JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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:
- A \(\mathrm{B}_z=6.2 \times 10^{-8} \mathrm{~T}\)
- B \(\mathrm{B}_z=3.1 \times 10^{-8} \mathrm{~T}\)
- C \(\mathrm{B}_z=1.55 \times 10^{-8} \mathrm{~T}\)
- D \(\mathrm{B}_z=9.3 \times 10^{-8} \mathrm{~T}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{B}_z=3.1 \times 10^{-8} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{E}=\mathrm{BC} \\ & 9.3=\mathrm{B} \times 3 \times 10^8 \\ & \mathrm{~B}=\frac{9.3}{3 \times 10^8}=3.1 \times 10^{-8} \mathrm{~T}\end{aligned}\)
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