Q 1 :

An alternating voltage of amplitude 40 V and frequency 4 kHz is applied directly across the capacitor of 12 μF. The maximum displacement current between the plates of the capacitor is nearly               [2024]

  • 12 A

     

  • 13 A

     

  • 10 A

     

  • 8 A

     

(1)     

          Displacement current is same as conduction current in capacitor.

           XC=1ωC=12πfC

           =12π×4×103×12×10-6=3.317Ω

           I=VXC=403.317=12A



Q 2 :

Electromagnetic waves travel in a medium with speed of 1.5×108 ms-1. The relative permeability of the medium is 2.0. The relative permittivity will be:      [2024]

  • 2

     

  • 1

     

  • 5

     

  • 4

     

(1)     

           Speed of light in vacuum=1μ0ϵ0=3×108

            Speed of light in medium=1μ00·μrr=32×108

           Also, given, ϵr=2, therefore, μr=2

 



Q 3 :

A plane EM wave is propagating along x direction. It has a wavelength of 4 mm. If electric field is in y direction with the maximum magnitude of 6O Vm-1, the equation for magnetic field is          [2024]

  • Bz=60sin[π2(x-3×108t)]k^T

     

  • Bz=2×10-7sin[π2×103(x-3×108t)]k^T

     

  • Bx=60sin[π2(x-3×108t)]i^T

     

  • Bz=2×10-7sin[π2(x-3×108t)]k^T

     

(2)     

             λ=4mm=4×10-3m

             K=2π4×10-3=π2×103·m-1

             ω=v×K=3×108×π2×103=3π2×1011rad/s

             Electric field y direction

             Propagation field x direction

             Mgnetic field z direction

             E^×B^=c^ and B0=E0c=2×10-7T

            Bz=2×10-7sin[π2×103(x-3×108t)]k^T



Q 4 :

The magnetic field in a plane electromagnetic wave is By=(3.5×10-7)sin(1.5×103x+0.5×1011t)T. The corresponding electric field will be             [2024]

  • Ey=1.17sin(1.5×103x+0.5×1011t)Vm-1

     

  • Ez=105sin(1.5×103x+0.5×1011t)Vm-1

     

  • Ez=1.17sin(1.5×103x+0.5×1011t)Vm-1

     

  • Ey=10.5sin(1.5×103x+0.5×1011t)Vm-1

     

(2)     

            E0=B0c

            EZ=3×108×(3.5×10-7)sin(1.5×103x+0.5×1011t)

           Ez=105sin(1.5×103x+0.5×1011t)Vm-1

 



Q 5 :

Match Column I with Column II

  Column I   Column II
A. Bdl=μ0ic+μ0ϵ0dΦEdt I. Gauss' law for electricity
B. Edl=dΦBdt II. Gauss' law for magnetism
C. EdA=Qϵ0 III. Faraday law
D. BdA=0 IV. Ampere-Maxwell law

 

Choose the correct answer from the options given below            [2024]

  • A-IV, B-I, C-III, D-II

     

  • A-II, B-III, C-I, D-IV

     

  • A-IV, B-III, C-I, D-II

     

  • A-I, B-II, C-III, D-IV

     

(3)          

              Ampere-Maxwell law, Bdl=μ0ic+μ0ϵ0dΦEdt

               Faraday law,  Edl=dΦBdt

               Gauss' law for electricity EdA=Qϵ0

               Gauss' law for magnetism BdA=0

 



Q 6 :

A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time) E=9.6j^V/m. The value of magnetic field at this point is           [2024]

  • 9.6j^T

     

  • 3.2×10-8i^T

     

  • 3.2×10-8k^T

     

  • 9.6×10-8k^T

     

(3)   

             EB=3×108

           B=E3×108=9.63×108

           B=3.2×10-8TB^

               =i^×j^=k^

            So, B=3.2×10-8k^T

 



Q 7 :

The electric field of an electromagnetic wave in free space is represented as E=E0cos(ωt-kz)i^.

The corresponding magnetic induction vector will be :            [2024]

  • B=E0ccos(ωt-kz)j^

     

  • B=E0ccos(ωt-kz)j^

     

  • B=E0ccos(ωt+kz)j^

     

  • B=E0ccos(ωt+kz)j^

     

(2)   

        Given E=E0cos(ωt-kz)i^

         E0=B0c,(B0=E0c)

        Also E and B are in same phase.

 



Q 8 :

Match List I with List II

  List I   List II
A Gauss's Law of Magnetostatics I Eda=1ϵ0ρdV
B Faraday's law of electromagnetic induction II Bda=0
C Ampere's Law III Edl=ddtBda
D Gauss's law of Electrostatics IV Bdl=μ0I

 

Choose the correct answer from the options given below:                  [2024]

  • A-I, B-III, C-IV, D-II

  • A-III, B-IV, C-I, D-II

  • A-IV, B-II, C-III, D-I

  • A-II, B-III, C-IV, D-I

     

(4)        

              (A) Gauss of magnetostatics Bda=0                  (II)

              (B) Faraday's law Edt=-ddtBda         (III)

              (C) Ampere's law Bdt=μ0I                                   (IV)

              (D) Gauss of electro Eda=1ϵ0ρdV              (I)

 



Q 9 :

Given below are two statements:

Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields.

Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface.

In the light of the above statements, choose the most appropriate answer from the options given below:          [2024]

  • Statement I is incorrect but Statement II is correct

     

  • Both Statement I and Statement II are correct.

     

  • Both Statement I and Statement II are incorrect.

     

  • Statement I is correct but Statement II is incorrect.

     

(2)     

 12ϵ0E2=B22μ0

 



Q 10 :

If frequency of electromagnetic wave is 60 MHz and it travels in air along z-direction, then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other, and the wavelength of the wave (in m) is:           [2024]

  • 2.5

     

  • 10

     

  • 5

     

  • 2

     

(3)     

         Speed=Wavelength×Frequency

            λ=Cf=3×10860×106=5m

 



Q 11 :

A parallel plate capacitor has a capacitance C=200 pF. It is connected to 230 V AC supply with an angular frequency 300 rad/s. The RMS value of conduction current in the circuit and displacement current in the capacitor respectively are            [2024]

  • 1.38μA and 1.38μA

     

  • 14.3μA and 143μA

     

  • 13.8μA and 138μA

     

  • 13.8μA and 13.8μA

     

(4)

 



Q 12 :

The electric field in an electromagnetic wave is given by E=i^40cosω(t-zc) NC-1.  The magnetic field induction of this wave is (in SI unit):      [2024]

  • B=j^40cosω(t-zc)

     

  • B=k^40ccosω(t-zc)

     

  • B=i^40ccosω(t-zc)

     

  • B=j^40ccosω(t-zc)

     

(4)

E=40cos(ωt-ωcz) NC-1i^

E is along +x direction

v is along +z direction

and Direction of c^ must be along E×B.

So direction of B will be along +y and magnitude of B will be E0c.

B0=E0c=40c



Q 13 :

The electric field of an electromagnetic wave in free space is

E=57 cos [7.5×106t5×103(3x+4y)](4i^3j^)N/C

The associated magnetic field in Tesla is:          [2025]

  • B=573×108 cos [7.5×106t5×103(3x+4y)](5k^)

     

  • B=573×108 cos [7.5×106t5×103(3x+4y)](k^)

     

  • B=573×108 cos [7.5×106t5×103(3x+4y)](5k^)

     

  • B=573×108 cos [7.5×106t5×103(3x+4y)](k^)

     

(3)

E = CB

B = 57×53×108

3i^+4j^ is direction of propagation

 (4i^3j^)×(k^)=4j^+3i^



Q 14 :

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3 Vm1. Then, the magnetic field vector of the wave at the point is          [2025]

  • Bz=9.3×108T

     

  • Bz=1.55×108T

     

  • Bz=6.2×108T

     

  • Bz=3.1×108T

     

(4)

E = BC

9.3=B×3×108

B=9.33×108=3.1×108T



Q 15 :

The magnetic field of an E.M. wave is given by

B=(32i^+12j^)30sin[ω(tzc)] (SI Units)

The corresponding electric field in S.I. units is:          [2025]

  • E=(12i^32j^)30c sin[ω(tzc)]

     

  • E=(34i^+14j^)30c cos[ω(tzc)]

     

  • E=(12i^+32j^)30c sin[ω(t+zc)]

     

  • E=(32i^12j^)30c sin[ω(t+zc)]

     

(1)

B=(32i^+12j^)30sin[ω(tzc)]

E×B=c and E0=B0c

Here E=(32(j^)+12i^)

       E0=30c

      E=(12i^32j^) 30c sin [ω(tzc)]



Q 16 :

A plane electromagnetic wave propagates along the +x direction in free space. The components of the electric field, E and magnetic field, B vectors associated with the wave in Cartesian frame are:          [2025]

  • Ey, Bx

     

  • Ey, Bz

     

  • Ex, By

     

  • Ez, By

     

(2)

Direction of wave propogation E×B



Q 17 :

If ε0 denotes the permittivity of free space and ΦE is the flux of the electric field through the area bounded by the closed surface, then dimension of (ε0dϕEdt) are that of:          [2025]

  • Electric field

     

  • Electric potential

     

  • Electric charge

     

  • Electric current

     

(4)

We know that formula for displacement current is given by

id=ε0dϕεdt

Dimension of ε0(dϕEdt) will be

Same as electric current

So [ε0dϕEdt]=A



Q 18 :

A parallel plate capacitor of area A = 16 cm2 and separation between the plates 10 cm, is charged by a DC current. Consider a hypothetical plane surface of area A0=3.2 cm2 inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6 A. At the same instant the displacement current through A0 is _______ mA.          [2025]



(1200)

id=ic

  Total displacement current = 6 A

Current density, Jd=IA=616

  The current through small area i'=Jd×A'

i'=616×3.2=1.12 A=1200 mA



Q 19 :

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5 μF. The dielectric constant of the medium between the capacitor plates is 1. It produces an instantaneous displacement current of 0.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be _______ Vs1.



(100)

q = CV

Differentiating, i=CdVdt

 dVdt=0.25×1032.5×106=100010=100



Q 20 :

Match List I with List II:                          [2023]

  List I   List II
A. Gauss’s Law in Electrostatics I. E·dl=-dϕBdt
B. Faraday’s Law II. B·dA=0
C. Gauss’s Law in Magnetism III. B·dl=μ0ic+μ0ε0dϕEdt
D. Ampere-Maxwell Law IV. E·dS=qε0

 

Choose the correct answer from the options given below:

  • A-I, B-II, C-III, D-IV

     

  • A-IV, B-I, C-II, D-III

     

  • A-III, B-IV, C-I, D-II

     

  • A-II, B-III, C-IV, D-I

     

(2)

Gauss's law of electrostatics

ϕ=E·ds=qε0

Faraday's law E·dl=-dϕBdt

Gauss's law of magnetism B·dA=0

Ampere's Maxwell law

B·dl=μ0iC+μ0ε0dϕEdt

Where iC: Conduction current

ε0dϕEdt: Displacement current



Q 21 :

A plane electromagnetic wave of frequency 20 MHz propagates in free space along x-direction. At a particular space and time, E=6.6j^ V/m. What is B at this point?              [2023]

  • -2.2×10-8i^ T

     

  • 2.2×10-8k^ T    

     

  • -2.2×10-8k^ T    

     

  • 2.2×10-8i^ T

     

(2)

E^=6.6j^,    ν=20 MHz,    c=3×108i^

|B|=|E|c=2.2×10-8 T

E^×B^=c^  B^=2.2×10-8k^ T



Q 22 :

Which of the following Maxwell’s equations is valid for time varying conditions but not valid for static conditions?                   [2023]

  • B·dl=μ0I

     

  • E·dl=0

     

  • E·dl=-ϕBt

     

  • D·dA=Q

     

(3)

Based on equations of Maxwell

 



Q 23 :

The source of time varying magnetic field may be                     [2023]

(A) a permanent magnet
(B) an electric field changing linearly with time
(C) direct current
(D) a decelerating charge particle
(E) an antenna fed with a digital signal

Choose the correct answer from the options given below:

  • (D) only

     

  • (C) and (E) only

     

  • (A) only

     

  • (B) and (D) only

     

(1)

Source of time varying magnetic field may be

→ accelerated or retarded charge which produces varying electric and magnetic fields.

→ An electric field varying linearly with time will not produce variable magnetic field as current will be constant

 



Q 24 :

In an electromagnetic wave, at an instant and at a particular position, the electric field is along the negative z-axis and magnetic field is along the positive x-axis. Then the direction of propagation of electromagnetic wave is           [2023]

  • at 45° angle from positive y-axis

     

  • negative y-axis

     

  • positive z-axis

     

  • positive y-axis

     

(2)

Direction of propagation of EM wave will be in the direction of E×B.



Q 25 :

To radiate EM signal of wavelength λ with high efficiency, the antennas should have a minimum size equal to            [2023]

  • λ2

     

  • λ4

     

  • 2λ

     

  • λ

     

(2)

Minimum length of antenna should be λ4.



Q 26 :

In a medium the speed of light wave decreases to 0.2 times to its speed in free space. The ratio of relative permittivity to the refractive index of the medium is x:1. The value of x is _______. (Given speed of light in free space =3×108 ms-1 and for the given medium μr=1)            [2023]



(5)

v=Cμ  μ=Cv=C0.2C=5

μ=εrμr  εr=μ2μr

εrμ=μμr=5



Q 27 :

Match List – I with List – II.                              [2026]

  List – I   List – II
  (Relation)   (Law)
A. E·dl=-ddtB·da I. Ampere’s circuital law
B. B·dl=μ0(I+ε0dΦEdt) II. Faraday’s laws of electromagnetic induction
C. E·da=1ε0VρdV III. Ampere–Maxwell law
D. B·dl=μ0I IV. Gauss’s law of electrostatics

 

Choose the correct answer from the options given below:

  • A–II, B–III, C–I, D–IV

     

  • A–II, B–III, C–IV, D–I

     

  • A–I, B–IV, C–III, D–II

     

  • A–IV, B–I, C–II, D–III

     

(2)

Theoretical

A-II, B-III, C-IV, D-I



Q 28 :

The electric field in a plane electromagnetic wave is given by : 

Ey=69sin [0.6×103x-1.8×1011t]V/m

The expression for magnetic field associated with this electromagnetic wave is ________ T.       [2026]

  • Bz=2.3×10-7sin [0.6×103x+1.8×1011t]

     

  • Bz=2.3×10-7sin [0.6×103x-1.8×1011t]

     

  • By=69sin [0.6×103x+1.8×1011t]

     

  • By=2.3×10-7sin [0.6×103x-1.8×1011t]

     

(2)

B^=c^×E^

  c^=i^  because phase of electric field is function of x.

  E^=j^  (given)

  B^=i^×j^=k^

|B|=|E|c=69×0.6×1031.8×1011=693×108

|B|=2.9×10-7

B2=2.9×10-7sin(0.6×103x-1.8×1011t)

(phase is same as that of electric field)



Q 29 :

The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by

Ey=20sin(3×106x-4.5×1014t)V/m,

(where x, t and other values have S.I. units). The dielectric constant of the medium is __________.

(Speed of light in free space 3×108 m/s)                   [2026]



(4)

n=CV

V=ωk=4.5×10143×106=32×108

n=2

n=μrεr    (μr=1)

2=εr

εr=4



Q 30 :

A plane electromagnetic wave is moving in free space with velocity c=3×108 m/s and its electric field is given as E=54sin(kz-ωt)j^ V/m, where j^ is the unit vector along y-axis. The magnetic field vector B  of the wave is:  [2026]

  • -1.8×10-7sin(kz-ωt)i^T

     

  • 1.4×10-7sin(kz-ωt)i^T

     

  • 1.4×10-7sin(kz-ωt)k^T

     

  • +1.8×10-7sin(kz-ωt)i^T

     

(1)

B^=C^×E^=k^×j^=-i^

  B=543×108sin(kz-ωt)(-i^)

=-1.8×10-7sin(kz-ωt)i^