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
where v is the velocity of the charged particle, and B is the magnitude of the magnetic field in the cyclotron. m, q are the mass and electric charge of a particle, respectively. The above equation yields
v = \frac{2\pi r}{T} = \frac{qBr}{m}
where T is the period of the charged particle. Since the relationship between the period and frequency is
\frac{2\pi}{T} = \omega
Thus,
\omega = \frac{qB}{m}
Now, let's look at the particles, Mass: m_p : m_d : m_\alpha = 1 : 2 : 4 Electric charge: 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}
\therefore \omega_p > \omega_d = \omega_\alpha

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

 

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