A proton moves through a magnetic field at 35.7 % 35.7% of the speed of light. At a location where the field has a magnitude of 0.00689 T 0.00689 T and the proton's velocity makes an angle of 149 ∘ 149∘ with the field, what is the magnitude of the magnetic force acting on the proton?

F = 6.08 ×10-¹⁴ N

Explanation:

The force experienced by a charged particle in a magnetic field is given by the formula

F = qvBSinΦ

Where F = the magnetic force on the charged particle

q = charge on the particle

v = magnitude velocity of the particle

B = magnitude of the magnetic field vector

Φ = angle between the magnetic field and the direction of motion or the velocity of the particle.

The calculation steps are in the attachment below.

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