Physics · 2019
JEE Main · 9 April 2019, Shift 1 · Q22
The magnetic field of a plane electromagnetic wave is given by: B = B_0 i [ cos ( kz -ω t )]+ B_1 j cos ( kz +ω t ) where B_0=3 × 10^-5 T and B_1=2 ×…
The magnetic field of a plane electromagnetic wave is given by :
$$\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{i}[\cos (\mathrm{kz}-\omega \mathrm{t})]+\mathrm{B}_1 \hat{j} \cos (\mathrm{kz}+\omega \mathrm{t})
$$
where $\displaystyle \mathrm{B}_0=3 \times 10^{-5} \mathrm{~T}$ and $\displaystyle \mathrm{B}_1=2 \times 10^{-6} \mathrm{~T}$. The rms value of the force experienced by a stationary charge $\displaystyle \mathrm{Q}=10^{-4} \mathrm{C}$ at $\displaystyle \mathrm{z}=0$ is closest to :
Official answer
(1)
0.$\displaystyle 6$ N
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.