Cyclotron Motion

Circular motion in a uniform magnetic field

Diagram (Fig. 10.5): a uniform field $\vec{B}$ directed into the plane of the paper, shown as a grid of crosses. A particle of charge $q$ moves with speed $v$ in the plane of the paper on a circle of radius $R$; the velocity is drawn tangential to the circle and the magnetic force $\vec{F}$ is drawn along the radius, pointing towards the centre.

Since $\vec{F}_m$ is always perpendicular to $\vec{v}$, and the speed is constant, the force is constant in magnitude and always directed towards the centre. It therefore acts as the centripetal force of a uniform circular motion:

$$qvB = \dfrac{mv^2}{R}$$

$$\therefore \ mv = p = qBR$$

This is the cyclotron formula. Rearranged,

$$R = \dfrac{mv}{qB} = \dfrac{p}{qB}$$

Period and cyclotron frequency

The particle covers the circumference $2\pi R$ at speed $v$, so

$$T = \dfrac{2\pi R}{v} = \dfrac{2\pi}{v}\left(\dfrac{mv}{qB}\right) = \dfrac{2\pi m}{qB}$$

$$f_c = \dfrac{1}{T} = \dfrac{qB}{2\pi m}$$

The essential point: $T$ and $f_c$ contain neither $v$ nor $R$. The frequency of revolution of a charged particle in a magnetic field is independent of its energy. The cyclotron is built on precisely this fact.

The half period, the time to complete one semicircle, is

$$t = \dfrac{T}{2} = \dfrac{\pi m}{qB}$$

10.3.1 The cyclotron accelerator

Particle accelerators supply MeV to GeV beams for studies of particle-matter interaction and for the medical treatment of certain tumours. The cyclotron was invented by Lawrence and Livingston in 1934 to study nuclear structure.

Construction and diagram (Fig. 10.6): two hollow, semicircular, disc-like metal chambers $D_1$ and $D_2$, called the dees, are placed with a narrow gap between their straight edges, the whole assembly inside a vacuum chamber. A uniform magnetic field $\vec{B}$ (up to about 1.5 T, from an electromagnet) is applied perpendicular to the plane of the dees — in the standard figure it is shown coming out of the plane of the paper. A high frequency alternating voltage, up to 10000 V at about 10 MHz, is applied between the two dees. A gas ionizing source at a point O (or P) in the gap supplies positive ions. Since the electric field in the gap is perpendicular to the magnetic field, the two are called crossed fields.

Working:

• The magnetic field puts the ion on a circular path; the high frequency electric field in the gap accelerates it.
• An ion emitted at the source is accelerated by the negative voltage of one dee, enters it, and describes a semicircle inside it in a time $T/2$. There is no electric field inside a dee, so the speed there is constant.
• On reaching the gap it is accelerated across by the potential difference, gaining kinetic energy $qV$, and then describes a semicircle of larger radius in the other dee.
• Because the applied voltage alternates, its sign reverses in step with the ion's circular motion, so the ion is accelerated at every crossing. Its energy grows and its radius grows, and the whole path is a spiral.
• At the largest radius it is deflected by a magnetic field and extracted through an exit slit.

Resonance condition

The frequency $f_a$ of the applied alternating voltage is adjusted so that the polarity of the dees reverses just as the ion arrives at the gap, after each semicircle. Since a semicircle always takes $T/2 = \pi m/qB$ whatever the radius,

$$f_a = f_c = \dfrac{qB}{2\pi m}$$

This is the resonance condition. Equivalently, the electric field between the dees must be reversed after every interval $t = \dfrac{\pi m}{qB}$ — which is what the board's numericals ask you to compute.

Energy of the extracted beam

At the exit radius $R_{\text{exit}}$, the cyclotron formula gives $v_{\text{exit}} = \dfrac{qB R_{\text{exit}}}{m}$, so

$$\text{K.E.} = \dfrac{1}{2}mv_{\text{exit}}^2 = \dfrac{q^2B^2R_{\text{exit}}^2}{2m}$$

The final energy is proportional to the square of the radius of the outermost circular path. Using $B = \dfrac{2\pi m f}{q}$ this can also be written

$$\text{K.E.} = 2\pi^2 m f^2 R^2$$

which is the form to use when a problem gives the dee radius and the oscillator frequency instead of $B$.

Standard results

• $qvB = \dfrac{mv^2}{R}$, so $p = mv = qBR$ and $R = \dfrac{mv}{qB}$
• $T = \dfrac{2\pi m}{qB}$, $f_c = \dfrac{qB}{2\pi m}$, both independent of $v$ and $R$
• Reversal interval $t = \dfrac{T}{2} = \dfrac{\pi m}{qB}$
• Resonance: $f_a = f_c$
• Energy gained per gap crossing $= qV$
• $\text{K.E.} = \dfrac{q^2B^2R_{\text{exit}}^2}{2m} = 2\pi^2 m f^2 R^2$
• Accelerators in India: 2 GeV electron Synchrotron at RRCAT Indore; 8-10 MeV Microtron at Savitribai Phule Pune University.