Question

A cyclotron is operated at an oscillator frequency of $\frac{80}{\pi }MHz$ and has a Dee of radius $R=60cm$. The magnitude of magnetic field $B$ (in Tesla) to accelerate deuteron is $\frac{x}{10}$ Tesla. Find x ? (charge on deuteron $e=1.6\times {10}^{-19}$ mass of deuteron $=3.24\times {10}^{-27}Kg)$

• A. 32.4
• B. 16.2
• C. 64
• D. 8

Answer 1

The correct option is A 32.4

Time period in cyclotron motion: $T=\frac{2\pi m}{qB}$
$\therefore \frac{1}{f}=\frac{2\pi m}{qB}$
where $f$ is frequency
$\therefore B=\frac{2\pi mf}{q}=\frac{2\pi \ast \frac{80}{\pi }\ast {10}^6\ast 3.24\ast {10}^{-27}}{1.6\ast {10}^{-19}}=\frac{32.4}{10}$
Given $B=\frac{x}{10}$
$\Rightarrow x=32.4$