Electro Magnetic Field MCQ PDF

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CHAPTER 6

TIME VARYING FIELDS AND MAXWELLS'S EQUATIONS

SAMPLE CHAPTER



342 Time Varying Fields and Maxwells's Equations Chap 6

GATE CLOUD Electromagnetics

Author: R K Kanodia & Ashish Murolia Edition: 1st

Publisher: JHUNJHUNUWALA ISBN: 9788192348384

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EXERCISE 6.1

MCQ 6.1.1 Match List I with List II and select the correct answer using the codes given below

(Notations have their usual meaning)

List-I List-II

a Ampere’s circuital law 1. D v : ρd =

bFaraday’s law

2.0B :d =

c Gauss’s law 3.t

E B

2

2#d =−

d Non existence of isolated

magneticharge

4.t

H J D

2

2#d = +

Codes :

a b c d

(A) 4 3 2 1

(B) 4 1 3 2

(C) 2 3 1 4(D) 4 3 1 2

MCQ 6.1.2 Magneto static fields is caused by

(A) stationary charges (B) steady currents

(C) time varying currents (D) none of these

MCQ 6.1.3 Let A be magnetic vector potential and E be electric field intensity at certain time

in a time varying EM field. The correct relation between E and A is

(A)t

E A

2

2=− (B)

t A

E

2

2=−

(C)t

E A2

2= (D)t

A E 2

2=

MCQ 6.1.4 A closed surface S defines the boundary line of magnetic medium such that the

field intensity inside it is B . Total outward magnetic flux through the closed

surface will be

(A) B S : (B) 0

(C) B S # (D) none of these



Chap 6 Time Varying Fields and Maxwells's Equations 343




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MCQ 6.1.5 A perfect conducting sphere of radius r is such that it’s net charge resides on the

surface. At any time t , magnetic field ( , )r t B inside the sphere will be

(A) 0

(B) uniform, independent of r (C) uniform, independent of t

(D) uniform, independent of both r and t

MCQ 6.1.6 The total magnetic flux through a conducting loop having electric field E 0=

inside it will be

(A) 0

(B) constant

(C) varying with time only

(D) varying with time and area of the surface both

MCQ 6.1.7 A cylindrical wire of a large cross section made of super conductor carries a current

I . The current in the superconductor will be confined.

(A) inside the wire (B) to the axis of cylindrical wire

(C) to the surface of the wire (D) none of these

MCQ 6.1.8 If B i denotes the magnetic flux density increasing with time and B d denotes the

magnetic flux density decreasing with time then which of the configuration is

correct for the induced current I in the stationary loop ?

MCQ 6.1.9 A circular loop is rotating about z -axis in a magnetic field cosB t B a y 0 ω= . The

total induced voltage in the loop is caused by

(A) Transformer emf (B) motion emf.

(C) Combination of (A) and (B) (D) none of these







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MCQ 6.1.10 A small conducting loop is released from rest with in a vertical evacuated cylinder

voltage induced in the falling loop is

(Assume earth magnetic field 10 T6= − at a constant angle of 10c below the

horizontal)

(A) zero (B) 1 mV

(C) 17.34 mV (D) 9.8 mV

MCQ 6.1.11 A square loop of side 1 m is located in the plane x 0= as shown in figure. A non-

uniform magnetic flux density through it is given as

B 4z t a x 3 2= ,

The emf induced in the loop at time sect 2= will be

(A) 16 volt (B) 4 volt−

(C) 4 volt (D) 2 volt−

MCQ 6.1.12 A very long straight wire carrying a current 5 AI = is placed at a distance of 2 m

from a square loop as shown in figure. If the side of the square loop is 1 m then the

total flux passing through the square loop will be

(A) 0.81 10 wb7#

− (B) 10 wb6−

(C) 4.05 10 wb7#

− (D) 2.0 10 wb7#

−







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MCQ 6.1.13 A straight conductor ab of length l lying in the xy plane is rotating about the

centre a at an angular velocity ω as shown in the figure.

If a magnetic field B is present in the space directed along a z then which of thefollowing statement is correct ?

(A) V ab is positive (B) V ab is negative

(C) V ba is positive (D) V ba is zero

MCQ 6.1.14 In a certain region magnetic flux density is given as 0.1 /Wb mt B a z 2= . An electric

loop with resistance 2 Ω and 4 Ω is lying in x -y plane as shown in the figure.

If the area of the loop is 1 m2 than, the voltage drop V 1 and V 2 across the two

resistances is respectively

(A) 66.7 mV and 33.3mV (B) 33.3mV and 66.7 mV

(C) 50 mV and 100 mV (D) 100 mV and 50 mV

MCQ 6.1.15 Assertion (A) : A small piece of bar magnet takes several seconds to emerge at

bottom when it is dropped down a vertical aluminum pipe where as an identicalunmagnetized piece takes a fraction of second to reach the bottom.

Reason (R) : When the bar magnet is dropped inside a conducting pipe, force

exerted on the magnet by induced eddy current is in upward direction.

(A) Both A and R are true and R is correct explanation of A.

(B) Both A and R are true but R is not the correct explanation of A.

(C) A is true but R is false.

(D) A is false but R is true.







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MCQ 6.1.16 A magnetic core of uniform cross section having two coils (Primary and secondary)

wound on it as shown in figure. The no. of turns of primary coil is 5000 and no. of

turns of secondary coil is 3000. If a voltage source of 12 volt is connected across the

primary coil then what will be the voltage across the secondary coil ?

(A) 72 volt (B) 7.2 volt

(C) 20 volt (D) 7.2 volt−

MCQ 6.1.17 Self inductance of a long solenoid having n turns per unit length will be proportional

to

(A) n (B) /n 1

(C) n 2 (D) /n 1 2

MCQ 6.1.18 A wire with resistance R is looped on a solenoid as shown in figure.

If a constant current is flowing in the solenoid then the induced current flowing in

the loop with resistance R will be

(A) non uniform (B) constant

(C) zero (D) none of these

MCQ 6.1.19 A long straight wire carries a current ( )cosI I t 0 ω= . If the current returns along

a coaxial conducting tube of radius r as shown in figure then magnetic field andelectric field inside the tube will be respectively.







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(A) radial, longitudinal (B) circumferential, longitudinal

(C) circumferential, radial (D) longitudinal, circumferential

MCQ 6.1.20 Assertion (A) : Two coils are wound around a cylindrical core such that the primary

coil has N 1 turns and the secondary coils has N 2 turns as shown in figure. If thesame flux passes through every turn of both coils then the ratio of emf induced in

the two coils is

V V

emf

emf

1

2 N N

1

2=

Reason (R) : In a primitive transformer, by choosing the appropriate no. of turns,

any desired secondary emf can be obtained.

(A) Both A and R are true and R is correct explanation of A.

(B) Both A and R are true but R is not the correct explanation of A.

(C) A is true but R is false.

(D) A is false but R is true.

MCQ 6.1.21 In a non magnetic medium electric field cosE E t 0 ω= is applied. If the permittivity

of medium is ε and the conductivity is σ then the ratio of the amplitudes of the

conduction current density and displacement current density will be

(A) /0µ ωε (B) /σ ωε

(C) /0σµ ωε (D) /ωε σ

MCQ 6.1.22 In a medium where no D.C. field is present, the conduction current density at

any point is given as J d 20 . /cos A mt a 1 5 10 y 8 2

#=

^ h. Electric flux density in the

medium will be

(A) 133.3 . /sin nC mt a 1 5 10 y 8 2

#^ h (B) 13.3 . /sin nC mt a 1 5 10 y 8 2

#^ h(C) 1.33 . /sin nC mt a 1 5 10 y

8 2#^ h (D) 1.33 . /sin nC mt a 1 5 10 y

8 2#− ^ h

MCQ 6.1.23 In a medium, the permittivity is a function of position such that 0d.

εε . If the

volume charge density inside the medium is zero then E :d is roughly equal to

(A) E ε (B) E ε−

(C) 0 (D) E :εd−







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MCQ 6.1.24 A conducting medium has permittivity, 4 0ε ε= and conductivity, 1.14 10 /s m8σ #=

. What will be the ratio of magnitude of displacement current and conduction

current in the medium at 50GHz ?

(A) .1 10 104# (B) .1 025 107

#

(C) .9 75 10 17#

− (D) .9 75 10 8#

−

MCQ 6.1.25 In free space, the electric field intensity at any point ( , , )r θ φ in spherical coordinate

system is given by

E sin cos

r t kr

a θ ω

= −

θ

^ hThe phasor form of magnetic field intensity in the free space will be

(A) sinr

k e a

jkr

0ωµθ

φ− (B) sin

r k

e a jkr

0ωµθ

− φ−

(C) r

k

e a

jkr 0ωµφ

−

(D)

sin

r

k

e a

jkr θφ

−

MCQ 6.1.26 Magnetic field intensity in free space is given as

H 0.1 /cos sin A my t bx a 15 6 10 z 9

#π π= −^ ^h hIt satisfies Maxwell’s equation when b equals to

(A) 46.5 /rad m! (B) 41.6 /rad m!

(C) 77.5 /rad m! (D) 60.28 /rad m!

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EXERCISE 6.2

MCQ 6.2.1 Two parallel conducting rails are being placed at a separation of 2 m as shown in

figure. One end of the rail is being connected through a resistor 10R Ω= and the

other end is kept open. A metal bar slides frictionlessly on the rails at a speed of

5 /m s away from the resistor. If the magnetic flux density 0.1 /Wb mB 2= pointing

out of the page fills entire region then the current I flowing in the resistor will be

(A) 0.01A (B) 0.01A−(C) 1 A (D) 0.1A−

Common Data for Question 2 - 3 :

A conducting wire is formed into a square loop of side 2 m. A very long straight

wire carrying a current 30 AI = is located at a distance 3 m from the square loop

as shown in figure.

MCQ 6.2.2 If the loop is pulled away from the straight wire at a velocity of 5 /m s then the

induced e.m.f. in the loop after . sec0 6 will be

(A) 5 voltµ (B) 2.5 voltµ

(C) 25 voltµ (D) 5 mvolt







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MCQ 6.2.3 If the loop is pulled downward in the parallel direction to the straight wire, such

that distance between the loop and wire is always 3 m then the induced e.m.f. in

the loop at any time t will be

(A) linearly increasing with t (B) always 0

(C) linearly decreasing with t (D) always constant but not zero.

MCQ 6.2.4 An infinitely long straight wire with a closed switch S carries a uniform current

4 AI = as shown in figure. A square loop of side 2 ma = and resistance 4R Ω= is

located at a distance 2 m from the wire. Now at any time t t 0= the switch is open

so the current I drops to zero. What will be the total charge that passes through

a corner of the square loop after t t 0= ?

(A) 277 nC (B) 693 nC

(C) 237 nC− (D) 139 nC

MCQ 6.2.5 A circular loop of radius 5 m carries a current 2 AI = . If another small circular

loop of radius 1 mm lies a distance 12 m above the large circular loop such that the

planes of the two loops are parallel and perpendicular to the common axis as shown

in figure then total flux through the small loop will be

(A) 1.62 fWb (B) 25.3 nWb

(C) 44.9 fWb (D) 45.4 pWb







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MCQ 6.2.6 A non magnetic medium at frequency 1.6 10 Hz f 8#= has permittivity 54 0ε ε=

and resistivity 0.77 mρ Ω −= . What will be the ratio of amplitudes of conduction

current to the displacement current ?

(A) 0.43 (B) 0.37

(C) 1.16 (D) 2.70

MCQ 6.2.7 Two voltmeters A and B with internal resistances RA and RB respectively is

connected to the diametrically opposite points of a long solenoid as shown in

figure. Current in the solenoid is increasing linearly with time. The correct relation

between the voltmeter’s reading V A and V B will be

(A) V V A B = (B) V V A B =−

(C)V V

RR

B

A

B

A= (D)V V

RR

B

A

B

A=−

Statement for Linked Question 8 - 9 :

Two parallel conducting rails are being placed at a separation of 5 m with a resistance

10R Ω= connected across it’s one end. A conducting bar slides frictionlessly on

the rails with a velocity of 4 /m s away from the resistance as shown in the figure.

MCQ 6.2.8 If a uniform magnetic field 2 TeslaB = pointing out of the page fills entire region

then the current I flowing in the bar will be

(A) 0 A (B) 40 A−

(C) 4 A (D) 4 A−







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MCQ 6.2.9 The force exerted by magnetic field on the sliding bar will be

(A) 4 N, opposes it’s motion

(B) 40 N, opposes it’s motion

(C) 40 N, in the direction of it’s motion(D) 0

MCQ 6.2.10 Two small resistor of 250 Ω each is connected through a perfectly conducting

filament such that it forms a square loop lying in x -y plane as shown in the figure

Magnetic flux density passing through the loop is given as

B 7.5 (120 30 )cos t a z cπ=− −

The induced current ( )I t in the loop will be

(A) . ( )sin t 0 02 120 30cπ − (B) . ( )sin t 2 8 10 120 303# cπ −

(C) . ( )sin t 5 7 120 30cπ− − (D) . ( )sin t 5 7 120 30cπ −


In a non uniform magnetic field 8 Teslax B a z 2= , two parallel rails with a separation

of 20 cm and connected with a voltmeter at it’s one end is located in x -y plane

as shown in figure. The Position of the bar which is sliding on the rails is given as

x 1 0.4t t 2= +^ hMCQ 6.2.11 Voltmeter reading at . sect 0 4= will be







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(A) 0.35 volt− (B) 0.35 volt

(C) 0.29 volt− (D) 1.6 volt−

MCQ 6.2.12 Voltmeter reading at 12 cmx = will be

(A) 12.27 mvolt (B) 14.64 mvolt−

(C) 23.4 mvolt (D) 23.4 mvolt−

MCQ 6.2.13 A rectangular loop of self inductance L is placed near a very long wire carrying

current i 1 as shown in figure (a). If i 1 be the rectangular pulse of current as shown

in figure (b) then the plot of the induced current i 2 in the loop versus time t will

be (assume the time constant of the loop, /L R&τ )

MCQ 6.2.14 Two parallel conducting rails is placed in a varying magnetic field 0.2cos t B a x ω=

. A conducting bar oscillates on the rails such that it’s position is given by

y 0.5 cos mt 1 ω= −^ h







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If one end of the rails are terminated in a resistance 5R Ω= , then the current i

flowing in the rails will be

(A) . sin cost t 0 01 1 2ω ω ω+^ h (B) . sin cost t 0 01 1 2ω ω ω− +^ h(C) . cos sint t 0 01 1 2ω ω ω+^ h (D) . sin sint t 0 05 1 2ω ω ω+^ h

MCQ 6.2.15 Electric flux density in a medium ( 10r ε = , 2r µ = ) is given as

D 1.33 . /sin C mt x a 3 10 0 2 y 8 2

# µ= −^ hMagnetic field intensity in the medium will be

(A) 10 . /sin A mt x a

3 10 0 2 y

5 8

# −

−

^ h (B) 2 . /sin A mt x a 3 10 0 2 y

8# −^ h

(C) 4 . /sin A mt x a 3 10 0 2 y 8

#− −^ h

(D) 4 . /sin A mt x a 3 10 0 2 y 8

# −^ hMCQ 6.2.16 In a non conducting medium ( )0σ = magnetic field intensity at any point is given

by H /cos A mt bx a 10 z 10= −^ h . The permittivity of the medium is ε 0.12 /nF m=

and permeability of the medium is µ 3 10 /H m5#= − . If no D.C. field is present in

medium, then value of b for which the field satisfies Maxwell’s equation is

(A) 600 /rad s−

(B) 600 /rad m

(C) 3.6 10 /rad m5#

(D) (A) and (B) both

MCQ 6.2.17 A current filament located on the x -axis in free space with in the interval

0.1 0.1 mx < <− carries current ( ) 8 AI t t = in a x direction. If the retarded vector

potential at point ( , , )P 0 0 2 be ( )t A then the plot of ( )t A versus time will be







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MCQ 6.2.18 In a non-conducting medium ( 0σ = , 1r r µ ε= = ), the retarded potentials are given

as V volty x ct = −^ h and /Wb my t A a c x

x = −^ h where c is velocity of waves in

free space. The field (electric and magnetic) inside the medium satisfies Maxwell’s

equation if

(A) 0J = only (B) 0v ρ = only

(C) 0J v ρ= = (D) Can’t be possible

MCQ 6.2.19 Electric field in free space in given as

E 5 sin cosy bx a 10 6 10 z 9

#π π= −^ ^h hIt satisfies Maxwell’s equation for ?b =

(A) 20 /rad m! π (B) /rad m300! π

(C) 10 /rad mπ (D) 30 /rad mπ


In a region of electric and magnetic fields E and B , respectively, the force

experienced by a test charge qC are given as follows for three different velocities.

Velocity m/sec Force, N

a x q a a y z +^ h a y q a y

a z 2q a a y z +^ h







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MCQ 6.2.20 What will be the magnetic field B in the region ?

(A) a x (B) a a x y −

(C) a z (D) a a y z −

MCQ 6.2.21 What will be electric field E in the region ?(A) a a x z − (B) a a y z −

(C) a a y z + (D) a a a y z x + −

MCQ 6.2.22 In Cartesian coordinates magnetic field is given by 2/x B a z =− . A square loop of

side 2 m is lying in xy plane and parallel to the y -axis. Now, the loop is moving in

that plane with a velocity 2v a x = as shown in the figure.

What will be the circulation of the induced electric field around the loop ?

(A)x x 2

16+^ h (B)

x 8

(C) x x 28+^ h (D)x x

16 2+

^ h


In a cylindrical coordinate system, magnetic field is given by

B sin

for m

for m

for m

t a

0 4

2 4 5

0 5

<

< <

>

z

ρ

ω ρ

ρ

=

Z

[

\

]]

]]

MCQ 6.2.23 The induced electric field in the region 4 m<ρ will be

(A) 0 (B) cos t a 2 ρω ω φ

(C) 2cos t a ω− φ (D)sin t

a 2

1ω φ

MCQ 6.2.24 The induced electric field at 4.5 mρ = is

(A) 0 (B) 17 cos t 18

ω ω−

(C) 4 cos t 9

ω ω (D) 17 cos t 4

ω ω−







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MCQ 6.2.25 The induced electric field in the region 5 m>ρ is

(A) cos t a 18ρ

ω ω− φ (B) cos t a

9ρ

ω ω−φ

(C) 9 cos t a ρ ω−

φ (D)

cos t a

9

ρ

ω ωφ

MCQ 6.2.26 In a certain region a test charge is moving with an angular velocity 2 / secrad along

a circular path of radius 2 m centred at origin in the x -y plane. If the magnetic flux

density in the region is 2 /Wb mB a z 2= then the electric field viewed by an observer

moving with the test charge is

(A) 8 /V ma ρ (B) 4 /V ma ρ

(C) 0 (D) 8 /V ma − ρ

MCQ 6.2.27 A 8 A current is flowing along a straight wire from a point charge situated at the

origin to infinity and passing through the point , ,2 2 2

^ h. The circulation of the

magnetic field intensity around the closed path formed by the triangle having thevertices , ,2 0 0^ h, , ,0 2 0^ h and , ,0 0 2^ h is equal to

(A) A87 (B) 3 A

(C) 7 A (D) 1 A

MCQ 6.2.28 Magnetic flux density, 0.1t B a z = Tesla threads only the loop abcd lying in the

plane xy as shown in the figure.

Consider the three voltmeters V 1, V 2 and V 3, connected across the resistance in the

same xy plane. If the area of the loop abcd is 1 m2

then the voltmeter readings areV 1 V 2 V 3

(A) 66.7 mV 33.3 mV 66.7 mV

(B) 33.3 mV 66.7 mV 33.3 mV

(C) 66.7 mV 66.7 mV 33.3 mV

(D) 33.3 mV 66.7 mV 66.7 mV







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A square wire loop of resistance R rotated at an angular velocity ω in the uniform

magnetic field 5 /mWb mB a y 2= as shown in the figure.

MCQ 6.2.29 If the angular velocity, 2 / secradω = then the induced e.m.f. in the loop will be

(A) 2 /sin V mµθ (B) 2 /cos V mµθ

(C) 4 /cos V mµθ (D) 4 /sin V mµθ

MCQ 6.2.30 If resistance, 40 mR Ω= then the current flowing in the square loop will be

(A) 0.2 sin mAθ (B) 0.1 sin mAθ

(C) 0.1 cos mAθ (D) 0.5 sin mAθ

MCQ 6.2.31 In a certain region magnetic flux density is given as sinB t B a y 0 ω= . A rectangular

loop of wire is defined in the region with it’s one corner at origin and one side along

z -axis as shown in the figure.

If the loop rotates at an angular velocity ω (same as the angular frequency of

magnetic field) then the maximum value of induced e.m.f in the loop will be

(A) B S 21

0 ω (B) B S 2 0 ω

(C) B S 0 ω (D) B S 4 0 ω







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MCQ 6.2.32 A 50 turn rectangular loop of area 64 cm2 rotates at 60 revolution per seconds in

a magnetic field 0.25 377 /sin Wb mt B 2= directed normal to the axis of rotation.

The rms value of the induced voltage is

(A) 2.13 volt (B) 21.33 volt

(C) 4.26 volt (D) 42.66 volt


Consider the figure shown below. Let 10 120 /cos Wb mB t 2π= and assume that the

magnetic field produced by ( )i t is negligible

MCQ 6.2.33 The value of v ab is

(A) 118.43 120cos t π− V (B) 118.43 120cos t π V

(C) 118.43 120sin t π− V (D) 118.43 120sin t π V

MCQ 6.2.34 The value of ( )i t is

(A) 0.47 120cos t π− A (B) 0.47 120cos t π A

(C) 0.47 120sin t π− A (D) 0.47 120sin t π A

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EXERCISE 6.3

MCQ 6.3.1 A magnetic field in air is measured to be B x y

x x y

y a a B y x 0 2 2 2 2=

+ −

+c m What

current distribution leads to this field ?

[Hint : The algebra is trivial in cylindrical coordinates.]

(A) , 0B

x y r J

z 10

02 2 !

µ=

+c m (B) , 0B

x y r J

z 20

02 2 !

µ=−

+c m(C) 0, 0r J != (D) , 0B

x y r J z 1

0

02 2 !

µ=

+c mMCQ 6.3.2 For static electric and magnetic fields in an inhomogeneous source-free medium

which of the following represents the correct form of Maxwell’s equations ?

(A) 0E :d = , 0B d # = (B) 0E :d = , 0B :d =

(C) 0E d # = , 0B d # = (D) 0E d # = , 0B :d =

MCQ 6.3.3 If C is closed curve enclosing a surface S , then magnetic field intensity H , the

current density J and the electric flux density D are related by

(A) d

t

d l H S J D

S C

$ :2

2= +

b l ## ## (B) d

t

d H l J D

S

S S

: :2

2= +

b l # ##

(C) d t

d l H S J D

S C : :

2

2= +b l # ## (D) d

t d H l J

D S

C S : :

2

2= +b l ## #

MCQ 6.3.4 The unit of H #d is

(A) Ampere (B) Ampere/meter

(C) Ampere/meter2 (D) Ampere-meter

MCQ 6.3.5 The Maxwell equationt

H J D

2

2#d = + is based on

(A) Ampere’s law (B) Gauss’ law

(C) Faraday’s law (D) Coulomb’s law

MCQ 6.3.6 A loop is rotating about they y -axis in a magnetic field ( )cosB t B a x 0 ω φ= + T

The voltage in the loop is

(A) zero

(B) due to rotation only

(C) due to transformer action only

(D) due to both rotation and transformer action

GATE 2009

GATE 2008

GATE 2007

GATE 2003

GATE 1998

GATE 1998







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MCQ 6.3.7 The credit of defining the following current is due to Maxwell

(A) Conduction current (B) Drift current

(C) Displacement current (D) Diffusion current

MCQ 6.3.8 A varying magnetic flux linking a coil is given by 1/3 t 3λΦ = . If at time 3 st = , theemf induced is 9 V, then the value of λ is.

(A) zero (B) 1 /Wb s2

(C) 1 /Wb s2− (D) 9 /Wb s2

MCQ 6.3.9 Assuming that each loop is stationary and time varying magnetic field B , induces

current I , which of the configurations in the figures are correct ?

(A) 1, 2, 3 and 4 (B) 1 and 3 only

(C) 2 and 4 only (D) 3 and 4 only

MCQ 6.3.10 Assertion (A) : For time varying field the relation V E d=− is inadequate.

Reason (R) : Faraday’s law states that for time varying field 0E d # =(A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is

the correct explanation of Assertion (A)

(B) Both Assertion (A) and Reason (R) are individually true but Reason (R) is

not the correct explanation of Assertion (A)

(C) Assertion (A) is true but Reason (R) is false

(D) Assertion (A) is false but Reason (R) is true

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MCQ 6.3.11 Who developed the concept of time varying electric field producing a magnetic field

?

(A) Gauss (B) Faraday

(C) Hertz (D) MaxwellMCQ 6.3.12 A single turn loop is situated in air, with a uniform magnetic field normal to

its plane. The area of the loop is 5 m2 and the rate of charge of flux density is

2 / /Wb m s2 . What is the emf appearing at the terminals of the loop ?

(A) 5 V− (B) 2 V−

(C) 0.4 V− (D) 10 V−

MCQ 6.3.13 Which of the following equations results from the circuital form of Ampere’s law ?

(A)t

E B

2

2#d =− (B) 0B :d =

(C) D : ρd = (D)t

H J D 2

2#d = +

MCQ 6.3.14 Assertion (A) : Capacitance of a solid conducting spherical body of radius a is

given by 4 a 0πε in free space.

Reason (R) : j H E J ωε#d = +

(A) Both A and R are individually true and R is the correct explanation of A.

(B) Both A and R are individually true but R is not the correct explanation of A.

(C) A is true but R is false

(D) A is false but R is true

MCQ 6.3.15 Two conducting thin coils X and Y (identical except for a thin cut in coil Y

) are placed in a uniform magnetic field which is decreasing at a constant rate.

If the plane of the coils is perpendicular to the field lines, which of the following

statement is correct ? As a result, emf is induced in

(A) both the coils

(B) coil Y only

(C) coil X only

(D) none of the two coils

MCQ 6.3.16 Assertion (A) : Time varying electric field produces magnetic fields.

Reason (R) : Time varying magnetic field produces electric fields.

(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is NOT the correct explanation of A

(C) A is true but R is false


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MCQ 6.3.17 Match List I (Electromagnetic Law) with List II (Different Form) and select the

correct answer using the code given below the lists :

List-I List-II

a. Ampere’s law 1. D v 4: ρ=

b. Faraday’s law 2. J t h

4:2

2=−

c. Gauss law 3.t

H J D

42

2# = +

d. Current 4.t

E B

42

2# =−

Codes :

a b c d

(A) 1 2 3 4(B) 3 4 1 2

(C) 1 4 3 2

(D) 3 2 1 4

MCQ 6.3.18 Two metal rings 1 and 2 are placed in a uniform magnetic field which is decreasing

with time with their planes perpendicular to the field. If the rings are identical

except that ring 2 has a thin air gap in it, which one of the following statements

is correct ?

(A) No e.m.f is induced in ring 1

(B) An e.m.f is induced in both the rings

(C) Equal Joule heating occurs in both the rings

(D) Joule heating does not occur in either ring.

MCQ 6.3.19 Which one of the following Maxwell’s equations gives the basic idea of radiation ?

(A)/

/

t

t

H D

E B

2 2

2 2

#

#

d

d

=

= 4 (B)

/

/

t

t

E B

D B :

2 2

2 2

#d

d

=−

=− 4

(C) 0

D

D

:

:

ρd

d

=

= 3 (D) / t

B

H D

:

2 2

ρ

#

d

d

=

= ^ h4MCQ 6.3.20 Which one of the following is NOT a correct Maxwell equation ?

(A)t

H D

J 2

2#d = + (B)

t E

H

2

2#d =

(C) D :d ρ= (D) 0B :d =

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MCQ 6.3.21 Match List I (Maxwell equation) with List II (Description) and select the correct

answer :

List I

a. 0d S B : = # b. d dv S D v

v : ρ= # #

c. d t

d l B

S E : $2

2=− # #

d. ( )d

t d l

D J S H : :

2

2=

+ # # List II

1. The mmf around a closed path is equal to the conduction current plus the

time derivative of the electric displacement current through any surfacebounded by the path.

2. The emf around a closed path is equal to the time derivative is equal to the

time derivative of the magnetic displacement through any surface bounded

by the path.

3. The total electric displacement through the surface enclosing a volume is

equal to total charge within the volume

4. The net magnetic flux emerging through any closed surface is zero.

Codes :

a b c d(A) 1 3 2 4

(B) 4 3 2 1

(C) 4 2 3 1

(D) 1 2 3 4

MCQ 6.3.22 The equation of continuity defines the relation between

(A) electric field and magnetic field

(B) electric field and charge density

(C) flux density and charge density

(D) current density and charge density

MCQ 6.3.23 What is the generalized Maxwell’s equationt

H J D

c 2

2#d = + for the free space ?

(A) 0H #d = (B) H J c #d =

(C)t

H D

2

2#d = (D) H D #d =

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MCQ 6.3.24 Magnetic field intensity is H 3 7 2y x a a a x y z = + + A/m. What is the current density

/A mJ 2 ?

(A) 2a y − (B) 7a z −

(C) 3a

x (D) 12a

y

MCQ 6.3.25 A circular loop placed perpendicular to a uniform sinusoidal magnetic field of

frequency 1ω is revolved about an axis through its diameter at an angular velocity

2ω rad/sec ( )<2 1ω ω as shown in the figure below. What are the frequencies for the

e.m.f induced in the loop ?

(A) and1 2ω ω

(B) , and1 2 2 2ω ω ω ω+

(C) , and2 1 2 2ω ω ω ω−

(D) and1 2 1 2ω ω ω ω− +

MCQ 6.3.26 Which one of the following is not a Maxwell’s equation ?

(A) j H E σ ωε#d = +^ h

(B) Q F E v B #= +^ h(C) d d

t d H l J S

D S

c s s : : :

2

2= + # # #

(D) d B S 0S

: = #

MCQ 6.3.27 Consider the following three equations :

1. t E B

d 2

2# =−

2.t

H J D

2

2#d = +

3. 0B :d =

Which of the above appear in Maxwell’s equations ?

(A) 1, 2 and 3 (B) 1 and 2

(C) 2 and 3 (D) 1 and 3

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MCQ 6.3.28 A straight current carrying conductor and two conducting loops A and B are shown

in the figure given below. What are the induced current in the two loops ?

(A) Anticlockwise in A and clockwise in B

(B) Clockwise in A and anticlockwise in B(C) Clockwise both in A and B

(D) Anticlockwise both in A and B

MCQ 6.3.29 Which one of the following equations is not Maxwell’s equation for a static

electromagnetic field in a linear homogeneous medium ?

(A) 0B :d = (B) 0D d # = v

(C) d I B l c

0: µ= # (D) A J 2

0µd =

MCQ 6.3.30 In free space, if 0v ρ = , the Poisson’s equation becomes(A) Maxwell’s divergence equation 0B :d =

(B) Laplacian equation 0V 2d =

(C) Kirchhoff’s voltage equation 0V Σ =

(D) None of the above

MCQ 6.3.31 Match List I with List II and select the correct answer using the codes given below

List I List II

a Continuity equation 1.

t

H J D

2

2#d = +

b Ampere’s law 2.t

J D

2

2=

c Displacement current 3.t

E B

2

2#d =−

d Faraday’s law 4.t

J v

2

2ρ#d =−

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Codes :

a b c d

(A) 4 3 2 1

(B) 4 1 2 3

(C) 2 3 4 1

(D) 2 1 4 3

MCQ 6.3.32 Match List I (Type of field denoted by A) with List II (Behaviour) and select the

correct answer using the codes given below :

List I List II

a A static electric field in a charge free region 1. 0A:d =

0A !#d

b A static electric field in a charged region 2. 0A: !d

0A#d =

c A steady magnetic field in a current carrying

conductor

3. 0A: !d

0A !#d

d A time-varying electric field in a charged medium

with time-varying magnetic field

4. 0A:d =

0A#d =

Codes :

a b c d

(A) 4 2 3 1

(B) 4 2 1 3

(C) 2 4 3 1

(D) 2 4 1 3

MCQ 6.3.33 Which one of the following pairs is not correctly matched ?

(A) Gauss Theorem : d dv D s D s v

: :d= # #

(B) Gauss’s Law : d dv D s v

: ρ= # #

(C) Coulomb’s Law : V dt

d m φ=−

(D) Stoke’s Theorem : ( )d d l s l s

: :ξ ξ#d= # #

MCQ 6.3.34 Maxwell equation ( / )t E B 2 2#d =− is represented in integral form as

(A) d t

d E l B l : :2

2=− # # (B) d

t d E l B s

s : :

2

2=− # #

(C) d t

d E l B l :2

2# =− # # (D) d

t d E l B l

s :

2

2# =− # #

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MCQ 6.3.35 The magnetic flux through each turn of a 100 turn coil is ( )t t 23 − milli-Webers

where t is in seconds. The induced e.m.f at 2t s= is

(A) 1 V (B) 1 V−

(C) 0.4 V (D) 0.4 V−

MCQ 6.3.36 Two conducting coils 1 and 2 (identical except that 2 is split) are placed in a

uniform magnetic field which decreases at a constant rate as in the figure. If the

planes of the coils are perpendicular to the field lines, the following statements are

made :

1. an e.m.f is induced in the split coil 2

2. e.m.fs are induced in both coils

3. equal Joule heating occurs in both coils

4. Joule heating does not occur in any coil

Which of the above statements is/are true ?

(A) 1 and 4 (B) 2 and 4

(C) 3 only (D) 2 only

MCQ 6.3.37 For linear isotropic materials, both E and H have the time dependence e j t ω and

regions of interest are free of charge. The value of H d # is given by

(A) E σ (B) j E ωε

(C) j E E σ ωε+ (D) j E E σ ωε−

MCQ 6.3.38 Which of the following equations is/are not Maxwell’s equations(s) ?

(A)t

J v

:2

2ρd =− (B) D v : ρd =

(C)t

E B :2

2d =− (D) d t

d H l E E s s

: :2

2σ ε= +b l # # Select the correct answer using the codes given below :

(A) 2 and 4 (B) 1 alone

(C) 1 and 3 (D) 1 and 4

MCQ 6.3.39 Assertion (A) : The relationship between Magnetic Vector potential A and the

current density J in free space is

( )A# #d d J 0µ=

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For a magnetic field in free space due to a dc or slowly varying current is A J 2

0µd =−

Reason (R) : For magnetic field due to dc or slowly varying current 0A:d = .


(B) Both A and R are true but R is NOT the correct explanation of A(C) A is true but R is false


MCQ 6.3.40 Given thatt

H J D

2

2#d = +

Assertion (A) : In the equation, the additional termt

D

2

2 is necessary.

Reason (R) : The equation will be consistent with the principle of conservation of

charge.


(B) Both A and R are true but R is NOT the correct explanation of A(C) A is true but R is false


MCQ 6.3.41 Consider coils , ,C C C 1 2 3 and C 4 (shown in the given figures) which are placed in the

time-varying electric field ( )t E and electric field produced by the coils ,C C 2 3l l and

C 4l carrying time varying current ( )I t respectively :

The electric field will induce an emf in the coils

(A) C 1 and C 2 (B) C 2 and C 3

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(C) C 1 and C 3 (D) C 2 and C 4

MCQ 6.3.42 A circular loop is rotating about the y -axis as a diameter in a magnetic field

sinB t B a Wb/mx 02ω= . The induced emf in the loop is

(A) due to transformer emf only(B) due to motional emf only

(C) due to a combination of transformer and motional emf

(D) zero

MCQ 6.3.43 Match List I (Law/quantity) with List II (Mathematical expression) and select the

correct answer :

List I List II

a. Gauss’s law 1. D : ρd =

b. Ampere’s law 2.t

E B

2

2#d =−

c. Faraday’s law 3. E H P #=

d. Poynting vector 4. q F E v B #= +^ h5.

t H J

D c

2

2#d = +

Codes :

a b c d(A) 1 2 4 3

(B) 3 5 2 1

(C) 1 5 2 3

(D) 3 2 4 1

***********

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SOLUTIONS 6.1

SOL 6.1.1 Option (D) is correct.

SOL 6.1.2 Option (B) is correct.

The line integral of magnetic field intensity along a closed loop is equal to the

current enclosed by it.

i.e. d H l : # I enc =

So, for the constant current, magnetic field intensity will be constant i.e.magnetostatic field is caused by steady currents.

SOL 6.1.3 Option (A) is correct.

From Faraday’s law the electric field intensity in a time varying field is defined as

E #d t

B

2

2=− where B is magnetic flux density in the EM field.

and since the magnetic flux density is equal to the curl of magnetic vector potential

i.e. B A#d=

So, putting it in equation (1), we get

E #d

t A

2

2#d=−

^ hor E #d

t A#d

2

2= −b l

Therefore, E t

A

2

2=−


Since total magnetic flux through a surface S is defined as

Φ d B S S

:= # From Maxwell’s equation it is known that curl of magnetic flux density is zero

B :d

0=

d B S S

: # ( )dv B 0v

:d= = # (Stokes Theorem)

Thus, net outwards flux will be zero for a closed surface.

SOL 6.1.5 Option (C) is correct.

From Faraday’s law, the relation between electric field and magnetic field is

E #d t

B

2

2=−







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Since the electric field inside a conducting sphere is zero.

i.e. E 0=

So the rate of change in magnetic flux density will be

t

B

2

2

( )E #d

=− 0=Therefore ( , )r t B will be uniform inside the sphere and independent of time.


From the integral form of Faraday’s law we have the relation between the electric

field intensity and net magnetic flux through a closed loop as

d E l : # dt d Φ=−

Since electric field intensity is zero (E 0= ) inside the conducting loop. So, the rate

of change in net magnetic flux through the closed loop is

dt

d Φ

0=i.e. Φ is constant and doesn’t vary with time.


A superconductor material carries zero magnetic field and zero electric field inside

it.

i.e. B 0= and 0E =

Now from Ampere-Maxwell equation we have the relation between the magnetic

flux density and electric field intensity as

B #d t

J E

0 0 02

2µ µ ε= +

So, J 0= (B 0= , 0E = )

Since the net current density inside the superconductor is zero so all the current

must be confined at the surface of the wire.


According to Lenz’s law the induced current I in a loop flows such as to produce

a magnetic field that opposes the change in ( )t B .

Now the configuration shown in option (A) and (B) for increasing magnetic flux

B i , the change in flux is in same direction to B i as well as the current I flowing

in the loop produces magnetic field in the same direction so it does not follow the

Lenz’s law.For the configuration shown in option (D), as the flux B d is decreasing with time

so the change in flux is in opposite direction to B d as well as the current I flowing

in the loop produces the magnetic field in opposite direction so it also does not

follow the Lenz’s law.

For the configuration shown in option (C), the flux density B d is decreasing with

time so the change in flux is in opposite direction to B d but the current I flowing

in the loop produces magnetic field in the same direction to B d (opposite to the

direction of change in flux density). Therefore this is the correct configuration.







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Induced emf in a conducting loop is given by

V emf dt d Φ=− where Φ is total magnetic flux passing through the loop.

Since, the magnetic field is non-uniform so the change in flux will be caused by itand the induced emf due to it is called transformer emf.

Again the field is in a y direction and the loop is rotating about z -axis so flux

through the loop will also vary due to the motion of the loop. This causes the emf

which is called motion emf. Thus, total induced voltage in the rotating loop is

caused by the combination of both the transformer and motion emf.


As the conducting loop is falling freely So, the flux through loop will remain

constant. Therefore, the voltage induced in the loop will be zero.

SOL 6.1.11 Option (B) is correct.The magnetic flux density passing through the loop is given as

B 4z t a x 3 2=

Since the flux density is directed normal to the plane x 0= so the total magnetic

flux passing through the square loop located in the plane x 0= is

Φ d B S := # ( )z t dydz 4z y

3 2

0

1

0

1

===

# # t 2= ( ( )d dydz S a x = )

Induced emf in a loop placed in magnetic field is defined as

V emf dt d Φ

=−

where Φ is the total magnetic flux passing through the loop. So the induced emfin the square loop is

V emf ( )dt

d t t 2

2

=− =− ( t 2Φ = )

Therefore at time t sec2= the induced emf is

V emf 4 volt=−


Magnetic flux density produced at a distance ρ from a long straight wire carrying

current I is defined as

B I

a

2

0

πρ

µ= φ

where a φ is the direction of flux density as determined by right hand rule. So the

flux density produced by straight wire at a distance ρ from it is

B I

a 2 n

0

πρµ

= (a n is unit vector normal to the loop)

Therefore the total magnet flux passing through the loop is

Φ d B S := # I

ad 2d

d a 0

πρµ

ρ=+ # (d ad S a n ρ= )

where d ρ is width of the strip of loop at a distance ρ from the straight wire. Thus,







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Φ I d

20

2

3

πµ

ρρ

= b l # lnI

2 230

πµ

= b l ( )( . )ln

25

1 50

πµ

=

( )( ) ( . )ln2 10 5 1 57#= − 4.05 10 Wb7

#= −

SOL 6.1.13

Option (A) is correct.Electric field intensity experienced by the moving conductor ab in the presence of

magnetic field B is given as

E v B #= where v is the velocity of the conductor

So, electric field will be directed from b to a as determined by right hand rule for

the cross vector. Therefore, the voltage difference between the two ends of the

conductor is given as

V ab d E l a

b

:=− # Thus, the positive terminal of voltage will be a and V ab will be positive.

SOL 6.1.14 Option (B) is correct.Given magnetic flux density through the square loop is

B 0.1 /Wb mt a z 2=

So, total magnetic flux passing through the loop is

Φ d B S := . 0.1t t 0 1 1= =^ ^h hThe induced emf (voltage) in the loop is given as

V emf 0.1 Voltdt d φ

=− =−

As determined by Lenz’s law the polarity of induced emf will be such that

V V 1 2+ V emf =−

Therefore, the voltage drop in the two resistances are respectively, V 1 ( ) . 33.3mVV

2 42

30 1

emf =+

− = =b land V 2 ( ) 66.7 mVV

2 44

emf =+

− =b lSOL 6.1.15 Option (A) is correct.

Consider a magnet bar being dropped inside a pipe as shown in figure.

Suppose the current I in the magnet flows counter clockwise (viewed from above)

as shown in figure. So near the ends of pipe, it’s field points upward. A ring

of pipe below the magnet experiences an increasing upward flux as the magnet

approaches and hence by Lenz’s law a current will be induced in it such as toproduce downward flux.

Thus, I ind must flow clockwise which is opposite to the current in the magnet.

Since opposite currents repel each other so, the force exerted on the magnet due

to the induced current is directed upward. Meanwhile a ring above the magnet

experiences a decreasing upward flux; so it’s induced current parallel to I and it

attracts magnet upward. And flux through the rings next to the magnet bar is

constant. So no current is induced in them.







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Thus, for all we can say that the force exerted by the eddy current (induced current

according to Lenz’s law) on the magnet is in upward direction which causes thedelay to reach the bottom. Whereas in the cases of unmagnetized bar no induced

current is formed. So it reaches in fraction of time.

Thus, A and R both true and R is correct explanation of A.


Voltage, V 1 N dt d

1Φ

=−

where Φ is total magnetic flux passing through it.

Again V 2 N dt d

2Φ

=−

Since both the coil are in same magnetic field so, change in flux will be same forboth the coil.

Comparing the equations (1) and (2) we get

V V

2

1 N N

2

1=

V 2 (12)V N 5000

30001

1

2Ν = = 7.2 volt=


The magnetic flux density inside a solenoid of n turns per unit length carrying


B nI 0µ=Let the length of solenoid be l and its cross sectional radius be r . So, the total

magnetic flux through the solenoid is

Φ ( )( )( )nI r nl 02µ π= (1)

Since the total magnetic flux through a coil having inductance L and carrying

current I is given as

Φ LI =

So comparing it with equation (1) we get,







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L n I l 02 2µ π=

and as for a given solenoid, radius r and length l is constant therefore

L n 2\

SOL 6.1.18

Option (C) is correct.The magnetic flux density inside the solenoid is defined as

B nI 0µ=

where n " no. of turns per unit length

I " current flowing in it.

So the total magnetic flux through the solenoid is

Φ d B S := # ( )( )nI a 02µ π=

where a " radius of solenoid

Induced emf in a loop placed in a magnetic field is defined as

V emf dt d Φ

=−

where Φ is the total magnetic flux passing through the loop. Since the resistance R

is looped over the solenoid so total flux through the loop will be equal to the total

flux through the solenoid and therefore the induced emf in the loop of resistance

will be

V emf a n dt dI 2

0π µ=−

Since current I flowing in the solenoid is constant so, the induced emf is

V emf 0=

and therefore the induced current in the loop will be zero.


It will be similar to the current in a solenoid.

So, the magnetic field will be in circumferential while the electric field is longitudinal


In Assertion (A) the magnetic flux through each turn of both coils are equal So

the net magnetic flux through the two coils are respectively

1Φ N 1Φ=

and 2Φ N 2Φ=

where Φ is the magnetic flux through a single loop of either coil and N 1, N 2 are the

total no. of turns of the two coils respectively.Therefore the induced emf in the two coils are

V emf1 dt d 1Φ

=− N dt d

1Φ=−

V emf2 dt d 2Φ

=− N dt d

2Φ

=−

Thus, the ratio of the induced emf in the two loops are

V V

emf

emf

1

2 N N

1

2=







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Now, in Reason (R) : a primitive transformer is similar to the cylinder core carrying

wound coils. It is the device in which by choosing the appropriate no. of turns, any

desired secondary emf can be obtained.

So, both the statements are correct but R is not the explanation of A.


Electric flux density in the medium is given as

D E ε= cosE t 0ε ω= ( cosE E t 0 ω= )

Therefore the displacement current density in the medium is

J d sint D

E t 02

2ωε ω= =−

and the conduction current density in the medium is

J c cosE E t 0σ σ ω= =

So, the ratio of amplitudes of conduction current density and displacement current

density is

J

J

d

c

ωεσ

=


The displacement current density in a medium is equal to the rate of change in

electric flux density in the medium.

J d t

D

2

2=

Since the displacement current density in the medium is given as

J d 20 . /cos A mt a 1 5 10 y 8 2

#= ^ hSo, the electric flux density in the medium is D dt C J d = + # (C " constant)

20 .cos t dt C a 1 5 10 y 8

#= +^ h # As there is no D.C. field present in the medium so, we get C 0= and thus,

D .

.sin t a

1 5 10

20 1 5 10y 8

8

#

#=

^ h 1.33 10 .sin t a 1 5 10 y

7 8##= − ^ h

133.3 . /sin nC mt a 1 5 10 y 8 2

#= ^ hSOL 6.1.23 Option (C) is correct.

Given the volume charge density, v ρ 0=So, from Maxwell’s equation we have

D :d v ρ=

D :d 0= (1)

Now, the electric flux density in a medium is defined as

D E ε= (where ε is the permittivity of the medium)

So, putting it in equation (1) we get,

( )E : εd 0=

or, ( ) ( )E E : :ε εd d+ 0=







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and since d

εε 0. 0&4 .ε (given)

Therefore, E :d 0.


The ratio of magnitudes of displacement current to conduction current in any

medium having permittivity ε and conductivity σ is given as

Conduction current

Displacement current

σωε

=

where ω is the angular frequency of the current in the medium.

Given frequency, f 50GHz=

Permittivity, ε .4 4 8 85 10012

# #ε= = −

Conductivity, σ 1.14 10 /s m8#=

So, ω f 2 2 50 10 100 109 9# # #π π π= = =

Therefore, the ratio of magnitudes of displacement current to the conduction

current is

I I

c

d .

.1 14 10

100 10 4 8 85 108

9 12

#

# # # #π=

−

.9 75 10 8#= −


Given the electric field intensity in time domain as

E sin cos

r t kr

a θ ω

= −

θ

^ hSo, the electric field intensity in phasor form is given as

E s sin

r e a

jkr θ= θ

−

and E s #d r kdr

rE a 1

s 2

= θ φ^ h sin jk

r e a

jkr θ= − φ

−^ hTherefore, from Maxwell’s equation we get the magnetic field intensity as

H s j r E s

0

#d

ω=− sin

r k

r e a

jkr

0ωθ

= φ−


In phasor form the magnetic field intensity can be written as

H s 0.1 /cos A my e a 15 jbx z π= −^ h

Similar as determined in MCQ 42 using Maxwell’s equation we get the relation

b15 2 2π +^ h 20 0ω π ε=

Here ω 6 109

#π=

So, b15 2 2π +^ h 3 106 10

8

9 2

#

#π= c m

b15 2 2π +^ h 400 2π=

b2 175 2π= & b 41.6 /rad m!=

***********







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SOLUTIONS 6.2


Induced emf. in the conducting loop formed by rail, bar and the resistor is given by

V emf dt d Φ=−

where Φ is total magnetic flux passing through the loop.

Consider the bar be located at a distance x from the resistor at time t . So the total

magnetic flux passing through the loop at time t is

Φ d B S := # Blx = (area of the loop is S lx = )

Now the induced emf in a loop placed in magnetic field is defined as

V emf dt d Φ

=−

where Φ is the total magnetic flux passing through the loop. Therefore the induced

emf in the square loop is

V emf ( )dt d

Blx =− Bl dt dx

=− ( Blx Φ = )

Since from the given figure, we have l 2 m= and B 0.1 /Wb m2=

and /dx dt = velocity of bar 5 /m s=

So, induced emf is

V emf ( . )( )( )0 1 2 5=− 1 volt=−

According to Lenz’s law the induced current I in a loop flows such as to produce

magnetic field that opposes the change in ( )t B . As the bar moves away from the

resistor the change in magnetic field will be out of the page so the induced current

will be in the same direction of I shown in figure.

Thus, the current in the loop is

I RV emf =− ( )101=− − 0.01A= (R 10Ω = )




B I

a 2

0

πρµ

= φ

where a φ is the direction of flux density as determined by right hand rule. So, the

magnetic flux density produced by the straight conducting wire linking through the







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loop is normal to the surface of the loop.

Now consider a strip of width d ρ of the square loop at distance ρ from the wire for

which the total magnetic flux linking through the square loop is given as

Φ d B S

S

:= #

( )I

ad 2

1a 0

πµ

ρ ρ=

ρ

ρ + # (area of the square loop is dS ad ρ= )

lnIa a

20

πµ

ρρ

= +b l

The induced emf due to the change in flux (when pulled away) is given as

V emf dt d Φ=− ln

Ia dt d a

20

πµ

ρρ

=− +b l; E

Therefore, V emf Ia

a dt d

dt d

21 10

πµ

ρρ

ρρ

=−+

−c m

Given dt d ρ

5 /velocity of loop m s= =

and since the loop is currently located at 3 m distance from the straight wire, so

after . sec0 6 it will be at

ρ 3 (0.6) v #= + ( velocity of the loopv " )

.3 0 6 5#= + 6 m=

So, V emf ( )

( ) ( )230 2

81 5

61 50 # #

πµ

=− −: D ( 2 , 30m Aa I = = )

25 10 volt7#= − 2.5 voltµ=


Since total magnetic flux through the loop depends on the distance from thestraight wire and the distance is constant. So the flux linking through the loop wil

be constant, if it is pulled parallel to the straight wire. Therefore the induced emf

in the loop is

V emf dt d Φ

=− 0= (Φ is constant)




B I

a

2

0

πρ

µ= φ

where a φ is the direction of flux density as determined by right hand rule.

Since the direction of magnetic flux density produced at the loop is normal to the

surface of the loop So, total flux passing through the loop is given by

Φ d B S S

:= # I

ad 2

0

2

4

πρµ

ρ=ρ =

c ^m h # (dS ad ρ= )

Ia d

20

2

4

πµ

ρρ

= # lnI

22

20

πµ

= ( )lnI

20

πµ

=

The current flowing in the loop is I loop and induced e.m.f. is V emf .







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So, V emf I Rdt d

loopΦ

= =−

( )dt

dQ R ( )ln

dt dI 20

πµ

=−

where Q is the total charge passing through a corner of square loop.

dt

dQ ( )ln

dt dI

420

πµ

=− ( 4R Ω= )

dQ ( )ln dI 4

20

πµ

=−

Therefore the total charge passing through a corner of square loop is

Q (2)ln dI 4

0

4

0

πµ

=− # ( )( )ln4

2 0 40

πµ

=− −

( )ln4

4 4 10 27

# #π

π=

−

2.77 10 C7#= − 277 nC=


Since the radius of small circular loop is negligible in comparison to the radius ofthe large loop. So, the flux density through the small loop will be constant and

equal to the flux on the axis of the loops.

So, B I

z R

Ra

2 2 2 / z 0

3 2

2µ=

+^ hwhere R " radius of large loop 5 m=

z " distance between the loops 12 m=

B ( )

a 2

2

12 5

5/ z

0

2 2 3 2

2#µ

#=+^ ^h h6 @

13 a

25z 3

0µ= ^ h

Therefore, the total flux passing through the small loop is Φ d B S := # r

1325

30 2

#µ

π= ^ h wherer is radius of small circular loop.

13

25 4 10 103

73 2# #

#π

π=−

−

^ ^h h 44.9 fWb=


Electric field in any medium is equal to the voltage drop per unit length.

i.e. E d

V =

where V " potential difference between two points.

d "

distance between the two points.The voltage difference between any two points in the medium is

V 2cosV ft 0 π=

So the conduction current density in the medium is given as

J c E σ= ( "σ conductivity of the medium)

E ρ

= ( "ρ resistivity of the medium)

2cos

d V

d V ft 0

ρ ρπ

= = ( )cosV V ft 20 π=







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or, J c d V 0ρ

=

and displacement current density in the medium is given as

J

d t

D

2

2=

t

E

2

2ε=

( )cos

t d

V ft 20

2

2ε

π=

; E ( )cosV V ft

20 π=

( )sind V

ft ft 2 20επ π= −6 @

or, J d 2

d f V 0π ε

=

Therefore, the ratio of amplitudes of conduction current and displacement current

in the medium is

I

I

d

c

J

J

d

C

= ( )/( )( )/( )

d f V

V d

2 0

0

π ε

ρ=

f 21

π ερ=

2 (1.6 10 ) (54 8.85 10 ) 0.77

18 12

π# # # # # #

= −

.2 7=


Total magnetic flux through the solenoid is given as

Φ nI 0µ=

where n is the no. of turns per unit length of solenoid and I is the current flowing

in the solenoid.

Since the solenoid carries current that is increasing linearly with time

i.e. I t \

So the net magnetic flux through the solenoid will be

Φ t \

or, Φ kt = where k is a constant

Therefore the emf induced in the loop consisting resistances RA, RB is

V emf dt d Φ=−

V emf k =−

and the current through R1 and R2 will be

I ind R Rk

1 2=−

+

Now according to Lenz’s law the induced current I in a loop flows such as to

produce a magnetic field that opposes the change in ( )t B .i.e. the induced current in the loop will be opposite to the direction of current in

solenoid (in anticlockwise direction).

So, V A I RR R

kRind A

A B

A= =−+

and V B I RR R

kRind B

A B

B =− =+b l

Thus, the ratio of voltmeter readings is

V V

B

A RR

B

A=−







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Induced emf in the conducting loop formed by rail, bar and the resistor is given by

V emf dt d Φ=−

where Φ is total magnetic flux passing through the loop.The bar is located at a distance x from the resistor at time t . So the total magnetic

flux passing through the loop at time t is

Φ d B S := # Blx = where l is separation between the rails

Now the induced emf in a loop placed in magnetic field is defined as

V emf dt d Φ=−

where Φ is the total magnetic flux passing through the loop. Therefore the induced

emf in the square loop is

V emf ( )dt

d Blx =− Bl

dt

dx =− ( Blx Φ = )

Since from the given figure, we have

l 5 m=

B 2 T=

and /dx dt " velocity of bar 4 /m s=

So, induced emf is

V emf ( )( )( )2 5 4=− volt40=−

Therefore the current in the bar loop will be

I R

V emf = 1040

=− 4 A=−

SOL 6.2.9 Option (B) is correct.As obtained in the previous question the current flowing in the sliding bar is

I 4 A=−

Now we consider magnetic field acts in a x direction and current in the sliding bar

is flowing in a z + direction as shown in the figure.

Therefore, the force exerted on the bar is

F Id l B #= # ( 4 ) (2 )dz a a z x 0

5

#= − # 8 z a y

05=− 6 @ 40 Na y =−







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i.e. The force exerted on the sliding bar is in opposite direction to the motion of

the sliding bar.


Given the magnetic flux density through the square loop is B 7.5 (120 30 )cos t a z cπ= −

So the total magnetic flux passing through the loop will be

Φ d B S S

:= # 7.5 (120 30 ) (1 1)( )cos t a a z z cπ #= − − −6 @ . ( )cos t 7 5 120 30cπ= −

Now, the induced emf in the square loop is given by

V emf dt d Φ

=− . ( )sin t 7 5 120 120 30# cπ π= −

The polarity of induced emf (according to Lenz’s law) will be such that induced

current in the loop will be in opposite direction to the current ( )I t shown in the

figure. So we have

( )I t R

V emf =−

. (120 30 )sin t 500

7 5 120# cπ

π=− − ( 250 250 )R 500 Ω = + =

. ( )sin t 5 7 120 30cπ=− −


As shown in figure the bar is sliding away from origin.

Now when the bar is located at a distance dx from the voltmeter, then, the vector

area of the loop formed by rail and the bar is d S (20 10 )( )dx a z

2#= −

So, the total magnetic flux passing through the loop is

Φ d B S S

:= # (8 )(20 10 )x dx a a z z

x 2 2

0#= − #

. .t t

3

1 6 1 0 4 2 3

=+^ h8 B

Therefore, the induced e.m.f. in the loop is given as

V emf . 3 . (1 1.2 )

dt d t t t

31 6 0 4 3 2 2Φ

# #=− =− + +^ h V emf 1.6 . . ( . )( . )0 4 0 4 1 1 2 0 44 2 2

#=− + +^ ^h h6 6@ @ ( . sect 0 4= )

0.35 volt=−

Since the voltmeter is connected in same manner as the direction of induced emf

(determined by Lenz’s law).

So the voltmeter reading will be

V 0.35 voltV emf = =−


Since the position of bar is give as

x .t t 1 0 4 2= +^ h

So for the position 12 cmx = we have







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.0 12 .t t 1 0 4 2= +^ hor, t . sec0 1193=

As calculated in previous question, the induced emf in the loop at a particular time

t is

V emf . . .t t t 1 6 0 4 1 1 23 2 2=− + +^ ^h h6 @So, at t . sec0 1193= ,

V emf . ( . ) . . . .1 6 0 1193 0 4 0 1193 1 1 2 0 11933 2 2=− + +^ ^ ^h h h7 6A @ .0 02344=− 23.4 mV=−

Since the voltmeter is connected in same manner as the direction of induced emf

as determined by Lenz’s law. Therefore, the voltmeter reading at 12 cmx = will be

V V emf = 23.4 mvolt=−


Consider the mutual inductance between the rectangular loop and straight wire be

M . So applying KVL in the rectangular loop we get,

M dt di 1 L

dt di

Ri 22= + ...(1)

Now from the shown figure (b), the current flowing in the straight wire is given as

i 1 ( ) ( )I u t I u t T 1 1= − − (I 1 is amplitude of the current)

or,dt di 1 ( ) ( )I t I t T 1 1δ δ = − − (2)

So, at t 0= dt di 1 I 1=

and MI 1 L

dt

di Ri 2

2= + (from equation (1))

Solving it we get

i 2 LM

I e /R L t 1= −^ h for t T 0 < <

Again in equation (2) at t T = we have

dt di 1 I 1=−

and MI 1− Ldt di

Ri 22= + (from equation (1))

Solving it we get

i 2 L

M I e / ( )R L t T 1=− − −^ h for t T >

Thus, the current in the rectangular loop is

i 2 L

M I e t T

LM

I e t T

0 < <

>

/

( / )( )

R L t

R L t T

1

1

=

−

−

− −

^ hZ

[

\

]]

]]

Plotting i 2 versus t we get







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Total magnetic flux passing through the loop formed by the resistance, bar and the

rails is given as:

Φ d B S S

:= #

B S := . 0.5(1 )cos t y a a 0 2 x x :ω= −6 6@ @ 0.1 . cos cost t 1 0 5 1 ω ω= − −^ h6 @ (y 0.5 cos mt 1 ω= −^ h )

. cos cost t 0 05 1ω ω= +^ h . cos cost t 0 05 2ω ω= +^ hSo, the induced emf in the loop is

V emf dt d Φ=−

and as determined by Lenz’s law, the induced current will be flowing in opposite

direction to the current i . So the current i in the loop will be

i R

V emf =− R dt

d 1 Φ=− −b l

. sin cos sint t t 50 05 2ω ω ω ω ω= − −6 @ 0. sin cost t 01 1 2ω ω ω=− +^ h


Given the electric flux density in the medium is

D 1.33 . /sin C mt x a 3 10 0 2 y 8 2

# µ= −^ hSo, the electric field intensity in the medium is given as

E D

ε= where ε is the permittivity of the medium

or, E D

r 0ε ε= .

. .sin t x a 10 8 85 10

1 33 10 3 10 0 2y 12

6 8

# #

# #

= −

−

−

^ h ( 10r ε = )

1.5 10 .sin t x a 3 10 0 2 y 4 8

##= −^ hNow, from maxwell’s equation we have

E #d t

B

2

2=−

or,t

B

2

2 E #d=−

x

E a

y z

2

2=−







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( 0.2) . .cos t x a 1 5 10 3 10 0 2 y 4 8

# ##=− − −^ ^h h 3 10 .cos t x a 3 10 0 2 y

3 8##= −^ h

Integrating both sides, we get the magnetic flux density in the medium as

B

3 10 .cos t x a

3 10 0 2 y

3 8##

= −^ h # .sin t x a

3 103 10 3 10 0 2 y 8

38

#

##= −^ h

10 .sin t x a 3 10 0 2 y 5 8

#= −− ^ h Tesla

Therefore the magnetic field intensity in the medium is

H B B

r 0µ µ µ= =

.sin t x

2 4 10

10 3 10 0 27

5 8

# #

#

π=

−−

− ^ h 2r µ =

Thus H 4 . /sin A mt x a 3 10 0 2 y 8

#= −^ hSOL 6.2.16 Option (D) is correct.

Given the magnetic field intensity in the medium is

H /cos A mt bx a 10 z 10= −^ h

Now from the Maxwell’s equation, we have

H #d t

D

2

2=

or,t

D

2

2 x

H a z

y 2

2=− sinb t bx a 10 y

10=− −^ h D sinb t bx dt C 1010= − − +^ h # where C is a constant.

Since no D.C. field is present in the medium so, we get C 0= and therefore,

D /cos C mbt bx a

1010 y 10

10 2= −^ hand the electric field intensity in the medium is given as

E D

ε=

.cosb

bx a 0 12 10 10

10 t y 9 10

10

# #= −− ^ h (ε 0.12 /nF m= )

Again From the Maxwell’s equation

E #d t

B

2

2=−

or,t

B

2

2 .

cosb bx a 1 2

10 t y

10#d=− −^ h: D

. sinb bx a 1 2 10 t z

2

10=− −^ hSo, the magnetic flux density in the medium is

B .

sinbbx dt a

1 210 t

z

210=− −^ h #

( . )

cosbt bx a

1 2 1010 z 10

210

#= −^ h (1)

We can also determine the value of magnetic flux density as :

B H µ=







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(3 10 )cos t bx a 10 z 5 10

#= −− ^ h (2)

Comparing the results of equation (1) and (2) we get,

( . )

b1 2 1010

2

# 3 10 5

#= −

b2 .3 6 105#=

b 600 /rad m!=


The magnetic vector potential for a direct current flowing in a filament is given as

A RI

dx a 4 x

0

πµ

= # Here current ( )I t flowing in the filament shown in figure is varying with time as

( )I t 8 At =

So, the retarded vector potential at the point P will be given as

A /

R

I t R c dx a

4 x 0

π

µ=

−^ h # where R is the distance of any point on the filamentary current from P as shownin the figure and c is the velocity of waves in free space. So, we have

R x 42= + and 3 10 /m sc 8#=

Therefore, A /

R

t R c dx a

48

.

.

x x

0

0 1

0 1

π

µ=

−

=−

^ h #

x

t dx

c dx

48

41

.

.

.

.0

20 1

0 1

0 1

0 1

πµ

=+

−−−

< F # #

8 10 lnt x x x 43 108 10

.

.

..7 2

0 1

0 1

8

7

0 10 1

#

##= + + −−

−

−

−^ h8 6B @

. .

. .

.lnt 8 10 0 1 4 01

0 1 4 01

0 53 10

7 15# #

= − +

+

−

− −

e o .t 8 10 0 53 108 15# #= −− −

or, A . /nWb mt a 80 5 3 10 x 7

#= − −^ h (1)

So, when A 0= t 6.6 10 6.6 secn9#= =−

and when t 0= A 5.3 10 /nWb m7#=− −

From equation (1) it is clear that A will be linearly increasing with respect to time

Therefore the plot of A versus t is







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Note : Time varying potential is usually called the retarded potential.


Given

Retarded scalar potential, V volty x ct = −^ hand retarded vector potential,

A /Wb my c

x

t a

x = −a kNow the magnetic flux density in the medium is given as

B A#d=

y

Aa

y z

2

2=− Teslat

c x

a z = −a k (1)

So, the magnetic field intensity in the medium is

H B

0µ= ( 0µ is the permittivity of the medium)

/A mt c x

a 1

z 0µ

= −a k (2)

and the electric field intensity in the medium is given as

E V t

A

2

2d=− −

x ct y y a a a y x x =− − − +^ h ct x a y = −^ h (3)

So, the electric flux density in the medium is

D E 0ε= ( 0ε is the permittivity of the medium)

/C mct x a y 02ε= −^ h (4)

Now we determine the condition for the field to satisfy all the four Maxwell’s

equation.

(i) D :d v ρ=

or, v ρ ct x a y 0: εd= −^ h6 @ (from equation (4)) 0=

It means the field satisfies Maxwell’s equation if 0v ρ = .

(ii) B :d 0=

Now, B :d t c x

a z :d= −a k9 C 0= (from equation (1))

So, it already, satisfies Maxwell’s equation

(iii) H #d t

J D

2

2= +







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Now, H #d x

H a z

y 2

2=−

c a a

1y y

0 0

0

µ µε

= = (from equation (2))

and from equation (4) we have

t

D

2

2 c a a y y 00

0εµ

ε= = (Since in free space c

1

0 0µ ε= )

Putting the two results in Maxwell’s equation, we get the condition

J 0=

(iv) E #d t

B

2

2=−

Now E #d x

E a a

y z z

2

2= =−

t

B

2

2 a z =

So, it already satisfies Maxwell’s equation. Thus, by combining all the results

we get the two required conditions as 0J = and 0v ρ = for the field to satisfy

Maxwell’s equation.


Given the electric field in time domain as

E 5 sin cosy bx a 10 6 10 z 9

#π π= −^ ^h hComparing it with the general equation for electric field intensity given as

E cosE t x a 0 z ω β = −^ hWe get, ω 6 109

#π=

Now in phasor form, the electric field intensity is

E s 5sin y e a 10 jbx z π= −^ h (1)

From Maxwell’s equation we get the magnetic field intensity as

H s j E

1s

0#d

ωµ=− ^ h j

y E

x E

a a sz x

sz y

0 2

2

2

2

ωµ= −< F

cos sin j

y e j b y e a a 50 10 5 10 jbx x y

jbx

0ωµ π π π= +− −^ ^h h6 @

Again from Maxwell’s equation we have the electric field intensity as

E s j H

1s

0#d

ωε= ^ h

j x

H

y H

a 1 sy sx

z 0 2

2

2

2

ωε= −< F

( )( ) ( ) ( )( ) ( )sin sin j b jb y e y e a 1 5 10 50 10 10 jbx jbx

z 20 0ω µ ε

π π π π= − +− −6 @

10sinb ye a

1

5 500

jbx

z 2 0 0

2 2

ω µ ε π π= +

−

6 @Comparing this result with equation (1) we get

b1 5 5002

0 0

2 2

ω µ ε π+^ h 5=

or, b 1002 2π+ 20 0ω µ ε=

b 1002 2π+ 6 103 10

19 2

8 2# ##

π= ^ ^h h ,c

6 10 190 0#ω π µ ε= =b l

b 1002 2π+ 400 2π=







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b2 300 2π=

b /rad m300! π=


The force experienced by a test charge q in presence of both electric fieldE

andmagnetic field B in the region will be evaluated by using Lorentz force equation as

F q v E B #= +^ hSo, putting the given three forces and their corresponding velocities in above

equation we get the following relations

q a a y z +^ h q E a B x #= +^ h (1)

q a y q E a B y #= +^ h (2)

2q a a y z +^ h q E a B z #= +^ h (3)

Subtracting equation (2) from (1) we get

a z a a B x y #= −^ h (4)

and subtracting equation (1) from (3) we get a y a a B z x #= −^ h (5)

Now we substitute B B B B a a a x x y y z z = + + in eq (4) to get

a z B B B B a a a a y z z y x z z x = − + −

So, comparing the , andx y z components of the two sides we get

B B x y + 1=

and B z 0=

Again by substituting B B B B a a a x x y y z z = + + in eq (5), we get

a y B B B B a a a a x y y x y z z y = − − +

So, comparing the , andx y z components of the two sides we get

B B x z + 1=and B y 0=

as calculated above B 0z = , therefore B 1x =

Thus, the magnetic flux density in the region is

B /Wb ma x 2= (B 1x = , B B 0y z = = )


As calculated in previous question the magnetic flux density in the region is

B /Wb ma x 2=

So, putting it in Lorentz force equation we get

F

q E V B #

= +^ hor, q a a y z +^ h q E a a x x #= +^ hTherefore, the electric field intensity in the medium is

E /V ma a y z = +


Given the magnetic flux density through the loop is

B 2/x a z =−

So the total magnetic flux passing through the loop is given as







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Φ d B S := # x

dxdy a a 2

z z y

y

x

x 22

:= − −++ b ^l h # #

lnx

x 2 2 2= +b ^l h ln

x x 4 2

= +b l

Therefore, the circulation of induced electric field in the loop is d E l

C

: # dt d Φ

=− lndt d

x x 4 2

=− +b l; E

x x dt

d x

x 2

4 2=−

++

b bl l

x x

x dt dx

24 2

2=−+

−b l

x x 28 2=+^ ^h h

x x 216=+^ h dt

dx v a 2 x = =b l

SOL 6.2.23

Option (A) is correct.As the magnetic flux density for 4<ρ is 0B = so, the total flux passing through

the closed loop defined by 4 mρ = is

Φ 0d B S := = # So, the induced electric field circulation for the region 4 m<ρ is given as

d E l C

: # 0dt d Φ

=− =

or, E 0= for 4 m<ρ


As the magnetic field for the region 4 m<ρ and 5 m>ρ is zero so we get the

distribution of magnetic flux density as shown in figure below.

At any distance ρ from origin in the region 4 5 m< <ρ , the circulation of induced

electric field is given as

d E l C

: # dt d Φ=−

dt d d B S :=− b l #

sindt d

t 2 42 2ω πρ π=− −^ h8 B







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cos t 2 162ω ω πρ π=− −^ hor, E 2πρ^ h cos t 2 162ω ω πρ π=− −^ h E

cos t

2

2 162

ρ

ρ ω ω=−

−^ hSo, the induced electric field intensity at 4.5 mρ = is

E .

( . ) cos t 4 52 4 5 162 ω ω=− −^ h

cos t 1817

ω ω=−


For the region 5 m>ρ the magnetic flux density is 0 and so the total magnetic flux

passing through the closed loop defined by 5 mρ = is

Φ d B S

0

5

:= # d d B S B S

4

5

0

4

: := + # # sin t d a S 0 2 z

4

5

:ω= + ^ h # sin t 2 5 42 2

ω π π= −^ ^ ^h h h8 B sin t 18π ω=

So, the circulation of magnetic flux density for any loop in the region 5 m>ρ is

d E l : # dt d ψ

=−

( )E 2πρ sindt d

t 18π ω=− ^ h cos t 18πω ω=−

So, the induced electric field intensity in the region 5 m>ρ is

E cos t a 218 πρπω ω= − φ

cos t a 9ρ

ω ω=− φ


Let the test charge be q coulomb So the force presence of experienced by the test

charge in the presence of magnetic field is

F q v B #= ^ h ...(i)

and the force experienced can be written in terms of the electric field intensity as

F q E =

Where E is field viewed by observer moving with test charge.Putting it in Eq. (i)

q E q v B #= ^ h E 2a a z #ωρ= φ^ ^h hwhere ω is angular velocity and ρ is radius of circular loop.

a 2 2 2= ρ^ ^ ^h h h 8 /V ma = ρ


Let the point change located at origin be Q and the current I is flowing out of the







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page through the closed triangular path as shown in the figure.

As the current I flows away from the point charge along the wire, the net charge

at origin will change with increasing time and given as

dt

dQ I =−

So the electric field intensity will also vary through the surface and for the varying

field circulation of magnetic field intensity around the triangular loop is defined as

d H l : # I I d enc c enc = +6 6@ @

where I c enc 6 @ is the actual flow of charge called enclosed conduction current and

I d enc 6 @ is the current due to the varying field called enclosed displacement current

which is given as

I d enc 6 @ dt d

d E S S

0 :ε= ^ h # dt d

d D S S

:= # (1)

From symmetry the total electric flux passing through the triangular surface is

d D S S

: # Q 8

=

So, I d enc 6 @ dt d Q

dt dQ I

8 81

8= = =−b l (from equation (1))

Where as I c enc 6 @ I =

So, the net circulation of the magnetic field intensity around the closed triangular

loop is

d H l C

: # I I d enc c enc = +6 6@ @

I

I 8=− + 78 8= ^ h 7 A= (I 8 A= )


The distribution of magnetic flux density and the resistance in the circuit are same

as given in section A (Q. 31) so, as calculated in the question, the two voltage drops

in the loop due to magnetic flux density 0.1t B a z = are

V 1 33.3mV=

and V 2 66.67 66.7mV mV= =

Now V 3 (voltmeter) which is directly connected to terminal cd is in parallel to







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both V 2 and V 1. It must be kept in mind that the loop formed by voltmeter V 3 and

resistance 2 Ω also carries the magnetic flux density crossing through it. So, in this

loop the induced emf will be produced which will be same as the field produced in

loop abcd at the enclosed fluxes will be same.

Therefore as calculated above induced emf in the loop of V 3 is

V emf 100 mV=

According to lenz’s law it’s polarity will be opposite to V 3 and so

V emf − V V 1 3= +

or, V 3 .100 33 3= − 66.7 mV=


The induced emf in a closed loop is defined as

V emf dt d Φ=−

where Φ is the total magnetic flux passing through the square loopAt any time t , angle between B and d S is θ since B is in a y direction so the total

magnetic flux passing through the square loop is

Φ d B S := # cosB S θ= ^ ^h h cos5 10 20 10 20 103 3 3

# # # # θ= − − −^ ^h h cos2 10 6

# θ= −

Therefore the induced emf in the loop is

V emf dt d Φ=− cos

dt d 2 10 6

# θ=− − ^ h sindt d 2 10 6

# θ θ= −

and asdt d θ = angular velocity 2 / secrad=

So, V emf sin2 10 26# θ= −^ ^h h 4 10 /sin V m6 θ#= − 4 /sin V mµθ=


As calculated in previous question the induced emf in the closed square loop is

V emf 4 /sin V mµθ=

So the induced current in the loop is

I R

V emf = where R is the resistance in the loop.

sin

40 10

4 103

6

#

#θ=

−

−

( 40 mR Ω=

) 0.1 sin mAθ=


The total magnetic flux through the square loop is given as

Φ d B S := # sin cosB t S 0 ω θ= ^ ^h hSo, the induced emf in the loop is

V emf dt d Φ

=− ( )( )sin cosdt d

B t S 0 ω θ=− 6 @







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sin cosB S dt d

t t 0 ω ω=− 6 @ ( t θ ω= )

2cosB S t 0 ω=−

Thus, the maximum value of induced emf is

V emf B S 0 ω=


As calculated in previous question the maximum induced voltage in the rotating

loop is given as

V emf B S 0 ω=

From the given data, we have

B 0 0.25 /Wb m2=

S 64 64 10cm m2 4 2#= = −

and ω 60 2 377 /secradπ#= = (In one revolution 2π radian is covered)

So, the r.m.s. value of the induced voltage is V

. .emf

r m s 6 @ V B S 2

12

1emf 0 ω= = .

21 0 25 64 10 3774

# # #= −^ h .0 4265=

Since the loop has 50 turns so net induced voltage will be 50 times the calculated

value.

i.e. V . .

emf r m s 6 @ .50 0 4265#= ^ h 21.33 volt=


e.m.f. induced in the loop due to the magnetic flux density is given as

V emf

t 2

2Φ=− cos

t

t 10 120 2

2

2π πρ=−

^ ^h h sin t 10 10 120 10 1202 2# #π π π=− −−^ ^ ^h h h

sin t 12 1202π π=

As determined by Lenz’s law the polarity of induced e.m.f will be such that b is at

positive terminal with respect to a .

i.e. V ba sinV t 12 120emf 2π π= =

or V ab sin t 12 1202π π=− 118.43 120sin Voltt π=−


As calculated in previous question, the voltage induced in the loop is

V ab sin t 12 1202π π=−

Therefore, the current flowing in the loop is given as

I t ^ h V 250

ab=− sin t 250

12 1202π π=

. sin t 0 47 120π=

***********







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SOLUTIONS 6.3


Given, the magnetic flux density in air as

B B x y

x a x y

y a y x 0 2 2 2 2

=+

−+

c m ...(1)

Now, we transform the expression in cylindrical system, substituting

x cosr φ= and y sinr φ=

a x cos sina a r φ φ= − φ

and a y sin cosa a r φ φ= + φ

So, we get B B a 0= φ

Therefore, the magnetic field intensity in air is given as

H a B B

0 0

0

µ µ= =

φ , which is constant

So, the current density of the field is

J 0H #d= = (since H is constant)


Maxwell equations for an EM wave is given as

B :d 0=

E :d v

ερ

=

E d # t

B

2

2=−

H d # t

D J

2

2= +

So, for static electric magnetic fields

B :d 0=

E :d /v ρ ε=

E d # 0= 0t

B 2

2 =b l H d # J =

t D 0

2

2=b l


H #d J t

D

2

2= + Maxwell Equations

d H S S

# :d h ## t

d J D

S S

:

2

2= +` j ## Integral form







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d H l : # t

d J D

S S

:2

2= +b l ## Stokes Theorem


From Maxwells equations we have

H d # t

J D

2

2= +

Thus, H d # has unit of current density J (i.e., /A m2)


This equation is based on Ampere’s law as from Ampere’s circuital law we have

d H l l

$ # I enclosed=

or, d H l l

$ # d J S S

:= # Applying Stoke’s theorem we get

d H S S

$#d^ h # d J S S

:= # H #d J =

Then, it is modified using continuity equation as

H #d t

J D

2

2= +


When a moving circuit is put in a time varying magnetic field induced emf have

two components. One due to time variation of magnetic flux density B and other

due to the motion of circuit in the field.

SOL 6.3.7

Option (C) is correct.From maxwell equation we have

H #d t

J D

2

2= +

The termt

D

2

2 defines displacement current.


Emf induced in a loop carrying a time varying magnetic flux Φ is defined as

V emf dt d Φ=−

9dt d t

31 3λ=−

b l 9 t 2λ=−

at time, 3 st = , we have

9 3 2λ=− ^ h

λ 1 /Wb s2=−


According to Lenz’s law the induced emf (or induced current) in a loop flows such

as to produce a magnetic field that opposed the change in B . The direction of the

magnetic field produced by the current is determined by right hand rule.







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Now, in figure (1), B directed upwarded increases with time where as the field

produced by current I is downward so, it obey’s the Lenz’s law.

In figure (2), B directed upward is decreasing with time whereas the field produced

by current I is downwards (i.e. additive to the change in B ) so, it doesn’t obey

Lenz’s law.

In figure (3), B directed upward is decreasing with time where as current I produces

the field directed upwards (i.e. opposite to the change in B ) So, it also obeys Lenz’s

law.

In figure (4), B directed upward is increasing with time whereas current I produces

field directed upward (i.e. additive to the change in B ) So, it doesn’t obey Lenz’s

law.

Thus, the configuration 1 and 3 are correct.


Faraday’s law states that for time varying field, E #d

t B

2

2=−

Since, the curl of gradient of a scalar function is always zero

i.e. V #d d h 0=

So, the expression for the field, E V d=− must include some other terms is

E V t

Ad

2

2=− −

i.e. A is true but R is false.


Faraday develops the concept of time varying electric field producing a magneticfield. The law he gave related to the theory is known as Faraday’s law.


Given, the area of loop

S 5 m2=

Rate of change of flux density,

t B

2

2 2 / /Wb m S2=

So, the emf in the loop is

V emf

t

d B S :2

2=− # 5 2= −

^ ^h h 10 V=−


The modified Maxwell’s differential equation.

H #d t

J D

2

2= +

This equation is derived from Ampere’s circuital law which is given as

d H l : # I enc =

d H S :#d h # d J S = # H #d J =







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Electric potential of an isolated sphere is defined as

C a 4 0πε= (free space)

The Maxwell’s equation in phasor form is written as

H #d j E E ωε σ= + j E J ωε= + J E σ=^ hSo A and R both are true individually but R is not the correct explanation of A.


If a coil is placed in a time varying magnetic field then the e.m.f. will induce in coil

So here in both the coil e.m.f. will be induced.


Both the statements are individually correct but R is not explanation of A.


Ampere’s lawH

#d

t J

D

2

2

= + 3a"

^ hFaraday’ law E #d

t B

2

2= 4b "^ h

Gauss law D :d v ρ= 1c "^ hCurrent continuity J :d

t 2

2ρ=− 2d "^ h


Since, the magnetic field perpendicular to the plane of the ring is decreasing with

time so, according to Faraday’s law emf induced in both the ring is

V emf t

d B S :2

2=− #

Therefore, emf will be induced in both the rings.


The Basic idea of radiation is given by the two Maxwells equation

H #d t

D

2

2=

E #d t

B

2

2=−


The correct maxwell’s equation are

H #d t

J D

2

2= + D : ρd =

E #d t

B

2

2=− 0B :d =


In List I

a. d S B : # 0=

The surface integral of magnetic flux density over the closed surface is zero or in

other words, net outward magnetic flux through any closed surface is zero. 4a "^ hb. d S D : # dv v

v

ρ= #







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Total outward electric flux through any closed surface is equal to the charge enclosed

in the region. 3b "^ hc. d l E :

#

t

d S B

2

2=−

# i.e. The line integral of the electric field intensity around a closed path is equal to

the surface integral of the time derivative of magnetic flux density 2c "^ hd. d H S : #

t d

D J a

2

2= +b l #

i.e. The line integral of magnetic field intensity around a closed path is equal to the

surface integral of sum of the current density and time derivative of electric flux

density. 1d "^ hSOL 6.3.22 Option (D) is correct.

The continuity equation is given as

J :d v ρ=−i.e. it relates current density J ^ h and charge density v ρ .


Given Maxwell’s equation is

H #d J t

D c

2

2= +

For free space, conductivity, 0σ = and so,

J c 0E σ= =

Therefore, we have the generalized equation

H #d t

D

2

2=


Given the magnetic field intensity,

H 3 7 2y x a a a x y z = + +

So from Ampere’s circuital law we have

J H #d=

y x

a a a

3 7 2

x

x

y

y

z

z =

2

2

2

2

2

2

a a a 0 2 0 0x y z = − − +^ ^ ^h h h 2a y =−

SOL 6.3.25 Option (A) is correct.The emf in the loop will be induced due to motion of the loop as well as the

variation in magnetic field given as

V emf t

d d S B

v B l #2

2=− + ^ h # #

So, the frequencies for the induced e.m.f. in the loop is 1ω and 2ω .


F Q E v B #= +^ h is Lorentz force equation.







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All of the given expressions are Maxwell’s equation.


The direction of magnetic flux due to the current ‘i ’ in the conductor is determinedby right hand rule. So, we get the flux through A is pointing into the paper while

the flux through B is pointing out of the paper.

According to Lenz’s law the induced e.m.f. opposes the flux that causes it. So again

by using right hand rule we get the direction of induced e.m.f. is anticlockwise in

A and clockwise in B .


A2

d J 0µ=−

This is the wave equation for static electromagnetic field.

i.e. It is not Maxwell’s equation.


Poission’s equation for an electric field is given as

V 2d v

ερ

=−

where, V is the electric potential at the point and v ρ is the volume charge density

in the region. So, for 0v ρ = we get,

V 2d 0=

Which is Laplacian equation.


Continuity equation J #d t v

2

2ρ=− 4a "^ h

Ampere’s law H #d t

J D

2

2= + 1b "^ h

Displacement current J t

D

2

2= 2c "^ h

Faraday’ law E #d t

B

2

2=− 3d "^ h


A static electric field in a charge free region is defined as

E :d 0= 4a "^ hand E #d 0=A static electric field in a charged region have

E :d 0v !

ερ

= 2b "^ hand E #d 0=

A steady magnetic field in a current carrying conductor have

B :d 0= 1c "^ h B #d 0J 0 !µ=

A time varying electric field in a charged medium with time varying magnetic field







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have

E #d t

B 02

2!=− 3d "^ h

E :d 0v !

ε

ρ=


V dt

d m Φ=−

It is Faraday’s law that states that the change in flux through any loop induces

e.m.f. in the loop.


From stokes theorem, we have

d E S # :d h # d E l := # (1)

Given, the Maxwell’s equation E #d ( / )t B 2 2=−

Putting this expression in equation (1) we get,

d E l : # t

d B S s

:2

2=− #


Induced emf in a coil of N turns is defined as

V emf N dt d Φ=−

where Φ is flux linking the coil. So, we get

V emf dt

d t t 100 23=− −

^ h t 100 3 22=− −^ h 100 1000 mV3 2 22=− − =−^_ h i (at 2 st = )

1 V=−


Since, the flux linking through both the coil is varying with time so, emf are

induced in both the coils.

Since, the loop 2 is split so, no current flows in it and so joule heating does not

occur in coil 2 while the joule heating occurs in closed loop 1 as current flows in it.

Therefore, only statement 2 is correct.

SOL 6.3.37

Option (C) is correct.The electric field intensity is

E e E j t

0= ω where E 0 is independent of time

So, from Maxwell’s equation we have

H #d t

J E

2

2ε= +

j e E E j t

0σ ε ω= + ω^ h j E E σ ωε= +


Equation (1) and (3) are not the Maxwell’s equation.




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From the Maxwell’s equation for a static field (DC) we have

B #d J 0µ=

A#d #d

^ h J 0µ=

A A2

:dd d−^ h J 0µ=

For static field (DC), A:d 0=

therefore we have, A2

d J 0µ=−

So, both A and R are true and R is correct explanation of A.


For a static field, Maxwells equation is defined as

H #d J =

and since divergence of the curl is zero

i.e. H :d #d h 0=

J :d 0=but in the time varying field, from continuity equation (conservation of charges)

J :d t

0v

2

2!

ρ=−

So, an additional term is included in the Maxwell’s equation.

i.e. H #d t

J D

2

2= +

wheret

D

2

2 is displacement current density which is a necessary term.

Therefore A and R both are true and R is correct explanation of A.


For any loop to have an induced e.m.f., magnetic flux lines must link with the coilObserving all the given figures we conclude that loop C 1 and C 2 carries the flux

lines through it and so both the loop will have an induced e.m.f.


Since, the circular loop is rotating about the y -axis as a diameter and the flux

lines is directed in a x direction. So, due to rotation magnetic flux changes and as

the flux density is function of time so, the magnetic flux also varies w.r.t time and

therefore the induced e.m.f. in the loop is due to a combination of transformer and

motional e.m.f. both.

SOL 6.3.43 Option (C) is correct.Gauss’s law D :d ρ= a 1"^ h

D2

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