam is polarised parallel to the interface, then the phase change experienced by the light upon reflection is

(a) ( )/2tan (b) ( )/1tan 1 (c) ( )/2tan 1 (d) ( )2tan 1

ee+ 0x

ddP2

d E+JGEJG





Ans: (c)

Since 0 011R I

E E = +

where ( )12/ 1

/ 1v c n iv c n i

= = = ++

( )/2 /20 0 0 02 22 4 4

ii

R I I Ii

i eE E E e Ei e

+ = = = + + +

where tan 2 = .

Thus phase change ( ) 12 2/ 2 tan cot tan = + = = =

Q46. A thin, infinitely long solenoid placed along the z - axis contains a magnetic flux . Which of the following vector potentials corresponds to the magnetic field at an arbitrary

point ( )zyx ,, ? (a) ( )

++= 0,2,2,, 2222 yxx

yxyAAA zyx

(b) ( )

++++= 0,2,2,, 222222 zyxx

zyxyAAA zyx

(c) ( )

++

++= 0,

2,

2,, 2222 yx

yxyxyxAAA zyx

(d) ( )

++= 0,2,2,, 2222 yxy

yxxAAA zyx

Ans: (a) 0B A= =JG JG JG Q47. An electromagnetically-shielded room is designed so that at a frequency 710= rad/s

the intensity of the external radiation that penetrates the room is 1% of the incident

radiation. If ( ) 161021 = m is the conductivity of the shielding material, its

minimum thickness should be (given that 3.210ln = ) (a) 4.60 mm (b) 2.30 mm (c) 0.23 mm (d) 0.46 mm

Ans: (b)





2 00

1 ln2

z II I e zI

= =

where 6 7 7 30 1 1100, 10 4 10 10 102 2 2

II

= = = =

( )31 ln 100 2.302 10z mm = =

Q48. A charged particle is at a distance d from an infinite conducting plane maintained at zero

potential. When released from rest, the particle reaches a speed u at a distance 2/d from

the plane. At what distance from the plane will the particle reach the speed u2 ?

(a) 6/d (b) 3/d (c) 4/d (d) 5/d

Ans: (b) 2 2

2 20

14 4

d x qF ma mdt d= = =

2

2 2

d x Adt x

= where2

016qAm = .

( )22 2 12dv A dv dv A dx d d Av v vdt x dt dt x dt dt dt x = = = 2

2v A C

x = + at , 0 Ax d v C

d = = = 1 12v A

x d = .

Thus 1 1 22/ 2

Au Ad d d

= = then 1 12 2

5du A x

x d = =

P

d

q+

0

x

qd

Documents

3. Electromagnetic Theory NET-JRF June 2011 - June 2014