Question
An alternating electric field of frequency $$\nu ,$$ is applied across the dees (radius = $$R$$ ) of a cyclotron that is being used to accelerate protons (mass = $$m$$). The operating magnetic field $$\left( B \right)$$ used in the cyclotron and the kinetic energy $$\left( K \right)$$ of the proton beam, produced by it, are given by
A.
$$B = \frac{{m\nu }}{e}\,{\text{and}}\,\,K = 2m{\pi ^2}{\nu ^2}{R^2}$$
B.
$$B = \frac{{2\pi m\nu }}{e}\,{\text{and}}\,\,K = {m^2}\pi \nu {R^2}$$
C.
$$B = \frac{{2\pi m\nu }}{e}\,{\text{and}}\,\,K = 2m{\pi ^2}{\nu ^2}{R^2}$$
D.
$$B = \frac{{m\nu }}{e}\,{\text{and}}\,\,K = {m^2}\pi \nu {R^2}$$
Answer :
$$B = \frac{{2\pi m\nu }}{e}\,{\text{and}}\,\,K = 2m{\pi ^2}{\nu ^2}{R^2}$$
Solution :
Frequency, $$\nu = \frac{{eB}}{{2\pi m}}$$
$$KE = \frac{1}{2}m{\nu ^2}\,\,{\text{and}}\,\,{\text{radius}}\,R = \frac{{m\nu }}{{eB}}$$
Here, velocity, $$\nu = \frac{{\pi R}}{{\frac{T}{2}}} = \frac{{2\pi R}}{T} = 2\pi R\nu $$
$$\therefore {\text{Radius,}}\,R = \frac{{m\left( {2\pi R\nu } \right)}}{{eB}}$$
Magnetic field, $$B = \frac{{2\pi m\nu }}{e}$$
Kinetic energy, $$K = \frac{1}{2}m{\left( {2\pi R\nu } \right)^2}$$
$$ = 2m{\pi ^2}{\nu ^2}{R^2}$$