JEE Main 2025 Jan 23 Shift 1, Physics Q18: Electromagnetic Waves

For an EM wave, $\vec{E}$, $\vec{B}$, and the direction of propagation $\hat{k}$ form a right-handed orthogonal set with $\hat{k}\times \hat{E}=\hat{B}$, and $|B_0|=E_0/c$.

From the phase $(3x+4y)$, $\hat{k}=\frac{3\hat{i}+4\hat{j}}{5}$. The unit vector of $\vec{E}$ is $\hat{E}=\frac{4\hat{i}-3\hat{j}}{5}$.

$\hat{k}\times \hat{E}=\frac{1}{25}(3\hat{i}+4\hat{j})\times (4\hat{i}-3\hat{j})=\frac{1}{25}[-9\hat{k}-16\hat{k}]=-\hat{k}$.

So $\hat{B}=-\hat{k}$. The peak field magnitude is $B_0=\frac{E_0}{c}=\frac{57}{3\times 10^8}$, but $\vec{E}$ here has magnitude $|4\hat{i}-3\hat{j}|\cdot 57=5\cdot 57$, so $\vec{B}$ carries the same factor of 5.

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