Q. The electric field of an electromagnetic wave in free space is $\vec E=57\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](4\hat i-3\hat j)\,N/C$. The associated magnetic field in Tesla is
A) $\vec B=\frac{57}{3\times10^8}\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](\hat k)$
B) $\vec B=-\frac{57}{3\times10^8}\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](\hat k)$
C) $\vec B=-\frac{57}{3\times10^8}\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](5\hat k)$
D) $\vec B=\frac{57}{3\times10^8}\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](5\hat k)$
Explanation
[Solution by Dr. Abubacker] The propagation direction is $\hat k_p=\frac{3\hat i+4\hat j}5$ and $\vec E\propto(4\hat i-3\hat j)=5\hat e$ with $\hat e=\frac{4\hat i-3\hat j}5$. $\hat k_p\times\hat e=\frac{3(-3)-4(4)}{25}\hat k=-\hat k$. The amplitude is $\frac{E_0}c$ with $E_0=57\times5$, so $\vec B=-\frac{57}{3\times10^8}\cos[\dots](5\hat k)$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.