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JEE MAIN 2019
09-04-2019 - S1
Question
The magnetic field of a plane electromagnetic wave is given by : $\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_0 \hat{\mathrm{i}}\left[\cos (\mathrm{kz}-\omega \mathrm{t})+\mathrm{B}_1 \hat{\mathrm{j}} \cos (\mathrm{kz}+\omega \mathrm{t})\right.$ where $\mathrm{B}_0=3 \times 10^{-5} \mathrm{~T}$ and $\mathrm{B}_1=2 \times 10^{-6} \mathrm{~T}$. The rms value of the force experienced by a stationary charge Q = $10^{-4}$ C at z =0 is closest to:
Select the correct option:
A
0.6 N
B
0.9 N
C
$3 \times 10^{-2} \mathrm{~N}$
D
0.1 N
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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