Question

A cyclotron is operating at a frequency of $12 \times 10^{6} Hz$. Mass of deuteron is $3.3 \times 10^{-27} kg$ and charge on deuteron is $1.6 \times 10^{-19} C .$ To accelerate deuterons the magnetic induction of the necessary magnetic field is:

• A. 0.016 T
• B. 0.16 T
• C. 16 T
• D. 1.6T

Answer 1

Answer: (D)

A cyclotron accelerates charged particles with a high-frequency, alternating voltage (potential difference). A perpendicular magnetic field causes the particles to go almost in a circle. Also a high frequency alternating voltage applied across the ' $D$ ' electrodes alternately attracts and repels charged particles. The centripetal force is provided by the transverse magnetic field $B$, and the force on a particle travelling in a magnetic field (which causes it to curve) is equal to $B q V$.
So, $ \frac{m v^{2}}{r}=B q v$
where, $m$ is mass, $q$ is charge, $v$ is velocity and $r$ is radius.
Also $v=r \omega $
$\therefore \omega=\frac{B q}{m}$
Since, $\omega =2 \pi n, $ we have
$2 \pi n =\frac{B q}{m} $
$\Rightarrow B =\frac{2 \pi n \times m}{q}$
$=\frac{2 \times \pi \times 12 \times 10^{6} \times 3.3 \times 10^{-27}}{1.6 \times 10^{-19}} $
$ \approx 1.6\, T$