Source: High school physics (Chinese)
Problem
An electron is fired at speed $v = 1.2\times 10^{7}$ m/s into a uniform magnetic field of magnitude $B = 0.02$ T. Find the magnitude of the magnetic force on the electron when the angle between $\vec{v}$ and $\vec{B}$ is:
- $\theta_1 = 30^\circ$
- $\theta_2 = 60^\circ$
The magnitude of the magnetic force is $F = evB\sin\theta$, where $e = 1.6\times 10^{-19}$ C.
Compute the prefactor: $evB = (1.6\times 10^{-19})(1.2\times 10^{7})(0.02) = 3.84\times 10^{-14}$ N.
For $\theta_1 = 30^\circ$: $F_1 = 3.84\times 10^{-14}\times \sin 30^\circ = 3.84\times 10^{-14}\times 0.5 = 1.92\times 10^{-14}$ N.
For $\theta_2 = 60^\circ$: $F_2 = 3.84\times 10^{-14}\times \sin 60^\circ \approx 3.84\times 10^{-14}\times 0.866 \approx 3.33\times 10^{-14}$ N.