Question

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.

Answer 1

Answer:

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