QJEEM24Phf00a: JEE Main 2024 Physics - Class 12 JEEM24Phf00a Solution: The magnetic field in a plane electromagnetic wave is B y = (…

For an electromagnetic wave propagating in free space, the relationship between the magnitudes of the electric field ($$E$$) and the magnetic field ($$B$$) can be described using the equation:

$E = cB$

where

• $c$ is the speed of light in vacuum, approximately $3.0 \times 10^8 \, \text{m/s}$ .
• $B$ is the magnitude of the magnetic field.

Given the magnetic field $B_y = (3.5 \times 10^{-7}) \sin (1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{T}$ , we can calculate the corresponding electric field magnitude using the formula above:

$E = (3.0 \times 10^8) \times (3.5 \times 10^{-7})$

$= 105 \, \text{Vm}^{-1}$

Thus, the magnitude of the electric field associated with the given magnetic field is $105 \, \text{Vm}^{-1}$ . The direction of the electric field is perpendicular to both the magnetic field and the direction of propagation. Given $B_y$ , this means $E$ will have components in the $x-z$ plane. Since electromagnetic waves are transverse, and given that the magnetic field is specified to be in the $y$ -direction, the corresponding electric field component must lie in a plane perpendicular to the $y$ -axis, which could be either the $x$ or the $z$ direction.

However, knowing electromagnetic wave properties, if the wave is propagating along the $x$ -axis and the magnetic field ( $B_y$ ) is along the $y$ -axis, then by right-hand rule, the electric field ( $E$ ) must be along the $z$ -axis to maintain the orthogonal relationship among the direction of propagation, electric field, and magnetic field vector directions.

Therefore, the correct option is:

$\text{Option A: } E_z = 105 \sin \left(1.5 \times 10^3 x + 0.5 \times 10^{11} t\right) \, \text{Vm}^{-1}$