A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, \overrightarrow E = 6.3\widehat j\,V/m. The corresponding magnetic field \overrightarrow B, at that point will be :

18.9 \times 10 - 8 \widehat k T

Electromagnetic Waves PYQ — JEE Main 2019 · 9 Jan · Shift 1 · Q51 with Solution

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Correct answer: B

Use the relation between $\vec E$ and $\vec B$ for an electromagnetic wave in free space

For a plane electromagnetic wave in free space,

$E = cB$

So,

$B = \frac{E}{c}$

Given:

$E = 6.3\ \text{V/m}, \qquad c = 3\times 10^8\ \text{m/s}$

Thus,

$B = \frac{6.3}{3\times 10^8}$

$B = 2.1\times 10^{-8}\ \text{T}$

For an electromagnetic wave,

$\vec E \times \vec B = \text{direction of propagation}$

The wave travels along the positive x-direction, i.e. along $\hat i$ .

Given:

$\vec E = 6.3\hat j\ \text{V/m}$

We need $\vec B$ such that

$\hat j \times \vec B = \hat i$

Using unit vectors,

$\hat j \times \hat k = \hat i$

Hence,

$\vec B \text{ is along } \hat k$

So,

$\vec B = 2.1\times 10^{-8}\hat k\ \text{T}$

Option B is:

$2.1\times 10^{-8}\hat k\ \text{T}$

This matches exactly.

The given frequency $50\ \text{MHz}$ is not needed here, because in free space the relation $E=cB$ is sufficient.

$\boxed{\vec B = 2.1\times 10^{-8}\hat k\ \text{T}}$

So the correct option is B.