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# A plane EM wave travelling along z-direction

Question:

A plane EM wave travelling along z-direction is described by

$E=E_{0} \sin (k z-\omega t) \hat{i}$ and $B=B_{0} \sin (k z-\omega t) \hat{j}$ show that,

(i) the average energy density of the wave is given by

$u_{a v}=\frac{1}{4} \epsilon_{0} E_{0}^{2}+\frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$

(ii) the time-averaged intensity of the wave is given by

$I_{a v}=\frac{1}{2} c \epsilon_{0} E_{0}^{2}$

Solution:

(i) The energy density due to electric field E is

uE = 1/2 ε0E2

The energy density due to magnetic field B is

uB = 1/2 B20

The average energy density of the wave is given by:

$u_{a v}=\frac{1}{4} \epsilon_{0} E_{0}^{2}+\frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$

(ii) We know that c=1/√μ0ε0

The time-averaged intensity of the wave is given as

$I_{a v}=\frac{1}{2} c \epsilon_{0} E_{0}^{2}$