Source: High school physics (Chinese)
Problem Sets:
Problem
A proton moves in a uniform magnetic field of magnitude $B = 0.5$ T along a circular orbit of radius $r = 80$ cm in a plane perpendicular to the field. The proton mass is $m = 1.67 \times 10^{-27}$ kg and its charge is $q = 1.6 \times 10^{-19}$ C.
- Find the speed of the proton.
- Find the kinetic energy of the proton.
- Find the period of the circular motion.
The magnetic force supplies the centripetal force: $qvB = \dfrac{mv^2}{r}$, giving $v = \dfrac{qBr}{m}$.
(1) $v = \dfrac{(1.6 \times 10^{-19})(0.5)(0.80)}{1.67 \times 10^{-27}} \approx 3.83 \times 10^{7}$ m/s.
(2) $E_k = \dfrac{1}{2}mv^2 = \dfrac{1}{2}(1.67 \times 10^{-27})(3.83 \times 10^{7})^2 \approx 1.22 \times 10^{-12}$ J.
(3) $T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB} = \dfrac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(0.5)} \approx 1.31 \times 10^{-7}$ s.