A mathematical representation of an electroma | vedclass

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Question

(C) The intensity $I$ of an electromagnetic wave is defined as the average energy per unit area per unit time,which is given by $I = \langle u \rangle

Vedclass · Accepted Answer

$I = \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_{max}^2$

Vedclass · Answer

(C) The intensity $I$ of an electromagnetic wave is defined as the average energy per unit area per unit time,which is given by $I = \langle u \rangle c$,where $\langle u \rangle$ is the average energy density and $c$ is the speed of light.The total energy density $u$ is the sum of electric and magnetic energy densities: $u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2\mu_0} B^2$.Since $E = cB$ and $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,the average energy density is $\langle u \rangle = \frac{1}{2} \epsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} B_{rms}^2 = \frac{1}{2} \epsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} (\frac{E_{rms}}{c})^2 = \epsilon_0 E_{rms}^2$.Given $E_{rms} = \frac{E_{max}}{\sqrt{2}}$,we have $\langle u \rangle = \epsilon_0 \frac{E_{max}^2}{2}$.Thus,$I = \langle u \rangle c = \frac{1}{2} \epsilon_0 c E_{max}^2$.Substituting $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we get $I = \frac{1}{2} \epsilon_0 \frac{1}{\sqrt{\mu_0 \epsilon_0}} E_{max}^2 = \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_{max}^2$.

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