A cyclotron.s oscillator frequency is 10 MHz and the operating magnetic field is 0.66 T. If the radius of its dees is 60 cm, then the kinetic energy of the proton beam produced by the accelerator is

Options

A. 9 MeV
B. 10 MeV
C. 7 MeV
D. 11 MeV

Answer

C

Explanation

Radius of circular path : R = `sqrt(2mk)/(qB)`
`k=(q^(2)B^(2)R^(2))/(2m)`
Cyclotron frequency is `v=(qB)/(2pim)`
or `q^(2)B^(2)=4pi^(2)m^(2)v^(2)`
`k=1/(2m)(4pi^(2)m^(2)v^(2))R^(2)`
`k=2pi^(2)mv^(2)R^(2)` …joule
`k=(2pi^(2)mv^(2)R^(2))/e...eV`
= `(2xx10xx(1.67xx10^(-27))xx(10xx10^(6))^(2)xx(0.6)^(2))/(1.6xx10^(-19))eV=7.2xx10^(6)eV=7.2MeV`