For an electromagnetic wave travelling in fre | vedclass

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(A) In an electromagnetic wave $(EMW)$,the average energy density associated with the electric field is given by $U_{e} = \frac{1}{4} \epsilon_{0} E_{

Vedclass · Accepted Answer

$U_{e} = U_{m}$

Vedclass · Answer

(A) In an electromagnetic wave $(EMW)$,the average energy density associated with the electric field is given by $U_{e} = \frac{1}{4} \epsilon_{0} E_{0}^{2}$.The average energy density associated with the magnetic field is given by $U_{m} = \frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$.Since $E_{0} = c B_{0}$ and $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}$,we have $E_{0}^{2} = c^{2} B_{0}^{2} = \frac{B_{0}^{2}}{\mu_{0} \epsilon_{0}}$.Substituting this into the expression for $U_{e}$,we get $U_{e} = \frac{1}{4} \epsilon_{0} \left( \frac{B_{0}^{2}}{\mu_{0} \epsilon_{0}} \right) = \frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}} = U_{m}$.Thus,the average energy densities due to electric and magnetic fields are equal.

For an electromagnetic wave travelling in free space,the relation between average energy densities due to electric $(U_{e})$ and magnetic $(U_{m})$ fields is

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