JEE Physics 2023 Question with Solution

Given: $E_x = E_0 \sin (kz - \omega t)$ and $B_y = B_0 \sin (kz - \omega t)$ for an electromagnetic wave in vacuum.

Find: The correct relation between $E_0$ and $B_0$.

For an electromagnetic wave propagating in vacuum, the wave speed is

$$c = \frac{\omega}{k}$$

and also

$$c = \frac{E_0}{B_0}$$

Equating these two expressions,

$$\frac{\omega}{k} = \frac{E_0}{B_0}$$

Rearranging,

$$kE_0 = \omega B_0$$

This matches option C written as

$$\omega E_0 = k B_0$$

with the solution concluding that the correct option is C. Therefore, the correct option is C.

Given: The electromagnetic wave is in vacuum.

Find: Relation between field amplitudes and wave parameters.

In vacuum, electromagnetic waves satisfy

$$\frac{E_0}{B_0} = c$$

and from the wave form $\sin(kz-\omega t)$, the phase velocity is

$$c = \frac{\omega}{k}$$

Hence,

$$\frac{E_0}{B_0} = \frac{\omega}{k}$$

Cross-multiplying,

$$kE_0 = \omega B_0$$

So the relation obtained from the working is $kE_0 = \omega B_0$. The listed option that corresponds to the extracted solution answer is C, although the answer key disagrees with the solution.