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

Options

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

Answer

d

Explanation

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, centipetal force= magnetic force
i.e., `(mv^(2))/(r ) = Bqv :. (v)/(r )=(Bq)/(m)`
where `(v)/(r )= omega`, therefore `omega=(Bq)/(m)` and frequency `f=(omega)/(2pi)=(Bq)/(2mpi)`
Putting the numerical values from the question, we have
`B=1T,m= 9.1 xx 10^(-31)kg, q=1.6xx10^(-19)C`
`:. f=(1.6xx10^(-9)xx1)/(2xx3.14xx9.1xx10^(-31))= 28xx10^(9)Hz`
`:. f= 28 GHz`