JEE Main · 2025 · 23 Jan 2025 · Shift 1 · EM Waves

The wave travels along the direction of its wave vector. From the phase $5\times 10^{-3}(3x+4y)$, the propagation direction is

$\hat{k}=\frac{3\hat{i}+4\hat{j}}{5}$

The electric field direction, from $\vec{E}\propto (4\hat{i}-3\hat{j})$, has unit vector

$\hat{E}=\frac{4\hat{i}-3\hat{j}}{5}$

For an electromagnetic wave, $\hat{B}=\hat{k}\times \hat{E}$:

$\hat{B}=\frac{(3\hat{i}+4\hat{j})\times (4\hat{i}-3\hat{j})}{25}=\frac{-9\hat{k}-16\hat{k}}{25}=-\hat{k}$

Since $(4\hat{i}-3\hat{j})$ is not a unit vector (its magnitude is 5), the true electric field amplitude is $57\times 5$, so the magnetic field amplitude is

$B_0=\frac{E_0}{c}=\frac{57\times 5}{3\times 10^8}$

Combining magnitude and direction:

$\vec{B}=-\frac{57}{3\times 10^8}\cos \left[7.5\times 10^{6}t-5\times 10^{-3}(3x+4y)\right](5\hat{k})$