91影视

The electron velocity is remains below the speed of light.

The start is from the rest.

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

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

$F_B=Bqv$ 鈥︹�� (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.

$F_c=\frac{m{v}^2}{R}$ 鈥︹��. (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" $F_E=qE$ 鈥︹��. (3)

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.

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

We know,

$\begin{array}{l}鈭�E鈰�dl=鈭�\frac{d蠒}{dt}\\ E\left(2蟺R\right)=鈭�\frac{d蠒}{dt}\\ E=鈭�\frac{1}{2蟺R}\frac{d蠒}{dt}\end{array}$

Substitute$鈭�\frac{1}{2蟺R}\frac{d蠒}{dt}$ for $E$into equation (4).

$\begin{array}{l}R\frac{dB}{dt}=鈭�\frac{1}{2蟺R}\frac{d蠒}{dt}\\ \frac{dB}{dt}=鈭�\frac{1}{2蟺R^2}\frac{d蠒}{dt}\\ B=鈭�\frac{1}{2}\frac{1}{蟺R^2}蠒+C\end{array}$

Here, $C$is constant.

At$t=0,\text{鈥�}B=0,\text{鈥�}C=0$

Therefore,

$B\left(R\right)=\frac{1}{2}\left(\frac{1}{蟺R^2}蠒\right)$

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