JEE Physics 2023 Question with Solution

Given: $\vec{E} = 6.6 \, \hat{j} \, \text{V/m}$, frequency $f = 20 \, \text{MHz}$, and the wave propagates along $\hat{i}$.

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

For an electromagnetic wave in free space,

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

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

Substituting the given value,

$$|\vec{B}| = \frac{6.6}{3 \times 10^8} = 2.2 \times 10^{-8} \, \text{T}$$

The direction is obtained from

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

where the propagation direction $\vec{C}$ is along $\hat{i}$. Since $\vec{E}$ is along $\hat{j}$, we need $\hat{j} \times \vec{B} = \hat{i}$, so $\vec{B}$ must be along $\hat{k}$.

Therefore,

$$\vec{B} = 2.2 \times 10^{-8} \, \hat{k} \, \text{T}$$

The solution working gives $\vec{B} = 2.2 \times 10^{-8} \, \hat{k} \, \text{T}$, which matches option C. The solution states option D, but that is inconsistent with the actual working and final answer expression.

Given: Electric field along $\hat{j}$ and propagation along $\hat{i}$.

Find: Which unit vector direction of $\vec{B}$ satisfies the electromagnetic wave orientation.

In free space, the three vectors are mutually perpendicular and satisfy

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

Here this becomes

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

Using unit-vector products,

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

Hence the magnetic field must be along $\hat{k}$, not along $\hat{i}$ and not along $-\hat{k}$.

With magnitude $2.2 \times 10^{-8} \, \text{T}$, the correct option is C.