An electron is accelerated from rest by a potential differ - ence of 350 V. It then enters a uniform magnetic field of magni - tude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

a) the speed of the electron is 1.11 * 10⁷ m/s

b) the radius of electron's path in the magnetic field is 3.16 * 10⁻⁴ m

Explanation:

a) Let's assume that we have an electron accelerated using a potential difference of V = 350, which gives the ion a speed of v. In order to find this speed we set the potential energy of the electron equal to its kinetic energy. Thus,

eV = 1/2 m v²

where

e is the charge of the electron m is the mass of the electron v is the speed of the electron

Thus,

v = √[2eV / m]

v = √[2 (1.6 * 10⁻¹⁹ C) (350 V) / 9.11 * 10⁻³¹ kg]

v = 1.11 * 10⁷ m/s

Therefore, the speed of the electron is 1.11 * 10⁷ m/s

b) Then the electron enters a region of uniform magnetic field, it moves in a circular path with a radius of:

r = mv / eB

where

m is the mass of the electron v is the speed of the electron e is the charge of the electron B is the magnetic field

Thus,

r = (9.11 * 10⁻³¹ kg) (1.11 * 10⁷ m/s) / (1.6 * 10⁻¹⁹ C) (200 * 10⁻³ T)

r = 3.16 * 10⁻⁴ m

Therefore, the radius of electron's path in the magnetic field is 3.16 * 10⁻⁴ m