A proton enters a uniform magnetic field of strength 0.5 T at 200 m/s. The magnetic field is oriented at an angle of 30° to the proton's velocity. What is the magnitude of the force that the proton experiences while it moves through the magnetic field?

The force experienced is 8 x 10⁻¹⁸ N

Explanation:

When any charge moves into the magnetic field, it experiences the force

The force can be calculated by the relation F = q v B sinθ

here q is the charge and v is its velocity.

B is magnetic field strength and θ is the angle between field and velocity vectors.

Thus F = 1.6 x 10⁻¹⁹ x 200 x 0.5 x sin 30

= 8 x 10⁻¹⁸ N