Question

In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s^–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10^–19 C, m_e= 9.1×10^–31 kg)

Answer 1

Magnetic field strength, B = 6.5 G = 6.5 × 10^–4 T

Speed of the electron, v = 4.8 × 10⁶ m/s

Charge on the electron, e = 1.6 × 10^–19 C

Mass of the electron, m_e = 9.1 × 10^–31 kg

Angle between the shot electron and magnetic field, θ = 90°

Magnetic force exerted on the electron in the magnetic field is given as:

F = evB sinθ

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron,

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

Hence, the radius of the circular orbit of the electron is 4.2 cm.