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प्रश्न
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, calculate the kinetic energy (in MeV) of the proton beam produced by the accelerator.
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उत्तर
Frequency of oscillators (v) = 10 MHz = 107 Hz
Mass of proton, `"m" = 1.67 xx 10^-27 "kg"`
Charge of proton = 1.6 x 10-19 C
Operating magnetic field is given by the relation
`"B" = (2pi"mv")/q = (2 xx 3.14 xx 1.67 xx 10^-27 xx 10^7)/(1.6 xx 10^-19)`
= 0.65 T
Radius of dees= 60 cm = 0.6 m
`"KE" = ("q"^2"B"^2"r"^2)/(2"m") = ((1.6 xx 10^-19)^2(0.65)^2(0.6)^2)/(2 xx 1.67 xx 10^-27 xx 1.6 xx 10^-13) "Mev"`.
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