Question

[The naive explanation for the pressure of light offered in Section 9.2.3 has its flaws, as you discovered if you worked Problem 9.11. Here’s another account, due originally to Planck.] A plane wave traveling through vacuum in the z direction encounters a perfect conductor occupying the region $\mathit{z}\mathbf{\ge }\mathbf{0}$, and reflects back:

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

(a) Find the accompanying magnetic field (in the region role="math" localid="1657454664985" $\mathbf{(}\mathit{z}\mathbf{>}\mathbf{0}\mathbf{)}$.

(b) Assuming inside the conductor, find the current K on the surface $\mathit{z}\mathbf{=}\mathbf{0}$, by invoking the appropriate boundary condition.

(c) Find the magnetic force per unit area on the surface, and compare its time average with the expected radiation pressure (Eq. 9.64).

Short answer

(a) The magnetic field is $B=\frac{E_0}{c}\left[\cos \right(kz-\omega t)+\cos (kz+\omega t\left)\right]\hat{y}$

(b) The current K on the surface is $K=\frac{2E_0}{{\mu }_{0}c}\cos \left(\omega t\right)\hat{x}$

(c) The magnetic force per unit area is $f={\epsilon }_0{E}_0^2$, and it is twice the pressure in Eq. 9.64.

Step-by-step solution

Expression for the electric field for : (z>0)

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

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

Here, k is the wave number, $\omega$is the angular frequency and t is the time.

Determine the accompanying magnetic field:

(a)

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

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

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

Determine the current K on the surfacez=0 :

(b)

It is known that $K\times (-\hat{Z})=\frac{1}{{\mu }_0}B$

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

$K\times (-\hat{z})=\frac{1}{{\mu }_0}\frac{E_0}{c}\left[\cos \right(kz-\omega t)+\cos (kz+\omega t\left)\right]\hat{y}$

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

Therefore, the current K on the surface is $K=\frac{2E_0}{{\mu }_{0}c}\cos \left(\omega t\right)\hat{x}$

Determine the magnetic force per unit area on the surface:

(c)

Write the expression for the force per unit area.

$f=K\times B_{\text{avg}}$

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

$f=\left(\frac{2E_0^2}{{\mu }_0{c}^2}\cos \left(\omega t\right)\hat{x}\right)\times \left[\cos \right(\omega t\left)\hat{y}\right]$

$f=2{\epsilon }_0{E}_0^2{\cos }^2\left(\omega t\right)\hat{z}$

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

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

$f={\epsilon }_0{E}_0^2$

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$f={\epsilon }_0{E}_0^2$,and it is twice the pressure in Eq. 9.64.