A cyclotron is made of sheet metal in the form of an empty tuna-fish can, set on a table with a flat-side down and then sliced from above through its center into two D-shaped pieces. The two "Dees" are then separated slightly so there is a small gap between them. A high-frequency alternating voltage is applied to the Dees, so they are always oppositely charged. At peak voltage there is therefore an electric field in the gap from the positive to the negative Dee that can accelerate charges across the gap. There is also a constant and uniform magnetic field applied vertically, i.e., perpendicular to the Dees, supplied by a large external electromagnet. Therefore after a charged particle has been accelerated across the gap it enters a Dee, where it follows a semicircular path due to the magnetic field and maintains constant speed because the electric field inside a Dee is negligible. By the time the charged particle has completed a semicircle it arrives back at the gap, but by now the charges on the two Dees have been reversed, so the particle is again accelerated in the gap, entering the previous Dee and then executing a larger semicircular path this time because it is moving faster. As the particle moves faster and faster the semicircular paths increase in radius, so in effect the particle moves in a spiraling path until it reaches the outer edge of the machine, where by then it has achieved a very large kinetic energy due to the multiple accelerations it has received by repeatedly passing through the gap. It is then deflected out of the cyclotron where it causes a high-energy collision with other particles at rest in the lab.

(a) Assuming that the charged particle is nonrelativistic, show that its kinetic energy by the time it reaches the outer radius \(R\) of the cyclotron is \(T=q^{2} B^{2} R^{2} / 2 m c^{2}\), where \(q\) and \(m\) are the particle's charge and mass, \(B\) is the magnetic field, and \(R\) is the outer radius of the cyclotron.

(b) If we want to accelerate protons to a kinetic energy of \(16 \mathrm{MeV}\), what must be the applied magnetic field \(B\) (in Gauss) if the diameter of the cyclotron is \(152 \mathrm{~cm}\) ? Protons have mass energy \(m c^{2}=938 \mathrm{MeV}\) and charge \(q=4.8 \times 10^{-10}\) esu. Note that \(1 \mathrm{eV}=1.602 \times 10^{-12} \mathrm{ergs}=1.602 \times 10^{-19}\) Joules. In Gaussian units \(B\) is measured in "Gauss," and in Standard International (SI) units \(B\) is measured in "Teslas," where 1 Tesla \(=10^{4}\) Gauss.