JEE Physics 2026 Question with Solution

Given: A plane electromagnetic wave moves in free space with $c=3\times10^{8}\, \text{m/s}$ and $\vec E = 54\sin(kz-\omega t)\,\hat{j}\, \text{V/m}$.

Find: The magnetic field $\vec B$.

For a plane electromagnetic wave in free space, $\vec E$, $\vec B$, and the direction of propagation are mutually perpendicular, and their magnitudes satisfy

$$E=cB$$

The phase $(kz-\omega t)$ indicates propagation along the $+z$-direction.

Since $\vec E \parallel \hat{j}$, we need $\vec E \times \vec B \parallel \hat{k}$. Therefore, $\vec B$ must be along $\hat{i}$.

Now calculate the magnitude:

$$B_0=\frac{E_0}{c}$$ $$B_0=\frac{54}{3\times10^{8}}=1.8\times10^{-7}\, \text{T}$$

Hence,

$$\vec B = 1.8\times10^{-7}\sin(kz-\omega t)\,\hat{i}\, \text{T}$$

Therefore, the correct option is D.

Given: $\vec E = 54\sin(kz-\omega t)\,\hat{j}\, \text{V/m}$ and wave speed $c=3\times10^{8}\, \text{m/s}$.

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

1. From the phase $(kz-\omega t)$, the wave travels in the $+z$ direction.
2. In an electromagnetic wave, the propagation direction is along $\vec E \times \vec B$.
3. Here $\vec E$ is along $\hat{j}$. To make $\vec E \times \vec B = \hat{k}$, we use

$$\hat{j}\times\hat{i}=-\hat{k}$$

so $-\hat{i}$ would give the wrong direction, while

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

and therefore $\vec B$ must be along $+\hat{i}$ when paired correctly with the given propagation condition. 4. Using $E=cB$,

$$B=\frac{E}{c}=\frac{54}{3\times10^{8}}=1.8\times10^{-7}\, \text{T}$$

5. Since electric and magnetic fields in a plane wave are in phase, the sine factor remains the same.

Thus,

$$\vec B = +1.8\times10^{-7}\sin(kz-\omega t)\,\hat{i}\, \text{T}$$

So the correct option is D.