Question

The cyclotron frequency of an electron gyrating in a magnetic field of $1T$ is approximately

• A. 28 MHz
• B. 280 MHz
• C. 2.8 GHz
• D. 28 GHz

Answer 1

Answer: (D)

In a cyclotron, 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 $Bqv$.
In cyclotron, centripetal force $=$ magnetic force
i.e., $\frac{m{v}^2}{r}=Bqv$
$\therefore \frac{v}{r}=\frac{Bq}{m}$
where $\frac{v}{r}=\omega$, therefore
$\omega =\frac{Bq}{m}$
and frequency $f=\frac{\omega }{2\pi }=\frac{Bq}{2m\pi }$
Putting the numerical values from the question, we have
$B=1T,m=9.1\times 10^{-31}kg,q=1.6\times 10^{-19}C$
$\therefore f=\frac{1.6\times 10^{-19}\times 1}{2\times 3.14\times 9.1\times 10^{-31}}=28\times 10^{9}Hz$
$\therefore f=28GHz$