Question

What is the magnetic force (in newtons) on a particle traveling in a 1.5 T magnetic field if q = 7.5 microcoulombs and v = 1.75 × 106 m/s at a 45° angle to the magnetic field? Answer with three signficant digits.

Answer 1

Given:

B(Magnetic field): 1.5 T

q= 7.5 microcoulombs

v= 1.75 x 10 ∧6 m/s

The angle ∅ between B and v is 45 °.

Now we know that F= qvB sin ∅

Substituting these values we get:

F= 7.5 x 10∧-6 x 1.75 x 10∧6 x 1.5 x sin 45

F= 16.752 N