The electric field of a plane electromagnetic | vedclass

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Question

(B) The amplitude of the electric field $(E_0)$ and magnetic field $(B_0)$ are related by the equation $E_0 / B_0 = c$,where $c$ is the speed of light

Vedclass · Accepted Answer

$8.86 	imes 10^{-12}$

Vedclass · Answer

(B) The amplitude of the electric field $(E_0)$ and magnetic field $(B_0)$ are related by the equation $E_0 / B_0 = c$,where $c$ is the speed of light.The average energy density of the magnetic field is given by $u_B = \frac{1}{4} \frac{B_0^2}{\mu_0}$.Substituting $B_0 = E_0 / c$,we get $u_B = \frac{1}{4} \frac{E_0^2}{\mu_0 c^2}$.Since $c^2 = 1 / (\mu_0 \varepsilon_0)$,we can rewrite the expression as $u_B = \frac{1}{4} \varepsilon_0 E_0^2$.Given $E_0 = 2 \, Vm^{-1}$ and $\varepsilon_0 = 8.854 	imes 10^{-12} \, F m^{-1}$,we have:$u_B = \frac{1}{4} 	imes 8.854 	imes 10^{-12} 	imes (2)^2$$u_B = 8.854 	imes 10^{-12} \approx 8.86 	imes 10^{-12} \, J m^{-3}$.

The electric field of a plane electromagnetic wave varies with time with an amplitude of $2 \, Vm^{-1}$ propagating along the $z$-axis. The average energy density of the magnetic field (in $J \, m^{-3}$) is:

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