Question

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?

Answer 1

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