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

Explanation

1. Angular frequency of a charged particle in cyclotron is,
`omega=(Bq)/m`
`therefore2pif=(Bq)/m`
`thereforeB=(2pifm)/q`
`thereforeB=((2)(3.14)(10xx10^(6))(1.67xx10^(-27)))/((1.6xx10^(-19)))`
`thereforeB=6.555xx10^(-1)=0.6555T`
2. Kinetic energy of proton,
`K=1/2mv^(2)`
`thereforeK=1/2mr^(2)omega^(2)`
`thereforeK=1/2mr^(2)(4pi^(2)f^(2))" "(becauseomega=2pif)`
`thereforeK=2pi^(2)mr^(2)f^(2)`
`thereforeK_(max)=2pi^(2)m(r_(max))^(2)f^(2)`
`thereforeK_(max)=2pi^(2)mR^(2)f^(2)`
`(becauser_(max)=R="radius of dee")`
`thereforeK_(max)=(2)(3.14)^(2)(1.67xx10^(-27))(0.6)^(2)(10xx10^(6))^(2)`
`thereforeK_(max)=1.186xx10^(-12)J`
`thereforeK_(max)=(1.86xx10^(-12))/(1.6xx10^(-19))eV=7.412xx10^(6)eV`
`thereforeK_(max)=(7.412xx10^(6))/(10^(6))MeV=7.412MeV`