The magnetic field of a plane electromagnetic | vedclass

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Question

(B) The total average energy density of an electromagnetic wave is $U_{avg} = \frac{B_0^2}{2 \mu_0}$.Given $B_0 = 400 \ \mu T = 400 	imes 10^{-6} \ T

Vedclass · Accepted Answer

$31.8 	imes 10^{-3} \ J \ m^{-3}$

Vedclass · Answer

(B) The total average energy density of an electromagnetic wave is $U_{avg} = \frac{B_0^2}{2 \mu_0}$.Given $B_0 = 400 \ \mu T = 400 	imes 10^{-6} \ T$ and $\mu_0 = 4 \pi 	imes 10^{-7} \ T \ m/A$.Substituting the values: $U_{avg} = \frac{(400 	imes 10^{-6})^2}{2 	imes 4 \pi 	imes 10^{-7}} = \frac{16 	imes 10^{-8}}{8 \pi 	imes 10^{-7}} = \frac{2 	imes 10^{-1}}{\pi} \approx 0.06366 \ J \ m^{-3} = 63.66 	imes 10^{-3} \ J \ m^{-3}$.In an electromagnetic wave,the average energy density is equally shared between the electric field and the magnetic field.Therefore,the average energy density corresponding to the electric field is $U_E = \frac{U_{avg}}{2} = \frac{63.66 	imes 10^{-3}}{2} = 31.83 	imes 10^{-3} \ J \ m^{-3}$.

The magnetic field of a plane electromagnetic wave is given by $B = (400 \ \mu T) \sin [ (4.0 \times 10^{15} \ s^{-1}) (t - \frac{x}{c}) ]$. The average energy density corresponding to the electric field is:

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