A cyclotron's oscillator frequency is `10MHz`. What should be the operating magnetic field fro accelerating protons? If the radius of its dees is `60cm`, what is the kinetic energy (in `MeV`) of the proton beam produced by the acceleration?
`(e-1.60xx10^-19C,m_p=1.67xx10^-27kg, 1MeV=1.6xx10^-13J)`

Explanation

Cyclotron:It is a device by which positely charged particles like protons,deutrons, etc.can be accelerated.Principle :A positively charged particle can be accelerated by making it to cross the same electric field repeatedly with the help of a magnetic field.
Working and theory:At a certain instant let `D_(1)` be positive and `D_(2)` be negative.A proton from an ion source will be accelerated towards `D_(2)` it describes a semi-cicular path with a constant speed and is acted upon only by the magnetic field.the radius of the circular path is given by
`qvB=(mv^(2))/r ` or `r=(mv)/(qB)`
Period of revolution, `T=(2pir)/v=(2pi)/v(mv)/(qB)=(2pim)/(qB)`
Frequency of revolution, `f=1/T=(qB)/(2pim)`
Clearly, frequency `f` is independent of both `v` and `r` and is called cyclotron frequency.If the frequency of applied `a.c.` is equal `f`,then every time the proton reaction of electric field is reversed and proton receives a push and finally it gains very high kinetic energy.The proton follows a spiral path.The accelerated proton are deflected towards the target.
Numerical:`f=10 MHz=10^(7)Hz,R=60cm=0.6 m,m_(p)=1.67xx10^(-27) kg`
The operating magnetic field for a accelerating protons is
`B=(2pim_(p)f)/e=(2xx3.14xx1.67xx10^(-27)xx10^(7))/(1.6xx10^(-19))=0.667 T`
Kinetic energy of the emerging beam.
`k_(max)=(e^(2)B^(2)R^(2))/(2m_(p))=((1.6xx10^(-19))^(2)xx(0.66)^(2)xx(0.6)^(2))/(2xx1.67xx10^(-27))=1.2xx10^(-12)J=(1.2xx10^(-12))/(1.602xx10^(-13))MeV=7.4MeV`.