Question

A point charge of 5.0 × 10–7 C moves to the right at 2.6 × 105 m/s in a magnetic field that is directed into the screen and has a field strength of 1.8 × 10–2 T.

What is the magnitude of the magnetic force acting on the charge?

0 N
2.3 × 10–3 N
23 N
2.3 × 1011 N

Answer 1

F = qvB sinθ. Here v is perpendicular to B (θ = 90°, sinθ = 1), so

F = (5.0×10^−7 C)(2.6×10^5 m/s)(1.8×10^−2 T)
= 2.34×10^−3 N ≈ 2.3×10^−3 N.

Answer: 2.3 × 10^−3 N.