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 $u = \frac{1}{2} \epsilon_0 E_{rms}^2 + \frac{1}{2} \frac{B_{rms}^2}{\mu_0}$. Since

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The energy density $u$ of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E_{rms}^2 + \frac{1}{2} \frac{B_{rms}^2}{\mu_0}$. Since the energy density of the electric field equals the energy density of the magnetic field,the total energy density is $u = \epsilon_0 E_{rms}^2$. Given $E = 50 \sin \frac{2 \pi}{\lambda}(ct - x)$,the peak electric field is $E_0 = 50 \ N/C$. The root mean square electric field is $E_{rms} = \frac{E_0}{\sqrt{2}}$. Thus,the total energy density is $u = \epsilon_0 \left(\frac{E_0}{\sqrt{2}}ight)^2 = \frac{1}{2} \epsilon_0 E_0^2$. The volume of the cylindrical region is $V = A 	imes l = 10 \ cm^2 	imes 50 \ cm = 10 	imes 10^{-4} \ m^2 	imes 50 	imes 10^{-2} \ m = 500 	imes 10^{-6} \ m^3 = 5 	imes 10^{-4} \ m^3$. The total energy $U = u 	imes V = \frac{1}{2} \epsilon_0 E_0^2 	imes V$. Substituting the values: $U = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes (50)^2 	imes 5 	imes 10^{-4}$. $U = 0.5 	imes 8.85 	imes 10^{-12} 	imes 2500 	imes 5 	imes 10^{-4} = 5.53 	imes 10^{-12} \ J$. Rounding to the nearest option,$U \approx 5.5 	imes 10^{-12} \ J$.

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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