Question

An electron is accelerated from rest through a potential difference of 300V. it then passes through a uniform 0.001-T magnetic field, oriented perpendicular to the electrons velocity. what is the magnitude of the magnetic force on the electron?

Answer 1

Given :

Potential difference, V = 300 V.

Magnetic Field, B = 0.001 T.

To Find :

The magnitude of the magnetic force on the electron.

Solution :

We know, for perpendicular orientation, force is given by :

`F = qVB\\\\F = 1.6 * 10^(-19) * 300* 0.001\ N\\\\F = 4.8* 10^(-20)\ N`

Hence, this is the required solution.