Q. The electric field in a plane electromagnetic wave is given by $E_z = 60 \cos (5x + 1.5 \times 10^9 t)$ V/m. Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
A) $B_y = 2 \times 10^{-7} \cos (5x + 1.5 \times 10^9 t)$ T
B) $B_x = 2 \times 10^{-7} \cos (5x + 1.5 \times 10^9 t)$ T
C) $B_z = 60\cos (5x + 1.5 \times 10^9 t)$ T
D) $B_y = 60\sin (5x + 1.5 \times 10^9 t)$ T
Explanation
[Solution by Claude - not from source] The wave travels along $-x$ ($kx+\omega t$), $\omega/k=3\times10^8$ m/s. $B_0=E_0/c=60/(3\times10^8)=2\times10^{-7}$ T. $\vec E\times\vec B$ must point along $-x$; $\hat z\times\hat y=-\hat x$, so $\vec B$ is along $y$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.