Question

A charged particle moves at 2.5 × 104 m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 × 10–2 T.

If the magnetic force is 7.5 × 10–2 N, what is the magnitude of the charge?

3.7 × 10–5 C
4.1 × 10–5 C
8.8 × 10–5 C
1.0 × 10–4 C

Answer 1

Use F = qvB sinθ, so q = F / (vB sinθ).

sin25° ≈ 0.4226.
vB sinθ = (2.5×10^4)(8.1×10^−2)(0.4226) ≈ 855.8

q = 7.5×10^−2 / 855.8 ≈ 8.8×10^−5 C.

Answer: 8.8 × 10^−5 C.