Consider a particle of charge q and mass m, free to move in the $\mathit{x}\mathit{y}$plane in response to an electromagnetic wave propagating in the z direction (Eq. 9.48—might as well set $\mathit{\delta }\mathbf{=}\mathbf{0}\mathbf{)}$).

(a) Ignoring the magnetic force, find the velocity of the particle, as a function of time. (Assume the average velocity is zero.)

(b) Now calculate the resulting magnetic force on the particle.

(c) Show that the (time) average magnetic force is zero.

The problem with this naive model for the pressure of light is that the velocity is ${\mathbf{90}}^{\mathbf{°}}$out of phase with the fields. For energy to be absorbed there’s got to be some resistance to the motion of the charges. Suppose we include a force of the form $\mathit{y}\mathit{m}\mathit{v}$, for some damping constant $\mathit{y}$.

(d) Repeat part (a) (ignore the exponentially damped transient). Repeat part (b), and find the average magnetic force on the particle.

Write the expression for the electric field.

$\mathit{F}\left(\omega tz,xt\right)\mathbf{=}{\mathit{E}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\left(kz-\right)\hat{}$ …… (1)

Here,tis the time,is the peak electric field,kis the wave number andis the angular frequency.

Write the expression for the magnetic field.

$\mathit{P}\left(\omega tz,yt\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{c}}{\mathit{E}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\left(kz-\right)\hat{}$ …… (2)

Here,cis the speed of light.

Write the expression for the frequency.

$\mathit{\omega }\mathbf{=}\mathit{c}\mathit{k}$ …… (3)

(a)

Write the expression for an electric force.

$F_0=qE$

Substitute $E=E_0\cos \left(kz-\omega t\right)\hat{x}$in the above expression.

$$\begin{gathered} F_0=q{E}_0\cos \left(kz-\omega t\right)\hat{x} \\ {F}_0=ma \\ {F}_0=m\frac{dv}{dt} \end{gathered}$$

Write the equation for the velocity.

role="math" localid="1655720398237" $$\begin{gathered} v=\frac{q{E}_0}{m}\hat{x}\int \cos \left(kz-\omega t\right)dt \\ =-\frac{q{E}_0}{m\omega }\sin \left(kz-wt\right)\hat{x}+C \end{gathered}$$

Here, $C=0$. Hence, the above equation becomes,

role="math" localid="1655720477833" $v=-\frac{q{E}_0}{m\omega }\sin \left(kz-\omega t\right)\hat{x}$

Therefore, the velocity of the particle is $v=-\frac{q{E}_0}{m\omega }\sin \left(kz-\omega t\right)\hat{x}$.

(b)

Write the expression for the magnetic force.

$F_m=q\left(v\times B\right)$

Here,qis the charge.

Substitute $v=-\frac{q{E}_0}{m\omega }\sin \left(kz-\omega t\right)\hat{x}$, and $B=\frac{1}{c}{E}_0\cos \left(kz-\omega t\right)\hat{y}$ in the above expression.

$$\begin{gathered} F_m=q\left(-\frac{q{E}_0}{m\omega }\right)\left(\frac{E_0}{c}\right)\sin \left(kz-\omega t\right)\cos \left(kz-\omega t\right)\left(\hat{x}\times \hat{y}\right) \\ {F}_m=-\frac{q^2{E}_0^2}{m\omega c}\sin \left(kz-wt\right)\cos \left(kz-\omega t\right)\hat{z} \end{gathered}$$ …… (4)

Therefore, the resulting magnetic force on the particle is $F_m=-\frac{q^2{E}_0^2}{m\omega c}\sin \left(kz-wt\right)\cos \left(kz-\omega t\right)\hat{z}$.

(c)

Integrate equation (4) to find the average magnetic force.

${\left(F_m\right)}_{avg}=-\frac{q^2{E}_0^2}{m\omega c}\hat{z}{\int }_0^T\sin \left(kz-\omega t\right)\cos \left(kz-\omega \omega t\right)dt$

Here, $T=\frac{2\mathrm{\pi }}{\omega }$is the period.

On further solving, the above equation becomes,

$$\begin{gathered} {\left(F_m\right)}_{avg}=-\frac{q^2{E}_0^2}{m\omega c}\hat{z}{\left(-\frac{1}{2\omega }{\sin }^2\left(kz-\omega t\right)\right)}_0^T \\ =\left(\frac{q^2{E}_0^2}{m\omega c}\hat{z}\right)\left(-\frac{1}{2\omega }\right)\left[{\sin }^2\left(kz-2\pi \right)-{\sin }^2\left(kz\right)\right] \\ =\left(\frac{q^2{E}_0^2}{m\omega c}\hat{z}\right)\left(-\frac{1}{2\omega }\right)\left[{\sin }^2\left(kz\right)-{\sin }^2\left(kz\right)\right] \\ =0 \end{gathered}$$

Therefore, it is proved that the (time) average magnetic force is zero.

(d)

Adding in the damping terms, form the required equation.

$$\begin{gathered} F=qE-\gamma mv \\ m\frac{dv}{dt}=q{E}_0\cos \left(kz-\omega t\right)\hat{x}-\gamma mv \\ \frac{dv}{dt}+\gamma v=\frac{q{E}_0}{m}\cos \left(kz-\omega t\right)\hat{x} \end{gathered}$$ …… (5)

The steady state solution has the following form.

$v=A\cos \left(kz-\omega t+\theta \right)\hat{x}$ …… (6)

Find the derivative of the above equation.

$\frac{dv}{dt}=A\omega \sin \left(kz-\omega t+\theta \right)\hat{x}$

Substitute $\frac{dv}{dt}=A\omega \sin \left(kz-\omega t+\theta \right)\hat{x}$and $v=A\cos \left(kz-\omega t+\theta \right)\hat{x}$in equation (5).

$$\begin{gathered} A\omega \sin \left(kz-\omega t+\theta \right)\hat{x}+\gamma A\cos \left(kz-\omega t+\theta \right)\hat{x}=\frac{q{E}_0}{m}\cos \left(kz-\omega t\right)\hat{x} \\ \left[\begin{array}{c}A\omega \sin \left(kz-\omega t+\theta \right)\hat{x}+\\ \gamma A\cos \left(kz-\omega +\theta \right)\hat{x}\end{array}\right]=\frac{q{E}_0}{m}\left[\begin{array}{c}\cos \theta \cos \left(kz-\omega t+\theta \right)+\\ \sin \theta \sin \left(kz-\omega t+\theta \right)\end{array}\right] \end{gathered}$$

Equate the sine terms.

$A\omega \frac{q{E}_0}{m}\sin \theta$ …… (7)

Equate the cosine terms.

$A\omega \frac{q{E}_0}{m}\cos \theta$ …… (8)

Square and add the equation (7) and (8).

$$\begin{gathered} A^2\left({\omega }^2+{\gamma }^2\right)={\left(\frac{q{E}_0}{m}\right)}^2 \\ A=\frac{q{E}_0}{m\sqrt{{\omega }^2+{\gamma }^2}} \end{gathered}$$

Substitute $A=\frac{q{E}_0}{m\sqrt{{\omega }^2+{\gamma }^2}}$in equation (6).

role="math" localid="1655722215918" $v=\frac{q{E}_0}{m\sqrt{{\omega }^2+{\gamma }^2}}\cos \left(kz-\omega t+\theta \right)\hat{x}$

Hence, the magnetic force will be,

$F_m=\frac{q{E}_0}{mc\sqrt{{\omega }^2+{\gamma }^2}}\cos \left(kz-\omega t+\theta \right)\cos \left(kz-\omega t\right)\hat{z}$

Write the equation to calculate the time average.

$\cos \left(kz-\omega t+\theta \right)=\cos \theta \cos \left(kz-\omega t\right)-\sin \theta \sin \left(kz-\omega t\right)$.

It is known that the average of $\cos \left(kz-\omega t\right)\sin \left(kz-\omega t\right)$is zero, so, the average magnetic force equation becomes,

${\left(F_m\right)}_{avg}=\frac{q^2{E}_0^2}{mc\sqrt{{\omega }^2+{\gamma }^2}}\hat{z}\cos \theta {\int }_0^T{\cos }^2\left(kz-\omega t\right)dt$

Substitute $\cos \theta =\frac{\gamma }{\sqrt{{\omega }^2+{\gamma }^2}}$in the above equation.

$$\begin{gathered} {\left(F_m\right)}_{avg}=\left(\frac{q^2{E}_0^2}{mc\sqrt{{\omega }^2+{\gamma }^2}}\hat{z}\right)\left(\frac{\gamma }{\sqrt{{\omega }^2+{\gamma }^2}}\right)\left(\frac{T}{2}\right) \\ {\left(F_m\right)}_{avg}=\left(\frac{q^2{E}_0^2}{mc\sqrt{{\omega }^2+{\gamma }^2}}\hat{z}\right)\left(\frac{\gamma }{\sqrt{{\omega }^2+{\gamma }^2}}\right)\left(\frac{\pi }{\omega }\right) \\ {\left(F_m\right)}_{avg}=\frac{\pi \gamma q^2{E}_0^2}{m\omega c\left({\omega }^2+{\gamma }^2\right)}\hat{z} \end{gathered}$$

Therefore, the velocity of the particle is $v=\frac{q{E}_0}{m\sqrt{{\omega }^2+{\gamma }^2}}\cos \left(kz-\omega t+\theta \right)\cos \left(kz-\omega t\right)\hat{z}$, the resulting magnetic force on the particle is $F_m=\frac{q{E}_0}{mc\sqrt{{\omega }^2+{\gamma }^2}}\cos \left(kz-\omega t+\theta \right)\cos \left(kz-\omega t\right)\hat{z}$and the (time) average magnetic force is ${\left(F_m\right)}_{avg}=\frac{\pi \gamma q^2{E}_0^2}{m\omega c\left({\omega }^2+{\gamma }^2\right)}\hat{z}$.