GRE Physics GR 1777 Problem Solution

021. 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 the magnetic Lorentz force.

$latex \\frac{mv^2}{r} = qBv &s=1$where $latex v$ is the velocity of the charged particle, and $latex B$ is the magnitude of the magnetic field in the cyclotron. $latex m$, $latex q$ are the mass and electric charge of a particle, respectively.

The above equation yields

$latex v = \\frac{2\\pi r}{T} = \\frac{qBr}{m} &s=1$where $latex T$ is the period of the charged particle. Since the relationship between the period and frequency is

$latex \\frac{2\\pi}{T} = \\omega &s=1$ 

Thus,

$latex \\omega = \\frac{qB}{m} &s=1$ 

Now, let\’s look at the particles, Mass: $latex m_p : m_d : m_\\alpha = 1 : 2 : 4$ Electric charge: $latex q_p : q_d : q_\\alpha : 1 : 1 : 2$

Therefore, their cyclotron frequencies in the same magnetic field,

$latex \\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} &s=1$$latex \\therefore \\omega_p > \\omega_d = \\omega_\\alpha$

Answer

(E) $latex \\omega_p > \\omega_d = \\omega_\\alpha$

References

https://en.wikipedia.org/wiki/Cyclotron_resonance http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/cyclot.html