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