Maximum speed in a cyclotron

In a cyclotron, the charged particle moves in a circular path under magnetic field $B$. The magnetic force provides the centripetal force: $qvB = \frac{mv^2}{r}$

So, $v = \frac{qBr}{m}$

Kinetic energy at the outer radius

The maximum kinetic energy is when the proton reaches the outer radius of the dee: $K = \frac{1}{2}mv^2$

Substituting $v = \frac{qBr}{m}$, $K = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2 = \frac{q^2B^2r^2}{2m}$

Substitute the given values

Given: $q = e = 1.6 \times 10^{-19}\,\text{C}$ $B = 1.0\,\text{T}$ $r = 60\,\text{cm} = 0.60\,\text{m}$ $m_p = 1.6 \times 10^{-27}\,\text{kg}$

Therefore, $K = \frac{(1.6 \times 10^{-19})^2 (1)^2 (0.60)^2}{2(1.6 \times 10^{-27})}$

Calculate step-by-step

First, $(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}$

and $(0.60)^2 = 0.36$

So numerator becomes: $2.56 \times 10^{-38} \times 0.36 = 0.9216 \times 10^{-38} = 9.216 \times 10^{-39}$

Denominator: $2 \times 1.6 \times 10^{-27} = 3.2 \times 10^{-27}$

Hence, $K = \frac{9.216 \times 10^{-39}}{3.2 \times 10^{-27}} = 2.88 \times 10^{-12}\,\text{J}$

Convert joules to eV

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

$K = \frac{2.88 \times 10^{-12}}{1.6 \times 10^{-19}} = 1.8 \times 10^7\,\text{eV}$

$K = 18\,\text{MeV}$

Match with options

$\boxed{18\,\text{MeV}}$

So the correct option is B.