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, me = 9.1×10⁻³¹ kg)

Magnetic field strength, B = 6.5 G = 6.5 x 10⁻⁴ T

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

Charge on the electron, e = 1.6 x 10^-19 C

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

Angle between the shot electron and magnetic field, 0 = 90°

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

F = evB sin0

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,

Fe = mv/r²

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

Fe = F

=>  mv/r²  =  evB sin0

=> r =  mv/ evB sin0

So,

r = (9.1 x 10 ^–31 x 4.8 x 10⁶ )/( 6.5 x 10⁻⁴ x 1.6 x 10^-19  x Sin 90º)

= 4.2 x 10⁻² m = 4.2 cm

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