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

`f=1.98* 10^7\ Hz`

Step-by-step 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=(qB)/(2\pi m)`

Put all the values,

`f=(1.6* 10^(-19)* 1.3)/(2\pi * 1.67* 10^(-27))\\\\f=1.98* 10^7\ Hz`

So, the oscillator frequency is
`1.98* 10^7\ Hz`.