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Let $\mathbf{E}_0$ and $\mathbf{B}_0$ denote the amplitude of electric and magnetic field of a plane electromagnetic wave in air. The magnitude of the average momentum transferred per unit area and per unit time to a totally absorbing surface is
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The correct answer is:
$\frac{1}{2} \varepsilon_0 E_0^2$
Intensity of electromagnetic wave,
$I=\frac{1}{2} \varepsilon_0 E_0^2 c=\frac{1}{2} \frac{B_0^2 c}{\mu_0}$
where, $B_0=$ electric field and $c=$ speed of light.
Now, momentum transfer per unit area per unit time, which is also called radiation pressure.
$\begin{gathered}
\frac{1}{A}\left(\frac{\Delta p}{\Delta t}\right)=\frac{I}{c}=\frac{\frac{1}{2} \varepsilon_0 E_0^2 c}{c} \\
\frac{I}{c}=\frac{1}{2} \varepsilon_0 E_0^2
\end{gathered}$
[where, $\Delta p=$ momentum transferred, $\Delta t=$ time and $A=$ area.]
$I=\frac{1}{2} \varepsilon_0 E_0^2 c=\frac{1}{2} \frac{B_0^2 c}{\mu_0}$
where, $B_0=$ electric field and $c=$ speed of light.
Now, momentum transfer per unit area per unit time, which is also called radiation pressure.
$\begin{gathered}
\frac{1}{A}\left(\frac{\Delta p}{\Delta t}\right)=\frac{I}{c}=\frac{\frac{1}{2} \varepsilon_0 E_0^2 c}{c} \\
\frac{I}{c}=\frac{1}{2} \varepsilon_0 E_0^2
\end{gathered}$
[where, $\Delta p=$ momentum transferred, $\Delta t=$ time and $A=$ area.]
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