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  4. The magnetic field of a plane electromagnetic wave is given by: vecB=B0 hati [cos (kz-ω t)]+B1 hatj cos (kz+ω t) Where B0 = 3× 10-5 T and B1 = 2× 10-6 T The rms value of the force (in newton) experienced by a stationary charge Q=10-4 C at z=0 is:

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Q. 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 (in newton) experienced by a stationary charge $Q=10^{-4}\,C$ at $z=0$ is:

Electromagnetic Waves

Solution:

$B_{0}=\sqrt{B_{0}^{2}+B_{1}^{2}}=\sqrt{30^{2}+2^{2}}\times10^{-6}$
$\approx30\times10^{-6}T $
$\therefore E_{0}=cB=3\times10^{8}\times30\times10^{-6}$
$=9\times10^{3} V /m$
$E_{rms}=\frac{E_{0}}{\sqrt{2}}=\frac{9}{\sqrt{2}}\times10^{3}V/ m$
Force on the charge,
$F =E_{rms} Q=\frac{9}{\sqrt{2}}\times10^{3}\times10^{-4}\simeq0.64 N$