Ncert Exercises & Solutions

8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} Hz$ and amplitude 48 V/m.

(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 electric field equals the average energy density of the magnetic field. [ $c = 3 \times 10^{8} {ms}^{- 1}$ ]

(a)
Hint: $\lambda=\frac{c}{\nu} $
Step: Find the Wavelength of the wave.
Frequency of the wave, $ν$ =
Amplitude of the electric field,
Wavelength is given by,

(b)
Hint: $c=\frac{E_{0}}{B_{0}} $
Step: Find amplitude of the magnetic field.

(c)
Hint: Recall energy density.
Step 1: Find energy density of the electric field.
$U_{E} = \frac{1}{2} \in_{0} E^{2}$
Step 2: Find energy density of the electric field.

$U_{B} = \frac{1}{2 μ_{0}} B^{2}$
Step 3: Relate electric and magnetic field.
E = cB
where $c = \frac{1}{\sqrt{\in_{0} μ_{0}}}$

Squaring on both sides, we get,

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