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The magnetic field of a plane electromagnetic wave is given by :

$\vec{B}=B_{0}\; \hat{i}\left [ cos\left ( kz-\omega t \right ) \right ]+B_{1}\hat{j}\; cos\left ( kz+\omega t \right )$

where $B_{0}=3\times 10^{-5}T$ and $B_{1}=2\times 10^{-6}T$ .

The rms value of the force experienced by a stationary charge $Q=10^{-4}C$ at $Z=0$ is closest to :

Option 1)
$0.6 \: N$

Option 2)
$0.1\: N$

Option 3)
$0.9\: N$

Option 4)
$3\times 10^{-2}\; N$

Answers (1)

best_answer

$B=B_{0}\; cos\left ( kx-\omega t \right )\hat{i}+B_{1}\; cos \left ( kz+\omega t \right )\hat{j}$

$B_{0}=3\times 10^{-5}T$

$B_{1}=2\times 10^{-6}T$

$Q=10^{-4}C$

$F_{rms}=Q\sqrt{\left ( \frac{CB_{0}}{\sqrt{2}} \right )^{2}+\left ( \frac{CB_{1}}{\sqrt{2}} \right )^{2}}$

$=10^{-4}\times \frac{3\times 10^{8}}{\sqrt{2}}\sqrt{\left ( 3\times 10^{-5} \right )^{2}+\left ( 2\times 10^{-6} \right )^{2}}$

$=10^{-4}\times \frac{3\times 10^{8}}{\sqrt{2}}\sqrt{9\times 10^{-10}+4\times 10^{-12}}$

$=10^{-4}\times \frac{3\times 10^{8}}{\sqrt{2}}\sqrt{900\times 10^{-12}+4\times 10^{-12}}$

$=\frac{3\times 10^{4}}{\sqrt{2}}\sqrt{90{4}}\times 10^{-6}$

$=0.63\; N$

Option 1)

$0.6 \: N$

Option 2)

$0.1\: N$

Option 3)

$0.9\: N$

Option 4)

$3\times 10^{-2}\; N$

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