JEE Physics 2024 Question with Solution

Given: A plane electromagnetic wave travels along the $X$-direction in free space and $\vec{E} = 9.6 \, \hat{j} \, \text{V/m}$.

Find: The magnetic field $\vec{B}$ at that point.

For an electromagnetic wave in free space,

$$E = cB$$

where $c = 3 \times 10^8 \, \text{m/s}$.

So, the magnitude of the magnetic field is

$$B = \frac{E}{c} = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8} \, \text{T}$$

Now determine the direction. The wave propagates along $\hat{i}$ and the electric field is along $\hat{j}$. Therefore the magnetic field must be perpendicular to both, so it is along $\hat{k}$.

Hence,

$$\vec{B} = 3.2 \times 10^{-8} \, \hat{k} \, \text{T}$$

Therefore, the correct option is A.

Given: Propagation direction is $+X$, and $\vec{E}$ is along $\hat{j}$.

Find: Direction and magnitude of $\vec{B}$.

In an electromagnetic wave, $\vec{E}$, $\vec{B}$, and the direction of propagation are mutually perpendicular, and

$$\vec{E} \times \vec{B}$$

gives the direction of wave propagation.

Since the wave travels along $\hat{i}$ and $\vec{E}$ is along $\hat{j}$, we need

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

This is satisfied by

$$\vec{B} \parallel \hat{k}$$

because $\hat{j} \times \hat{k} = \hat{i}$.

Using

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

we get

$$B = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8} \, \text{T}$$

Thus,

$$\vec{B} = 3.2 \times 10^{-8} \, \hat{k} \, \text{T}$$

So the correct option is A.