91影视

Write the expression for the electric field.

$\mathit{F}\left(蝇tz,xt\right)\mathbf{=}{\mathit{E}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\left(kz-\right)\hat{}$ 鈥︹�� (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.

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

Here,cis the speed of light.

Write the expression for the frequency.

$\mathit{蝇}\mathbf{=}\mathit{c}\mathit{k}$ 鈥︹�� (3)

(a)

Write the expression for an electric force.

$F_0=qE$

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

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

Write the equation for the velocity.

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

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

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

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

(b)

Write the expression for the magnetic force.

$F_m=q\left(v脳B\right)$

Here,qis the charge.

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

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

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

(c)

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

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

Here, $T=\frac{2\mathrm{蟺}}{蝇}$is the period.

On further solving, the above equation becomes,

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

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

(d)

Adding in the damping terms, form the required equation.

$$\begin{gathered} F=qE-纬mv \\ m\frac{dv}{dt}=q{E}_0\cos \left(kz-蝇t\right)\hat{x}-纬mv \\ \frac{dv}{dt}+纬v=\frac{q{E}_0}{m}\cos \left(kz-蝇t\right)\hat{x} \end{gathered}$$ 鈥︹�� (5)

The steady state solution has the following form.

$v=A\cos \left(kz-蝇t+胃\right)\hat{x}$ 鈥︹�� (6)

Find the derivative of the above equation.

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

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

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

Equate the sine terms.

$A蝇\frac{q{E}_0}{m}\sin 胃$ 鈥︹�� (7)

Equate the cosine terms.

$A蝇\frac{q{E}_0}{m}\cos 胃$ 鈥︹�� (8)

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

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

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

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

Hence, the magnetic force will be,

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

Write the equation to calculate the time average.

$\cos \left(kz-蝇t+胃\right)=\cos 胃\cos \left(kz-蝇t\right)-\sin 胃\sin \left(kz-蝇t\right)$.

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

${\left(F_m\right)}_{avg}=\frac{q^2{E}_0^2}{mc\sqrt{{蝇}^2+{纬}^2}}\hat{z}\cos 胃{鈭�}_0^T{\cos }^2\left(kz-蝇t\right)dt$

Substitute $\cos 胃=\frac{纬}{\sqrt{{蝇}^2+{纬}^2}}$in the above equation.

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

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