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In a Plane Electromagnetic Wave, the Electric Field Oscillates Sinusoidally at a Frequency of 2.0 × 10^10 Hz and Amplitude 48 V m^−1. What is the Wavelength of the Wave? - Physics

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m−1.

(a) What is the wavelength of the wave?

(b) What is the amplitude of the oscillating magnetic field?

(c) Show that the average energy density of the field equals the average energy density of the field. [= 3 × 108 m s−1.]

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Solution

Frequency of the electromagnetic wave, ν = 2.0 × 1010 Hz

Electric field amplitude, E0 = 48 V m−1

Speed of light, c = 3 × 108 m/s

(a) Wavelength of a wave is given as:

`lambda=c/v`

`=(3xx10^8)/(2xx10^10)=0.015m`

(b) Magnetic field strength is given as:

`B_0=E_0/c`

`=48/(3xx10^8)=1.6xx10^-7 T`

(c) Energy density of the electric field is given as:

`U_E=1/2in_0E^2`

And, energy density of the magnetic field is given as:

`U_B=1/(2mu_0)B^2`

Where,

0 = Permittivity of free space

μ0 = Permeability of free space

We have the relation connecting E and B as:

E = cB … (1)

Where,

`c=1/(sqrt(in_0mu_0))` … (2)

Putting equation (2) in equation (1), we get

`E=1/sqrt(in_0mu_0)`

Squaring both sides, we get

`E=1/(in_0mu_0) B^2`

`in_0E^2=B^2/mu_0`

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 8 Electromagnetic Waves
Q 10 | Page 286
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