JEE Main 2024 — Electromagnetic Waves Question with Solution

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:

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