Source: High school physics (Chinese)
Problem Sets:
Problem
An electron with kinetic energy $E_k = 10$ eV moves in a circle in a plane perpendicular to a uniform magnetic field of magnitude $B = 10^{-4}$ T. The electron's charge is $e = -1.6 \times 10^{-19}$ C and its mass is $m = 9.1 \times 10^{-31}$ kg.
- Find the orbital radius $R$ of the electron.
- Find the cyclotron period $T$ of the electron.
- Looking along the direction of $\vec{B}$, does the electron rotate clockwise?
Convert the kinetic energy to joules: $E_k = 10 \text{ eV} = 1.6 \times 10^{-18}$ J. The electron's speed follows from $E_k = \tfrac{1}{2}mv^2$:
$$v = \sqrt{\dfrac{2E_k}{m}} = \sqrt{\dfrac{2(1.6 \times 10^{-18})}{9.1 \times 10^{-31}}} \approx 1.88 \times 10^{6} \text{ m/s}.$$(1) The magnetic force supplies the centripetal force: $|e|vB = \dfrac{mv^2}{R}$, so
$$R = \dfrac{mv}{|e|B} = \dfrac{(9.1 \times 10^{-31})(1.88 \times 10^{6})}{(1.6 \times 10^{-19})(10^{-4})} \approx 0.107 \text{ m} \approx 10.7 \text{ cm}.$$(2) Period: $T = \dfrac{2\pi m}{|e|B} = \dfrac{2\pi (9.1 \times 10^{-31})}{(1.6 \times 10^{-19})(10^{-4})} \approx 3.57 \times 10^{-7}$ s.
(3) For a negative charge, the cyclotron angular velocity $\vec{\omega} = -\dfrac{q\vec{B}}{m}$ is parallel to $\vec{B}$. Looking along $\vec{B}$, the vector $\vec{\omega}$ points away from the viewer; by the right-hand rule this corresponds to clockwise rotation. So yes, the electron rotates clockwise when viewed along $\vec{B}$.