Q. A plane $EM$ wave of frequency $30\ MHz$ travels in free space along the $x$-direction. The electric field component of the wave at a particular point of space and time $E = 6\ V/m$ along $y$-direction. Its magnetic field component $B$ at this point would be
A) $2 \times 10^{- 8}$ T along $z$-direction
B) $6 \times 10^{- 6}$ T along $x$-direction
C) $2 \times 10^{- 8}$ T along $y$-direction
D) $6 \times 10^{- 8}$ T along $z$-direction
Explanation
(a) Here, $E = 6V/m,\ c = 3 \times 10^{8}ms^{- 1}$
$B = \frac{E}{C} = \frac{6V/m}{3 \times 10^{8}ms^{- 1}} = 2 \times 10^{- 8}T$
$E$ is along the $y$-direction and the plane $e.m.$ wave propagate along $x$-direction. Therefore, $B$ should be in a direction perpendicular to both $x$ and $y$-axis. Using vector algebra $\overline{E} \times \overline{B}$ should be along $x$-direction. Since $\left( + \widehat{j} \right) \times \left( + \widehat{k} \right) = \widehat{i},\ B$ is along the $z$-direction.
Thus, magnetic field component $B$ would be $2 \times 10^{- 8}T$ along $z$-direction
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.