In an electromagnetic wave,E = 50 sin(omega t | vedclass

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(A) The intensity $I$ (average power per unit area) of an electromagnetic wave is given by $I = \frac{1}{2} \varepsilon v E_p^2$,where $E_p$ is the pe

Vedclass · Accepted Answer

$1.65$

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(A) The intensity $I$ (average power per unit area) of an electromagnetic wave is given by $I = \frac{1}{2} \varepsilon v E_p^2$,where $E_p$ is the peak electric field amplitude and $v$ is the speed of the wave in the medium.Given $E_p = 50 \ V/m$,$\mu = 4\mu_0$,and $\varepsilon = \varepsilon_0$.The speed of the wave in the medium is $v = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{1}{\sqrt{4\mu_0\varepsilon_0}} = \frac{1}{2\sqrt{\mu_0\varepsilon_0}} = \frac{c}{2}$.Substituting $c = 3 	imes 10^8 \ m/s$,we get $v = \frac{3 	imes 10^8}{2} = 1.5 	imes 10^8 \ m/s$.Now,calculating the intensity $I = \frac{1}{2} \varepsilon_0 v E_p^2 = \frac{1}{2} 	imes (8.854 	imes 10^{-12}) 	imes (1.5 	imes 10^8) 	imes (50)^2$.$I = 0.5 	imes 8.854 	imes 10^{-12} 	imes 1.5 	imes 10^8 	imes 2500$.$I \approx 1.66 \ W/m^2$. Using the approximation $\varepsilon_0 \approx 8.8 	imes 10^{-12}$,we get $I = 0.5 	imes 8.8 	imes 10^{-12} 	imes 1.5 	imes 10^8 	imes 2500 = 1.65 \ W/m^2$.

In an electromagnetic wave,$E = 50 \sin(\omega t - kx)$. If $\mu = 4\mu_0$ and $\varepsilon = \varepsilon_0$,then the average power per unit area is $-$ (in $W/m^2$)

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