An electron moves perpendicular to a magnetic field of 0.5 T with a speed of 2 × 106 m/s. If the magnitude of the electronic charge is 1.6 × 10−19 C, what is the magnetic force acting on the electron?
1.6 × 10−13 N
The magnetic (Lorentz) force on a moving charge is \(F = qvB\sin\theta\).
Motion is perpendicular to the field, so θ = 90° and sin θ = 1.
\(F = (1.6 \times 10^{-19})(2 \times 10^{6})(0.5)\).
\(F = 1.6 \times 10^{-19} \times 10^{6} = 1.6 \times 10^{-13}\ \text{N}\).
Hence, the answer is 1.6 × 10−13 N.
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