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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 - Physics

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Question

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.

Numerical

Solution

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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2014-2015 (March) Ajmer Set 2

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