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(A) In a cyclotron,the frequency of the alternating electric field $f$ is equal to the cyclotron frequency $f_c = \frac{eB}{2\pi m}$.From this,the mag

Vedclass · Accepted Answer

$\frac{2 \pi m f}{e}, 2 \pi^2 m f^2 R^2$

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(A) In a cyclotron,the frequency of the alternating electric field $f$ is equal to the cyclotron frequency $f_c = \frac{eB}{2\pi m}$.From this,the magnetic field $B$ is given by $B = \frac{2\pi mf}{e}$.The maximum velocity $v$ of the proton at the exit radius $R$ is given by $v = \omega R = (2\pi f)R$.The kinetic energy $K.E.$ is given by $K.E. = \frac{1}{2}mv^2$.Substituting $v = 2\pi f R$,we get $K.E. = \frac{1}{2}m(2\pi f R)^2 = \frac{1}{2}m(4\pi^2 f^2 R^2) = 2\pi^2 mf^2 R^2$.Thus,the magnetic field is $\frac{2\pi mf}{e}$ and the kinetic energy is $2\pi^2 mf^2 R^2$.

An alternating electric field of frequency $f$ is applied across the radius $R$ of a cyclotron to accelerate protons (mass $m$). The operating magnetic field $B$ used and kinetic energy $(K.E.)$ of the proton beam produced by it are respectively ($e=$ charge on proton).

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