A long solenoid has 5000 turns per metre and carries a current of 2 A . An electron is projected from the axis of the solenoid with a velocity of 5 10^5 m/s in a direction perpendicular to the axis. The magnitude of the magnetic force experienced by the electron is: (Charge of an electron e = 1.6 10⁻¹⁹ C )

Step-by-step solution

First, calculate the magnetic field inside the long solenoid. The field is uniform and directed parallel to the axis of the solenoid. B = ₀ n I B = (4 10⁻⁷ T m/A ) (5000 m ⁻¹) (2 A ) B = 4 10⁻³ T Since the electron is projected perpendicular to the axis of the solenoid, the angle between its velocity vector v and the magnetic field vector B is 90^ . The magnitude of the magnetic force (Lorentz force) on the electron is given by: F = qvB (90^ ) F = (1.6 10⁻¹⁹ C ) (5 10^5 m/s ) (4 10⁻³ T ) 1 F = (8.0 10⁻¹⁴) (4 10⁻³)