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?

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

powerful computers now, not then

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