An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.285 T. If the kinetic energy of the electron is 2.10 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron

The speed of the electron is 6.79 x 10⁵ m/s

The radius of the circular path is 1.357 x 10⁻⁵ m

Explanation:

Given;

magnetic field, B = 0.285 T

energy of electron, E = 2.10 x 10⁻¹⁹ J

The kinetic energy of the electron is calculated as;

Where;

is the mass of electron = 9.11 x 10⁻³¹ kg

V is the speed of the electron

The radius of the circular path is given by;

where;

q is the charge of the electron = 1.6 x 10⁻¹⁹ C