# 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? - Physics

Numerical

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 E field equals the average energy density of the B field. [c = 3 × 108 m s−1]

#### Solution

Frequency of the electromagnetic wave, v = 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"

= (3 xx 10^8)/(2 xx 10^10)

= 0.015 m

(b) Magnetic field strength is given as:

"B"_0 = "E"_0/"c"

= 48/(3 xx 10^8)

= 1.6 × 10−7 T

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

"U"_"E" = 1/2in_0"E"^2

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

"U"_"B"= 1 /(2μ_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_0μ_0)) ….......(2)

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

"E" = 1/sqrt(in_0μ_0)"B"

Squaring both sides, we get

"E"^2 = 1/(in_0μ_0)"B"^2

in_0"E"^2 = "B"^2/μ_0

1/2in_0"E"^2 = 1/2 "B"^2/μ_0

UE = UB

Concept: Electromagnetic Waves
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 8 Electromagnetic Waves
Exercise | Q 8.10 | Page 286
NCERT Class 12 Physics Textbook
Chapter 8 Electromagnetic Waves
Exercise | Q 10 | Page 286

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