A long solenoid of length 0.5 m has 500 turns and carries a current of 2 A . A proton is projected with a speed of 5 10^6 m/s perpendicular to the axis of the solenoid. Calculate the magnitude of the magnetic force experienced by the proton.

Step-by-step solution

The magnetic field inside a long solenoid is uniform and directed along its axis. Number of turns per unit length, n = N L = 500 0.5 = 1000 m ⁻¹ . Magnetic field, B = ₀ n I = (4 10⁻⁷) 1000 2 = 8 10⁻⁴ T . Since the proton is projected perpendicular to the axis, its velocity is perpendicular to the magnetic field ( = 90^ ). Force on the proton, F = qvB (90^ ) F = (1.6 10⁻¹⁹ C ) (5 10^6 m/s ) (8 10⁻⁴ T ) F = 64 10⁻¹⁷ N = 6.4 10⁻¹⁶ N . Answer: 6.4 10⁻¹⁶ N