Chapter 8, Problem 6P

(a)

To determine

Prove that the electron beam must be ejected out of the field in parallel path to the input beam as mentioned in the Figure given in the textbook.

Explanation of Solution

Calculation:

Refer to the mentioned Figure given in the textbook.

Consider the general expression to calculate the force using Newton’s second law of motion.

$F=ma=-eu\times B$        (1)

Here,

$m$ is the mass of the particle,

$a$ is the acceleration,

$u$ is the velocity, and

$B$ is the electric field intensity.

Rewrite the equation (1) as follows,

$\begin{array}{l}m\frac{du}{dt}=-e\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ u_x& u_y& u_z\\ 0& 0& B_{\text{o}}\end{array}\right|\\ m\frac{d}{dt}\left(u_x,u_y,u_z\right)=-e\left[\left(B_{\text{o}}{u}_y\right)a_x-\left(B_{\text{o}}{u}_x\right)a_y+\left(0\right)a_z\right]\\ m\frac{d}{dt}\left(u_x,u_y,u_z\right)=-e\left(B_{\text{o}}{u}_y,-B_{\text{o}}{u}_x,0\right)\\ m\frac{d}{dt}\left(u_x,u_y,u_z\right)=-e\left(B_{\text{o}}{u}_y,-B_{\text{o}}{u}_x,0\right)\end{array}$

Reduce the equation as follows,

$\frac{d}{dt}\left(u_x,u_y,u_z\right)=-\frac{e}{m}\left(B_{\text{o}}{u}_y,-B_{\text{o}}{u}_x,0\right)$        (2)

From equation (2), the velocities are,

$\frac{d{u}_x}{dt}=-\frac{e{B}_{\text{o}}}{m}{u}_y$

$\frac{d{u}_x}{dt}=-\omega u_y\left\{\because \omega =\frac{e{B}_{\text{o}}}{m}\right\}$        (3)

$\frac{d{u}_y}{dt}=\frac{e{B}_{\text{o}}}{m}{u}_x$

$\frac{d{u}_y}{dt}=\omega u_x\left\{\because \omega =\frac{e{B}_{\text{o}}}{m}\right\}$        (4)

$\frac{d{u}_z}{dt}=0\to u_z=C=0$        (5)

From equation (3),

$\begin{array}{l}\frac{d^2{u}_x}{d{t}^2}=-\omega \frac{d{u}_y}{dt}\\ {\ddot{u}}_x=-\omega {\dot{u}}_y\\ {\ddot{u}}_x=-{\omega }^2{u}_x\left\{\because {\dot{u}}_y=\omega u_x\right\}\\ {\ddot{u}}_x+{\omega }^2{u}_x=0\end{array}$

On solving the above expression,

$u_x=A\cos \omega t+B\sin \omega t$        (6)

From equation (3),

$\begin{array}{l}{\dot{u}}_x=-\omega u_y\\ u_y=-\frac{1}{\omega }{\dot{u}}_x\\ u_y=-\frac{1}{\omega }\frac{d}{dx}\left(A\cos \omega t+B\sin \omega t\right)\left\{\because u_x=A\cos \omega t+B\sin \omega t\right\}\\ u_y=-\frac{1}{\omega }\left(-A\omega \sin \omega t+B\omega \cos \omega t\right)\end{array}$

Reduce the equation as follows,

$u_y=A\sin \omega t-B\cos \omega t$        (7)

Substitute 0 for t in equation (6) and (7).

$u_x=u_{\text{o}},u_y=0\to A=u_{\text{o}},B=0$

Therefore, equation (6) and (7) becomes,

$\begin{array}{l}{u}_x=u_{\text{o}}\cos \omega t\\ \frac{dx}{dt}=u_{\text{o}}\cos \omega t\left\{\because u_x=\frac{dx}{dt}\right\}\end{array}$

$x=\frac{u_{\text{o}}}{\omega }\sin \omega t+c_1$        (8)

$\begin{array}{l}{u}_y=u_{\text{o}}\sin \omega t\\ \frac{dy}{dt}=u_{\text{o}}\sin \omega t\end{array}$

$y=-\frac{u_{\text{o}}}{\omega }\cos \omega t+c_2$        (9)

At $t=0$, $x=y=0\to c_1=0,c_2=\frac{u_{\text{o}}}{\omega }$. Therefore, the equation (8) and (9) becomes,

$x=\frac{u_{\text{o}}}{\omega }\sin \omega t$        (10)

$y=-\frac{u_{\text{o}}}{\omega }\cos \omega t+\frac{u_{\text{o}}}{\omega }$

$y=\frac{u_{\text{o}}}{\omega }\left(1-\cos \omega t\right)$        (11)

From equation (10) and (11),

${\left(\frac{u_{\text{o}}}{\omega }\right)}^2\left(co{s}^2\omega t+{\sin }^2\omega t\right)={\left(\frac{u_{\text{o}}}{\omega }\right)}^2=x^2+{\left(y-\frac{u_{\text{o}}}{\omega }\right)}^2$

From the above expression, the electron must move in a circle which is centered at $\left(0,\frac{u_{\text{o}}}{\omega }\right)$ with radius $\frac{u_{\text{o}}}{\omega }$. In the referred Figure, the field does not exist throughout the circular region, therefore electron passes through a semi-circle and ejected out horizontally.

Conclusion:

Thus, the electron beam must be ejected out of the field in parallel path to the input beam is proved.

(b)

To determine

Find an expression for the exit distance (d) above the entry point as shown in the mentioned Figure given in the textbook.

Answer to Problem 6P

The exit distance (d) above the entry point is $\frac{2u_{\text{o}}m}{e{B}_{\text{o}}}$.

Explanation of Solution

Calculation:

Refer to the mentioned Figure given in the textbook. The distance d is twice the radius of the semi-circle. Therefore,

$\begin{array}{c}d=\frac{2u_{\text{o}}}{\omega }\\ =\frac{2u_{\text{o}}}{\left(\frac{e{B}_{\text{o}}}{m}\right)}\left\{\because \omega =\frac{e{B}_{\text{o}}}{m}\right\}\\ =\frac{2u_{\text{o}}m}{e{B}_{\text{o}}}\end{array}$

Conclusion:

Thus, the exit distance (d) above the entry point is $\frac{2u_{\text{o}}m}{e{B}_{\text{o}}}$.