An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field.

(a)Calculate the speed of the electron?

(b)Find the radius of its path in the magnetic field.

• The potential difference is$350V$ .
• The magnetic field is $200mT$.

We use the principle of conservation of energy in which potential energy is equal to kinetic energy to find velocity. To find the radius, we have to write the equilibrium of magnetic force and centrifugal force.

Formula:

$PE=qV$

$KE=0.5m{v}^2$

$F_b=qvB$

$F_c=\frac{m{v}^2}{r}$

We use conservation of energy to find the speed of the electron as:

$PE=KE$

Substitute the values, and we get,

$\begin{array}{c}qV=0.5m{v}^2\\ (1.6\times {10}^{-19})\times 350=0.5\times 9.11\times {10}^{-31}\times v^2\\ v=1.11\times {10}^7\text{\hspace{0.17em}m/s}\end{array}$

Thus, the speed of the electron is .$1.11\times {10}^7\text{\hspace{0.17em}m/s}$

We know that the electron moves in a circular path in a magnetic field because magnetic force gets balanced by the centrifugal force. So:

$qvB=\frac{m{v}^2}{r}$

By rearranging,

$\begin{array}{c}r=\frac{mv}{qB}\\ =\frac{9.11\times {10}^{-31}\times 1.11\times {10}^7}{1.6\times {10}^{-19}\times 200\times {10}^{-3}}\\ =3.16\times {10}^{-4}m\end{array}$

Thus, the radius of the path is $3.16\times {10}^{-4}m$.