An alternating electric field of frequency v, 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:

  • Option 1)

    B=\frac{mv}{e}\: \: and\: \: K=2m\pi^{2}\nu^{2} R^{2}

  • Option 2)

    B=\frac{2\pi mv}{e}\: \: and\: \: K=m^{2}\pi\nu R^{2}

  • Option 3)

    B=\frac{2\pi mv}{e}\: \: and\: \: K=2m\pi^{2}\nu^{2} R^{2}

  • Option 4)

    B=\frac{mv}{e}\: \: and\: \: K=m^{2}\pi\nu R^{2}

 

Answers (1)

As discussed in @8243 

We know that

T=\frac{2\pi m }{qB} = \frac{2\pi m}{eB}

\frac{1}{T}=\frac{eB}{2\pi m} = B = \frac{2\pi mf}{e}

\frac{mv^{2}}{R} = evB=v=\frac{eBR}{m}= \frac{e2\pi mfR}{2me}=2\pi f

K=\frac{1}{2}mv^{2}=\frac{1}{2}m (2\pi fR)^{2} (\because f=v)

K=2m\pi^{2} f^{2}R^{2} or, K=2\pi ^{2}mv^{2}R^{2}

 


Option 1)

B=\frac{mv}{e}\: \: and\: \: K=2m\pi^{2}\nu^{2} R^{2}

Option 2)

B=\frac{2\pi mv}{e}\: \: and\: \: K=m^{2}\pi\nu R^{2}

Option 3)

B=\frac{2\pi mv}{e}\: \: and\: \: K=2m\pi^{2}\nu^{2} R^{2}

Option 4)

B=\frac{mv}{e}\: \: and\: \: K=m^{2}\pi\nu R^{2}

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