A plane electromagnetic wave of frequency 20 MHz travels in free space along the $+x$ direction. At a…

A plane electromagnetic wave of frequency 20 MHz travels in free space along the $+x$ direction. At a particular point in space and time, the electric field vector of the wave is $\mathrm{E}_y=9.3 \mathrm{Vm}^{-1}$. Then, the magnetic field vector of the wave at that point is
  1. $\mathrm{B}_z=6.2 \times 10^{-8} \mathrm{~T}$
  2. $\mathrm{B}_z=3.1 \times 10^{-8} \mathrm{~T}$
  3. $\mathrm{B}_z=1.55 \times 10^{-8} \mathrm{~T}$
  4. $\mathrm{B}_z=9.3 \times 10^{-8} \mathrm{~T}$

Solution

$\begin{aligned} & \mathrm{E}=\mathrm{BC} \\ & 9.3=\mathrm{B} \times 3 \times 10^8 \\ & \mathrm{~B}=\frac{9.3}{3 \times 10^8}=3.1 \times 10^{-8} \mathrm{~T}\end{aligned}$

Asked in: JEE Main 2025 (23 Jan Shift 2)

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