91Ó°ÊÓ

Write the expression for electric field for $(z>0)$

$\mathit{E}\mathbf{(}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}{\mathit{E}}_{\mathbf{0}}\mathbf{\left[}\mathit{c}\mathit{o}\mathit{s}\mathbf{\right(}\mathit{k}\mathit{z}\mathbf{-}\mathit{Ó\neg }\mathit{t}\mathbf{)}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{k}\mathit{z}\mathbf{+}\mathit{Ó\neg }\mathit{t}\mathbf{\left)}\mathbf{\right]}\hat{\mathbf{x}}$ …… (1)

Here, k is the wave number, $\mathrm{Ó\neg }$is the angular frequency and t is the time.

(a)

Since, $(E×B)$points in the direction of propagation, write the equation for the magnetic field.

$B=\frac{E_0}{c}\left[\cos \right(kz-\mathrm{Ó\neg }t)+\cos (kz+\mathrm{Ó\neg }t\left)\right]\hat{y}$

Therefore, the magnetic field is $B=\frac{E_0}{c}\left[\cos \right(kz-\mathrm{Ó\neg }t)+\cos (kz+\mathrm{Ó\neg }t\left)\right]\hat{y}$

(b)

It is known that $K×(-\hat{Z})=\frac{1}{{\mathrm{μ}}_0}B$

Substitute $B=\frac{E_0}{c}\left[\cos \right(kz-\mathrm{Ó\neg }t)+\cos (kz+\mathrm{Ó\neg }t\left)\right]\hat{y}$in the above equation.

$K×(-\hat{z})=\frac{1}{{\mathrm{μ}}_0}\frac{E_0}{c}\left[\cos \right(kz-\mathrm{Ó\neg }t)+\cos (kz+\mathrm{Ó\neg }t\left)\right]\hat{y}$

$$\begin{gathered} =\frac{E_0}{{\mathrm{μ}}_{0}c}\left[2\cos \right(\mathrm{Ó\neg }t\left)\right]\hat{y} \\ K=\frac{2E_0}{{\mathrm{μ}}_{0}c}\cos \left(\mathrm{Ó\neg }t\right)\hat{x} \end{gathered}$$

(c)

Write the expression for the force per unit area.

$f=K×B_{\text{avg}}$

Substitute $K=\frac{2E_0}{{\mathrm{μ}}_{0}c}\cos \left(\mathrm{Ó\neg }t\right)\hat{x}$expression and $B_{avg}=\left(\cos \mathrm{Ó\neg }t\right)\hat{y}$in the above expression.

$f=\left(\frac{2E_0^2}{{\mathrm{μ}}_0{c}^2}\cos \left(\mathrm{Ó\neg }t\right)\hat{x}\right)×\left[\cos \right(\mathrm{Ó\neg }t\left)\hat{y}\right]$

$f=2{\mathrm{Î\mu }}_0{E}_0^2{\cos }^2\left(\mathrm{Ó\neg }t\right)\hat{z}$

It is known that the time average of ${\cos }^2\left(\mathrm{Ó\neg }t\right)\text{is}\frac{1}{2}$Hence, the above equation becomes,

$f=2{\mathrm{Î\mu }}_0{E}_0^2\left(\frac{1}{2}\right)\hat{z}$

$f={\mathrm{Î\mu }}_0{E}_0^2$

This is twice the pressure in Eq. 9.64, but that was for a perfect absorber, whereas this is a perfect reflector.

Therefore, the magnetic force per unit area is$f={\mathrm{Î\mu }}_0{E}_0^2$,and it is twice the pressure in Eq. 9.64.