Chapter 27 Problem 66

Problem

model cyclotron

The cyclotron is a device used to accelerate elementary particles such as protons to high speeds. Particles starting at point A with some initial velocity travel in circular orbits in the magnetic field $B$. The particles are accelerated to higher speeds each time they pass in the gap between the metal “dees,” where there is an electric field $E$. (There is no electric field within the hollow metal dees.) The electric field changes direction each half-cycle, due to an ac voltage $V=V_{0}sin2\pi ft$, so that the particles are increased in speed at each passage through the gap.

(a) Show that the frequency $f$ of the voltage must be $f=Bq/2\pi m$, where $q$ is the charge on the particles and $m$ their mass.

(b) Show that the kinetic energy of the particles increases by $2qV_{0}$ each revolution, assuming that the gap is small.

(c) If the radius of the cyclotron is 0.50 m and the magnetic field strength is 0.60 T, what will be the maximum kinetic energy of accelerated protons in MeV?

solution

(a)

The frequency of the voltage must match the frequency of circular motion of the particles, so that the electric field is synchronized with the circular motion

We have already solved the cyclotron in the class

$r={\frac {mv}{qB}}$

$T={\frac {2\pi r}{v}}$

$f={\frac {v}{2\pi r}}={\frac {v}{2\pi {\frac {mv}{qB}}}}={\frac {Bq}{2\pi m}}$

this does not depend on the radius, so the same driving frequency throughout the acceleration can be used

(b)

For a small gap, the electric field across the gap will be approximately constant and uniform as the particles cross the gap. If the motion and the voltage are synchronized so that the maximum voltage occurs when the particles are at the gap, the particles receive an energy increase of $K=qV_{0}$. Since particle passes through two gaps, energy increase per revolution is $2qV_{0}$

(c)

The maximum kinetic energy will occur at the outside of the cyclotron.

$K_{\textrm {max}}={\frac {1}{2}}mv_{\textrm {max}}^{2}={\frac {1}{2}}m\left({\frac {qBr_{\textrm {max}}}{m}}\right)^{2}={\frac {1}{2}}{\frac {q^{2}B^{2}r_{\textrm {max}}^{2}}{m}}$

${\frac {1}{2}}{\frac {(0.5{\textrm {m}})^{2}(0.6{\textrm {T}})^{2}(1.6\times 10^{-19}{\textrm {C}})^{2}}{(1.67\times 10^{-27}{\textrm {kg}})}}=6.898\times 10^{-13}{\textrm {J}}=4.3{\textrm {MeV}}$