The oscillating magnetic field in a plane electromagnetic wave is given by B_y=5 \times 10^-6 \sin 1000 \pi(5 x-4 \times 10^8 t) T . The amplitude of electric field will be :

Electromagnetic Waves PYQ — JEE Main 2022 · 26 Jul · Shift 2 · Q45 with Solution

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Correct answer: D

Given magnetic field

$B_y = 5 \times 10^{-6} \sin\left[1000\pi(5x-4\times 10^8 t)\right]\, \text{T}$

So, the amplitude of magnetic field is

$B_0 = 5 \times 10^{-6}\, \text{T}$
Relation between electric and magnetic field amplitudes in an electromagnetic wave

For a plane electromagnetic wave,

$E_0 = v B_0$

where $v$ is the speed of the wave.
Find wave speed from the wave equation

The phase is

$1000\pi(5x-4\times 10^8 t)$

Compare with the standard form:

$\sin(kx-\omega t)$

Expanding,

$1000\pi(5x-4\times 10^8 t) = 5000\pi x - 4\times 10^{11}\pi t$

Hence,

$k = 5000\pi$ $\omega = 4\times 10^{11}\pi$

Therefore,

$v = \frac{\omega}{k} = \frac{4\times 10^{11}\pi}{5000\pi} = 8\times 10^7\, \text{m/s}$
Now calculate electric field amplitude

$E_0 = vB_0 = (8\times 10^7)(5\times 10^{-6})$

$E_0 = 40\times 10^1 = 4\times 10^2\, \text{V/m}$
Match with options

$E_0 = 4\times 10^2\, \text{V/m}$

This corresponds to Option D.

Final Answer: Option D