Question

A point charge of 5.7 µC moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mT, as shown in the diagram.

A field of circles labeled B. There is a positively charged particle with a velocity vector pointing 37 degrees north of east labeled v.
What is the magnitude of the magnetic force acting on the charge?

6.6 × 10–3 N
4.9 × 10–3 N
4.9 × 103 N
6.6 × 103 N

Answer 1

F = qvB sinθ. Here q = 5.7×10⁻⁶ C, v = 4.5×10⁵ m/s, B = 3.2×10⁻³ T, θ = 37°.

sin37° ≈ 0.6018, so
F = (5.7×10⁻⁶)(4.5×10⁵)(3.2×10⁻³)(0.6018) ≈ 4.9×10⁻³ N.

Answer: 4.9 × 10⁻³ N.