JEE Main 2024 — Electromagnetic Waves Question with Solution

To determine the magnetic field induction of the given electromagnetic wave, we need to use the relationship between the electric field $$\overrightarrow{\mathrm{E}}$$ and the magnetic field $$\overrightarrow{\mathrm{B}}$$ in an electromagnetic wave. For an electromagnetic wave propagating in vacuum, the following relation holds:

$$\overrightarrow{\mathrm{B}}=\frac{\overrightarrow{\mathrm{E}}\times \hat{\mathrm{k}}}{\mathrm{c}}$$

where:

• $$\overrightarrow{\mathrm{E}}$$ is the electric field.
• $$\hat{\mathrm{k}}$$ is the unit vector in the direction of propagation of the wave.
• $$\mathrm{c}$$ is the speed of light in a vacuum.

Given the electric field:

$$\overrightarrow{\mathrm{E}}=\hat{i}40\cos \omega (\mathrm{t}-z/\mathrm{c}){\mathrm{NC}}^{-1}$$

The wave is propagating in the $$z$$-direction, so $$\hat{\mathrm{k}}=\hat{z}$$. The unit vector $$\hat{\mathrm{i}}$$ represents the $$x$$-direction.

The magnetic field induction is given by:

$$\overrightarrow{\mathrm{B}}=\frac{(\hat{\mathrm{i}}40\cos \omega (\mathrm{t}-z/\mathrm{c}))\times \hat{\mathrm{z}}}{\mathrm{c}}$$

The cross product $$\hat{\mathrm{i}}\times \hat{\mathrm{z}}$$ yields $$\hat{\mathrm{j}}$$ (the unit vector in the $$y$$-direction):

$$\overrightarrow{\mathrm{B}}=\hat{\mathrm{j}}\frac{40\cos \omega (\mathrm{t}-z/\mathrm{c})}{\mathrm{c}}$$

Therefore, the magnetic field induction is:

$$\overrightarrow{\mathrm{B}}=\hat{\mathrm{j}}\frac{40}{\mathrm{c}}\cos \omega (\mathrm{t}-z/\mathrm{c})$$

The correct answer is:

Option A:

$$\overrightarrow{\mathrm{B}}=\hat{j}\frac{40}{\mathrm{c}}\cos \omega (\mathrm{t}-z/\mathrm{c})$$