The electric field of a plane electromagnetic | vedclass

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Question

(B) In an electromagnetic wave,the average energy density of the electric field $\langle u_E \rangle$ is equal to the average energy density of the ma

Vedclass · Accepted Answer

$8.85 	imes 10^{-12}$

Vedclass · Answer

(B) In an electromagnetic wave,the average energy density of the electric field $\langle u_E angle$ is equal to the average energy density of the magnetic field $\langle u_B angle$.The formula for the average energy density of the electric field is given by $\langle u_E angle = \frac{1}{4} \varepsilon_0 E_0^2$.Given,amplitude $E_0 = 2 \, V/m$ and permittivity of free space $\varepsilon_0 = 8.85 	imes 10^{-12} \, F/m$.Substituting the values:$\langle u_B angle = \langle u_E angle = \frac{1}{4} 	imes (8.85 	imes 10^{-12}) 	imes (2)^2$$\langle u_B angle = \frac{1}{4} 	imes 8.85 	imes 10^{-12} 	imes 4$$\langle u_B angle = 8.85 	imes 10^{-12} \, J/m^3$.

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

Similar Questions

The electric field part of an electromagnetic wave in a medium is represented by:
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$E_z = 0.$
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