The electric field of plane electromagnetic w | vedclass

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Question

$\left\langle {{{\rm{u}}_{\rm{B}}}} \right\rangle  = \left\langle {{{\rm{u}}_{\rm{E}}}} \right\rangle $

$=\frac{1}{4} \varepsilon_{0} \mathrm{

Vedclass · Accepted Answer

$8.85 	imes 10^{-12}$

Vedclass · Answer

$\left\langle {{{m{u}}_{m{B}}}} ightangle  = \left\langle {{{m{u}}_{m{E}}}} ightangle $

$=\frac{1}{4} \varepsilon_{0} \mathrm{E}_{0}^{2}=\frac{1}{4} 	imes 8.85 	imes 10^{-12}(2)^{2} $

$=8.85 	imes 10^{-12} \mathrm{\,J} / \mathrm{m}^{3} $

The electric field of plane electromagnetic wave of amplitude $2\,V/m$ varies with time, propagating along $z-$ axis. The average energy density of magnetic field (in $J/m^3$ ) is

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