Q 1 :    

In the given electromagnetic wave Ey=600sin(ωt-kx)Vm-1, intensity of the associated light beam is (in W/m2)

(Given ε0=9×10-12C2N-1m-2)               [2024]

  • 729

     

  • 243

     

  • 972

     

  • 486

     

(4)   

 Intensity, I=12ϵ0E02cI=92×36×3=486W/m2

 



Q 2 :    

A plane electromagnetic wave propagating in x-direction is described by

Ey=(200Vm-1)sin[1.5×107t-0.05x];

The intensity of the wave is

(Use ε0=8.85×10-12C2N-1m-2)              [2024]

  • 35.4 Wm-2

     

  • 53.1 Wm-2

     

  • 26.2 Wm-2

     

  • 106.2 Wm-2

     

(2)   

         I=12ε0E02×c

          I=53.1 W/m2

 



Q 3 :    

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×1010Hz and an amplitude of 50Vm-1. The total average energy density of the electromagnetic field of the wave is [Use ε0=8.85×10-12C2/Nm2]                       [2024]

  • 1.106×10-8Jm-3

     

  • 2.212×10-10Jm-3

     

  • 4.425×10-8Jm-3

     

  • 2.212×10-8Jm-3

     

(1)   

          UE=12ϵ0E2

            UE=12×8.85×10-12×(50)2=1.106×10-8J/m3

 



Q 4 :    

Due to presence of an em-wave whose electric component is given by E = 100 sin (ωtkx) NC1, a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as          [2025]

  • 25 sin (ωtkxNC1

     

  • 200 sin (ωtkxNC1

     

  • 400 sin (ωtkxNC1

     

  • 50 sin (ωtkxNC1

     

(3)

Speed of the light,

c=E0B0  E0=cB0

E0=(3×108)(2×108)

E0=6 Vm1

As, B = along y-axis

      v = along negative x-axis

Hence, E0 = along z-axis



Q 5 :    

The unit of 2Iε0c is:

(I = intensity of an electromagnetic wave, c : speed of light)          [2025]

  • Vm

     

  • NC

     

  • Nm

     

  • NC1

     

(4)

Speed of light c=ωk=4×1085

                           =0.8×108 m/sec

So,  E0=cB0=0.8×108×5×106

              =400 V/m=4×102