Source: High school physics (Chinese)
Problem Sets:
Problem
A cyclotron has dees of maximum radius $R = 60$ cm. It is used to accelerate protons from rest to a kinetic energy of $4.0$ MeV. The proton mass is $m = 1.67 \times 10^{-27}$ kg, its charge is $q = 1.6 \times 10^{-19}$ C, and $1$ eV $= 1.6 \times 10^{-19}$ J.
- Find the required magnetic induction $B$.
- If the voltage between the dees during each gap crossing is $U = 2.0 \times 10^{4}$ V, find the time required to accelerate the proton to the above energy.
(1) From $E_k = \dfrac{q^2 B^2 R^2}{2m}$:
$$B = \dfrac{1}{qR}\sqrt{2mE_k}.$$With $E_k = 4.0$ MeV $= 6.4 \times 10^{-13}$ J:
$$B = \dfrac{1}{(1.6 \times 10^{-19})(0.6)}\sqrt{2(1.67 \times 10^{-27})(6.4 \times 10^{-13})} \approx 0.48 \text{ T}.$$(2) Each gap crossing increases the energy by $\Delta E = qU = 2.0 \times 10^{4}$ eV. The total number of crossings required is
$$n = \dfrac{E_k}{qU} = \dfrac{4.0 \times 10^{6} \text{ eV}}{2.0 \times 10^{4} \text{ eV}} = 200.$$Between successive crossings the proton performs a half-revolution of duration $T/2 = \pi m / (qB)$. The total acceleration time (gap-crossing time negligible) is
$$t = n \cdot \dfrac{T}{2} = \dfrac{n \pi m}{qB} = \dfrac{200 \pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(0.48)} \approx 1.37 \times 10^{-5} \text{ s}.$$