QJEEM24Ph333b: JEE Main 2024 Physics - Class 12 JEEM24Ph333b Solution: The electric field in an electromagnetic wave is given by . The …

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:

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