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 $BqV$.
So, $\frac{m{v}^2}{r}=Bqv$
where, $m$ is mass, $q$ is charge, $v$ is velocity and $r$ is radius.
Also $v=r\omega$
$\therefore \omega =\frac{Bq}{m}$
Since, $\omega =2\pi n,$ we have
$2\pi n=\frac{Bq}{m}$
$\Rightarrow B=\frac{2\pi n\times m}{q}$
$=\frac{2\times \pi \times 12\times 10^6\times 3.3\times 10^{-27}}{1.6\times 10^{-19}}$
$\approx 1.6T$