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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 (B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by
A
$B = \frac{m\nu}{e}$ and $K = m^2\pi\nu R^2$
B
$B = \frac{m\nu}{e}$ and $K = 2m\pi^2\nu^2R^2$
C
$B = \frac{2\pi m\nu}{e}$ and $K = m^2\pi\nu R^2$
D
$B = \frac{2\pi m\nu}{e}$ and $K = 2m\pi^2\nu^2R^2$
Detailed Solution
The applied frequency must equal the cyclotron frequency: $\nu = \frac{eB}{2\pi m}$ ...(i)
$B = \frac{2\pi m\nu}{e}$
At the edge of the dees: $R = \frac{mv}{eB} = \frac{\sqrt{2mK}}{eB}$
$K = \frac{R^2e^2B^2}{2m}$; substituting B: $K = \frac{R^2e^2}{2m}\cdot\frac{4\pi^2m^2\nu^2}{e^2} = 2m\pi^2\nu^2R^2$
$B = \frac{2\pi m\nu}{e}$
At the edge of the dees: $R = \frac{mv}{eB} = \frac{\sqrt{2mK}}{eB}$
$K = \frac{R^2e^2B^2}{2m}$; substituting B: $K = \frac{R^2e^2}{2m}\cdot\frac{4\pi^2m^2\nu^2}{e^2} = 2m\pi^2\nu^2R^2$