Physics | MHT CET 2018 | Discussion Page

An alternating electric field of frequency v is applied across the dees (radius R) of a cyclotron to accelerate protons (mass m). The operating magnetic field B used and KE of the proton beam produced by it are respectively(where, e= charge on proton).

$\frac{2\pi mv}{e},2\pi^{2}mv^{2}R^{2}$

$\frac{2\pi^{2} mv}{e^{2}},4\pi^{2}mv^{2}R^{2}$

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

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

Answer:

Option A

Explanation:

In cyclotron frequncy, $v=\frac{eB}{2\pi m}$

or $B=\frac{2\pi m.v}{e}$

Also, relation for radius (R)is

$R=\frac{ m.v}{e B}$ or $v=\frac{eBR}{m}$

$\therefore$ $K E =\frac{1}{2}mv^{2}=\frac{1}{2}m\times\frac{ e^{2}B^{2}R^{2}}{m^{2}}=\frac{ e^{2}B^{2}R^{2}}{2m}$

= $\frac{ e^{2}R^{2}}{2m}\times\frac{4\pi^{2}m^{2}v^{2}}{e^{2}}=2\pi^{2}m R^{2}v^{2}$

Discussion Forum

Answer:

Explanation: