Question

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.

If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?

9.1 × 10–5 m/s
1.3 × 10–4 m/s
7.6 × 103 m/s
1.1 × 104 m/s

Answer 1

Use F = qvB sinθ so v = F / (q B sinθ).

sin 35° ≈ 0.574, so
qB sinθ = (8.4×10⁻⁴)(6.7×10⁻³)(0.574) ≈ 3.23×10⁻⁶.

v ≈ (3.5×10⁻²) / (3.23×10⁻⁶) ≈ 1.08×10⁴ m/s.

Answer: 1.1 × 10⁴ m/s.