In a cyclotron, a magnetic field of `2*4T` is used to accelerate protons. How rapidly should the electric field between the dees be reversed? The mass and the charge of protons are `1*67xx10^(-27)kg` and `1*6xx10^(-19)C` respectively.

Answer

`1*37xx10^8s`, `3*64xx10^7Hz`

Explanation

Here, `B=2*4T`, `m=1*67xx10^(-27)kg`,
`q=1*6xx10^(-19)C`.
Time taken by proton to complete semicircular path
`t=T/2=(2pim)/(2Bq)=(pim)/(Bq)=(3*14xx1*67xx10^(-27))/(2*4xx1*6xx10^(-19))`
`=1*37xx10^-8s`
The direction of electric field should be reversed after every `1*37xx10^-8s`.
Time period of oscillation of electric field is
`T=2t=2xx1*37xx10^-8s`
Frequency of oscillation of electric field is
`T=2t=2xx1*37xx10^-8s`
Frequency of oscillation of applied electric field
`v=1/R=(1)/(2xx1*37xx10^-8)=3*64xx10^7Hz`