The electric field of an electromagnetic wave in free space is $\vec E = 57\cos\left[7.5\times10^6t - 5\times10^{-3}(3x+4y)\right](4\hat i - 3\hat j)\ \text{N/C}$. The associated magnetic field in tesla is
Answer: (C) $\vec B = -\dfrac{57}{3\times10^8}\cos\left[7.5\times10^6t - 5\times10^{-3}(3x+4y)\right](5\hat k)$
The phase is $\omega t - \vec k\cdot\vec r$ with $\vec k \propto 3\hat i + 4\hat j$, so the wave travels along $\hat n = \dfrac{3\hat i + 4\hat j}{5}$.
For an EM wave, $\vec B = \dfrac{1}{c}\,\hat n\times\vec E$:
$$\hat n\times(4\hat i - 3\hat j) = \frac{1}{5}(3\hat i + 4\hat j)\times(4\hat i - 3\hat j) = \frac{1}{5}(-9 - 16)\hat k = -5\hat k$$
$$\vec B = -\frac{57}{3\times10^8}\cos\left[7.5\times10^6t - 5\times10^{-3}(3x+4y)\right](5\hat k)$$
(Check: $|\vec E|$ amplitude is $57\times5$, and $|\vec B| = |\vec E|/c$.)
Solution by Sreeraj P, M.Sc Physics