In the given electromagnetic wave E_y = 600 s | vedclass

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Question

(A) The intensity $I$ of an electromagnetic wave is given by the formula:$I = \frac{1}{2} \epsilon_0 E_0^2 c$Given:$E_0 = 600 \ V/m$$\epsilon_0 = 9

Vedclass · Accepted Answer

$486$

Vedclass · Answer

(A) The intensity $I$ of an electromagnetic wave is given by the formula:$I = \frac{1}{2} \epsilon_0 E_0^2 c$Given:$E_0 = 600 \ V/m$$\epsilon_0 = 9 	imes 10^{-12} \ C^2 N^{-1} m^{-2}$$c = 3 	imes 10^8 \ m/s$Substituting the values:$I = \frac{1}{2} 	imes (9 	imes 10^{-12}) 	imes (600)^2 	imes (3 	imes 10^8)$$I = \frac{1}{2} 	imes 9 	imes 10^{-12} 	imes 360000 	imes 3 	imes 10^8$$I = \frac{1}{2} 	imes 9 	imes 36 	imes 3 	imes 10^{-12} 	imes 10^4 	imes 10^8$$I = \frac{1}{2} 	imes 972 	imes 10^0$$I = 486 \ W/m^2$

In the given electromagnetic wave $E_y = 600 \sin (\omega t - kx) \ Vm^{-1}$,the intensity of the associated light beam is (in $W/m^2$); (Given $\epsilon_0 = 9 \times 10^{-12} \ C^2 N^{-1} m^{-2}$ and $c = 3 \times 10^8 \ m/s$)

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