A plane electromagnetic wave of frequency 50 | vedclass

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Question

(A) The average energy density of the electric field is given by $K_{E} = \frac{1}{4} \varepsilon_0 E_0^2$.The average energy density of the magnetic

Vedclass · Accepted Answer

$K_{E} = K_{B}$

Vedclass · Answer

(A) The average energy density of the electric field is given by $K_{E} = \frac{1}{4} \varepsilon_0 E_0^2$.The average energy density of the magnetic field is given by $K_{B} = \frac{B_0^2}{4 \mu_0}$.In an electromagnetic wave,the relationship between the amplitudes of the electric and magnetic fields is $E_0 = c B_0$,where $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.Substituting $E_0 = c B_0$ into the expression for $K_{E}$:$K_{E} = \frac{1}{4} \varepsilon_0 (c B_0)^2 = \frac{1}{4} \varepsilon_0 \left(\frac{1}{\mu_0 \varepsilon_0}\right) B_0^2 = \frac{B_0^2}{4 \mu_0}$.Comparing the two expressions,we find that $K_{E} = K_{B}$.

$A$ plane electromagnetic wave of frequency $50 \ MHz$ travels in free space. If the average energy densities in the electric field and magnetic field are $K_{E}$ and $K_{B}$ respectively,then the correct option in the following is

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