A proton moves with a speed of 5 .0 × 10 6 m s - 1 along the x - axis. It enters a region where there is a magnetic field of magnitude 2.0 T directed at an angle of 30 ° to the x - axis and lying in the x y - plane. The magnitude of the magnetic force on the proton is

Step-by-step solution

Given, The speed of proton, v = 5.0 × 10 6   m   s - 1 The magnetic field, B = 2.0   T Angle θ = 30 o Charge on the proton, q = 1.6 × 10 - 19   C When the proton enters in magnitude field, it experiences a magnetic force F = q v × B F = q v B sin ⁡ θ F = 1.6 × 10 - 19 × 5 × 10 6 × 2 × sin ⁡ 30 ° = 8.0 × 10 - 13    N