A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be: [use $$m_p = 1.6 \times 10^{-27}$$ kg, $$e = 1.6 \times 10^{-19}$$ C]

A cyclotron with magnetic field $$B = 1.0$$ T and radius $$R = 60$$ cm = 0.6 m accelerates protons. Find the kinetic energy in MeV.

Since for a charged particle in a cyclotron at the maximum radius the magnetic force equals the centripetal force, we have

$$qvB = \frac{mv^2}{R} \implies mv = qBR$$

From this relation one obtains

$$v = \frac{qBR}{m}$$

Substituting the expression for $$v$$ into the kinetic energy formula yields

$$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{qBR}{m}\right)^2 = \frac{q^2B^2R^2}{2m}$$

Substituting $$q = e = 1.6 \times 10^{-19}$$ C, $$B = 1.0$$ T, $$R = 0.6$$ m, and $$m_p = 1.6 \times 10^{-27}$$ kg into this expression gives

$$KE = \frac{(1.6 \times 10^{-19})^2 \times (1.0)^2 \times (0.6)^2}{2 \times 1.6 \times 10^{-27}}$$

Numerator: $$(1.6)^2 \times 10^{-38} \times 0.36 = 2.56 \times 0.36 \times 10^{-38} = 0.9216 \times 10^{-38}$$

Denominator: $$3.2 \times 10^{-27}$$

Therefore,

$$KE = \frac{0.9216 \times 10^{-38}}{3.2 \times 10^{-27}} = 0.288 \times 10^{-11} = 2.88 \times 10^{-12} \text{ J}$$

Since $$1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$$ and $$1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$$,

$$KE = \frac{2.88 \times 10^{-12}}{1.6 \times 10^{-13}} = 18 \text{ MeV}$$

Answer: Option B: 18