A plane electromagnetic wave with electric fi | vedclass

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Question

(B) For a plane electromagnetic wave incident on a completely absorbing surface,the momentum $p$ delivered is given by the relation $p = \frac{U}{c}$,

Vedclass · Accepted Answer

$\frac{U B_0}{E_0}$

Vedclass · Answer

(B) For a plane electromagnetic wave incident on a completely absorbing surface,the momentum $p$ delivered is given by the relation $p = \frac{U}{c}$,where $U$ is the energy transferred and $c$ is the speed of light in vacuum.We know that for an electromagnetic wave,the relationship between the electric field amplitude $E_0$ and magnetic field amplitude $B_0$ is given by $c = \frac{E_0}{B_0}$.Substituting this value of $c$ into the momentum equation:$p = \frac{U}{E_0 / B_0} = \frac{U B_0}{E_0}$.Thus,the magnitude of the total momentum delivered is $\frac{U B_0}{E_0}$.

$A$ plane electromagnetic wave with electric field amplitude $E_0$ and magnetic field amplitude $B_0$ is incident on a surface. If the total energy transferred to the surface in time $t$ is $U$,then the magnitude of the total momentum delivered to the surface for complete absorption is

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