Physics GRE GR 1777, Problem 021 Solution

21. Electrodynamics (Cyclotron)

Solution
The cyclotron frequency is the frequency of a charged particle moving perpendicular to the direction of a constant magnetic field B.

Since the motion of a charged particle in the cyclotron is circular (because of a uniform magnetic field), we can calculate the cyclotron frequency by using the fact that the centripetal force equals to the magnetic Lorentz force.

$\frac{mv^2}{r} = qBv$$\frac{mv^2}{r} = qBv$
where $v$$v$ is the velocity of the charged particle, and $B$$B$ is the magnitude of the magnetic field in the cyclotron. $m$$m$, $q$$q$ are the mass and electric charge of a particle, respectively.

The above equation yields

$v = \frac{2\pi r}{T} = \frac{qBr}{m}$$v = \frac{2\pi r}{T} = \frac{qBr}{m}$
where $T$$T$ is the period of the charged particle.

Since the relationship between the period and frequency is

$\frac{2\pi}{T} = \omega$$\frac{2\pi}{T} = \omega$

Thus,

$\omega = \frac{qB}{m}$$\omega = \frac{qB}{m}$

Now, let's look at the particles,

Mass: $m_p : m_d : m_\alpha = 1 : 2 : 4$$m_p : m_d : m_\alpha = 1 : 2 : 4$
Electric charge: $q_p : q_d : q_\alpha : 1 : 1 : 2$$q_p : q_d : q_\alpha : 1 : 1 : 2$

Therefore, their cyclotron frequencies in the same magnetic field,

$\omega_a : \omega_d : \omega_\alpha : \frac{q_a B}{m_a} : \frac{q_d B}{m_d} = \frac{q_\alpha B}{m_\alpha} = \frac{1}{1} : \frac{1}{2} : \frac{2}{4}$$\omega_a : \omega_d : \omega_\alpha : \frac{q_a B}{m_a} : \frac{q_d B}{m_d} = \frac{q_\alpha B}{m_\alpha} = \frac{1}{1} : \frac{1}{2} : \frac{2}{4}$

$\therefore \omega_p > \omega_d = \omega_\alpha$$\therefore \omega_p > \omega_d = \omega_\alpha$

(E) $\omega_p > \omega_d = \omega_\alpha$$\omega_p > \omega_d = \omega_\alpha$
https://en.wikipedia.org/wiki/Cyclotron_resonance
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/cyclot.html