Question

Electrons undergoing cyclotron motion can be sped up by increasing the magnetic field; the accompanying electric field will impart tangential acceleration. This is the principle of the betatron. One would like to keep the radius of the orbit constant during the process. Show that this can be achieved by designing a magnet such that the average field over the area of the orbit is twice the field at the circumference (Fig. 7.53). Assume the electrons start from rest in zero field, and that the apparatus is symmetric about the center of the orbit. (Assume also that the electron velocity remains well below the speed of light, so that nonrelativistic mechanics applies.) [Hint: Differentiate Eq. 5.3 with respect to time, and use .$F=ma=qE$]

Short answer

It is proved that the average field over the area of the orbit is twice the field at the circumference.

Step-by-step solution

Write the given data from question.

The electron velocity is remains below the speed of light.

The start is from the rest.

The force on electron is as follows:$\begin{array}{l}qE=ma\\ F=qE\end{array}$

Determine the formula to prove that the average field over the area of the orbit is twice the field at the circumference.

The expression for magnetic force on the electron is given as follows.

$F_B=Bqv$ …… (1)

Here,$B$ is the magnetic field,$q$ is the charge and $v$ is the velocity.

The expression for centripetal force acting on the electron is given as follows.

$F_c=\frac{m{v}^2}{R}$ ……. (2)

Here,$m$ is the mass of the electron and $R$ is the radius.

The expression for the electric force on the electron is given by,

role="math" localid="1658239700853" $F_E=qE$ ……. (3)

Prove that the average field over the area of the orbit is twice the field at the circumference.

The magnetic and centripetal force on the electron is equal.

$\begin{array}{l}Bqv=\frac{m{v}^2}{R}\\ qBR=mv\end{array}$

Differentiate the above equation with respect to $t$.

$qR\frac{dB}{dt}=m\frac{dv}{dt}$

Substitute$a$for $\frac{dv}{dt}$into above equation.

$qR\frac{dB}{dt}=ma$

Substitute $qE$for $ma$into above equation.

$\begin{array}{l}qR\frac{dB}{dt}=qE\\ R\frac{dB}{dt}=E\end{array}$ ……. (4)

We know,

$\begin{array}{l}\int E\cdot dl=-\frac{d\varphi }{dt}\\ E\left(2\pi R\right)=-\frac{d\varphi }{dt}\\ E=-\frac{1}{2\pi R}\frac{d\varphi }{dt}\end{array}$

Substitute$-\frac{1}{2\pi R}\frac{d\varphi }{dt}$ for $E$into equation (4).

$\begin{array}{l}R\frac{dB}{dt}=-\frac{1}{2\pi R}\frac{d\varphi }{dt}\\ \frac{dB}{dt}=-\frac{1}{2\pi R^2}\frac{d\varphi }{dt}\\ B=-\frac{1}{2}\frac{1}{\pi R^2}\varphi +C\end{array}$

Here, $C$is constant.

At$t=0,B=0,C=0$

Therefore,

$B\left(R\right)=\frac{1}{2}\left(\frac{1}{\pi R^2}\varphi \right)$

Hence,the average field over the area of the orbit is twice the field at the circumference.