The electric field in an electromagnetic wave | vedclass

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Question

(D) The energy density $u$ of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E_0^2$,where $E_0$ is the amplitude of the electric fiel

Vedclass · Accepted Answer

$5.5 	imes 10^{-12} \ J$

Vedclass · Answer

(D) The energy density $u$ of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E_0^2$,where $E_0$ is the amplitude of the electric field.Given $E_0 = 50 \ N/C$,$\epsilon_0 = 8.85 	imes 10^{-12} \ F/m$.The volume $V$ of the cylindrical region is $A 	imes l = 10 \ cm^2 	imes 50 \ cm = 10 	imes 10^{-4} \ m^2 	imes 0.5 \ m = 5 	imes 10^{-4} \ m^3$.The total energy $U$ is $u 	imes V = \frac{1}{2} \epsilon_0 E_0^2 V$.$U = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes (50)^2 	imes (5 	imes 10^{-4})$.$U = 0.5 	imes 8.85 	imes 10^{-12} 	imes 2500 	imes 5 	imes 10^{-4}$.$U = 5.53 	imes 10^{-12} \ J \approx 5.5 	imes 10^{-12} \ J$.

The electric field in an electromagnetic wave is given by $E = 50 \sin \frac{2 \pi}{\lambda}(ct - x) \ N/C$. Find the energy contained in a cylindrical region of cross-section $10 \ cm^2$ and length $50 \ cm$ along the $x$-axis.

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