The fields are E (z, t)=E_{0} \cos (k z-\omega t) \hat{ x }, B (z, t)=\frac{1}{c} E_{0} \cos (k z-\omega t) \hat{ y }, \text { with } \omega=c k .
(a) The electric force is F _{e}=q E =q E_{0} \cos (k z-\omega t) \hat{ x }=m a =m \frac{d v }{d t} , so
v =\frac{q E_{0}}{m} \hat{ x } \int \cos (k z-\omega t) d t=-\frac{q E_{0}}{m \omega} \sin (k z-\omega t) \hat{ x }+ C.
\text { But } v _{\text {ave }}= C = 0 , \text { so } v =-\frac{q E_{0}}{m \omega} \sin (k z-\omega t) \hat{ x } .
b) The magnetic force is
F _{m}=q( v \times B )=q\left(-\frac{q E_{0}}{m \omega}\right)\left(\frac{E_{0}}{c}\right) \sin (k z-\omega t) \cos (k z-\omega t)(\hat{ x } \times \hat{ y })=-\frac{q^{2} E_{0}^{2}}{m \omega c} \sin (k z-\omega t) \cos (k z-\omega t) \hat{ z }.
(c) The (time) average force is \left( F _{m}\right)_{ ave }=-\frac{q^{2} E_{0}^{2}}{m \omega c} \hat{ z } \int_{0}^{T} \sin (k z-\omega t) \cos (k z-\omega t) d t, \text { where } T=2 \pi / \omega is the period. The integral is -\left.\frac{1}{2 \omega} \sin ^{2}(k z-\omega t)\right|_{0} ^{T}=-\frac{1}{2 \omega}\left[\sin ^{2}(k z-2 \pi)-\sin ^{2}(k z)\right]=0, \text { so }\left( F _{m}\right)_{ ave }= 0 .
(d) Adding in the damping term,
F =q E -\gamma m v =q E_{0} \cos (k z-\omega t) \hat{ x }-\gamma m v =m \frac{d v }{d t} \Rightarrow \frac{d v }{d t}+\gamma v =\frac{q E_{0}}{m} \cos (k z-\omega t) \hat{ x }.
The steady state solution has the form v =A \cos (k z-\omega t+\theta) \hat{ x }, \quad \frac{d v }{d t}=A \omega \sin (k z-\omega t+\theta) \hat{ x } . Putting this in, and using the trig identity \cos u=\cos \theta \cos (u+\theta)+\sin \theta \sin (u+\theta) ,
A \omega \sin (k z-\omega t+\theta)+\gamma A \cos (k z-\omega t+\theta)=\frac{q E_{0}}{m}[\cos \theta \cos (k z-\omega t+\theta)+\sin \theta \sin (k z-\omega t+\theta)].
Equating like terms:
A \omega=\frac{q E_{0}}{m} \sin \theta, A \gamma=\frac{q E_{0}}{m} \cos \theta \Rightarrow \tan \theta=\frac{\omega}{\gamma}, A^{2}\left(\omega^{2}+\gamma^{2}\right)=\left(\frac{q E_{0}}{m}\right)^{2} \Rightarrow A=\frac{q E_{0}}{m \sqrt{\omega^{2}+\gamma^{2}}} .
So
v =\frac{q E_{0}}{m \sqrt{\omega^{2}+\gamma^{2}}} \cos (k z-\omega t+\theta) \hat{ x }, \theta \equiv \tan ^{-1}(\omega / \gamma) ; F _{m}=\frac{q^{2} E_{0}^{2}}{m c \sqrt{\omega^{2}+\gamma^{2}}} \cos (k z-\omega t+\theta) \cos (k z-\omega t) \hat{ z } .
To calculate the time average, write \cos (k z-\omega t+\theta)=\cos \theta \cos (k z-\omega t)-\sin \theta \sin (k z-\omega t) . We already know that the average of \cos (k z-\omega t) \sin (k z-\omega t) is zero, so
\left( F _{m}\right)_{ ave }=\frac{q^{2} E_{0}^{2}}{m c \sqrt{\omega^{2}+\gamma^{2}}} \hat{ z } \cos \theta \int_{0}^{T} \cos ^{2}(k z-\omega t) d t .
The integral is T / 2=\pi / \omega, \text { and } \cos \theta=\gamma / \sqrt{\omega^{2}+\gamma^{2}} \text { (see figure), so }\left( F _{m}\right)_{\text {ave }}=\frac{\pi \gamma q^{2} E_{0}^{2}}{m \omega c\left(\omega^{2}+\gamma^{2}\right)} \hat{ z } .
