Question

In a certain cyclotron a proton moves in a circle of radius 0.530 m. The magnitude of the magnetic field is 1.30 T. (a) What is the oscillator frequency

Answer 1

Answer:

$f=1.98\times 10^7\ Hz$

Explanation:

Given that,

The radius of circle, r = 0.53 m

The magnitude of the magnetic field, B = 1.3 T

We need to find the oscillator frequency. It is given by :

$f=\dfrac{qB}{2\pi m}$

Put all the values,

$f=\dfrac{1.6\times 10^{-19}\times 1.3}{2\pi \times 1.67\times 10^{-27}}\\\\f=1.98\times 10^7\ Hz$

So, the oscillator frequency is $1.98\times 10^7\ Hz$.