Q131 / 200 · NEET 2012 (Screening) · Physics

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

  • A
    $B=\frac{m v}{e}$ and $K=2 m \pi^2 v^2 R^2$
  • B
    $B=\frac{2 \pi m v}{e}$ and $K=m^2 \pi v R^2$
  • Correct answer
    C
    $B=\frac{2 \pi m v}{e}$ and $K=2 m \pi^2 v^2 R^2$
  • D
    $B=\frac{m v}{e}$ and $K=m^2 \pi v R^2$

Solution

Frequency $v=\frac{e B}{2 \pi m}$ $\mathrm{KE}=\frac{1}{2} m v^2$ and radius $R=\frac{m v}{e B}$ Here, velocity $v=\frac{\pi R}{T / 2}=\frac{2 \pi R}{T}=2 \pi R v$ $\therefore \quad$ Radius $R=\frac{m(2 \pi R v)}{e B}$ Magnetic field $B=\frac{2 \pi m v}{e}$ Kinetic energy $K=\frac{1}{2} m(2 \pi R v)^2=2 m \pi^2 v^2 R^2$

Hence, the correct answer is option: 'C'.

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