Please provide steps and explanations. Thank you. A cyclotron with dee radius 50

Question

Please provide steps and explanations. Thank you.

A cyclotron with dee radius 50.5 cm is operated at an oscillator frequency of 10.5 MHz to accelerate proton:s (a) What magnitude B of magnetic field is required to achieve resonance? (b) At that field magnitude, what is the kinetic energy of a proton emerging from the cyclotron? MeV Suppose, instead, that B 1.40 T (c) What oscillator frequency is required to achieve resonance now? MHz (d) At that frequency, what is the kinetic energy of an emerging proton? Mev

Explanation / Answer

for circular motion in magnetic field,

Fb = m a_c

q v B = m v^2 / r

v/ r = q B / m

(A) f = v / (2 pi r)

2 pi 10.5 x 10^6 = (q B / m)

2 pi 10.5 x 10^6 = (1.6 x 10^-19 x B) / (1.67 x 10^-27)

B = 0.69 T  

(B) f = v / (2 pi r )

10.5 x 10^6 = v / (2 x pi x 0.505)

v = 3.33 x 10^7 m/s

KE = m v^2 / 2 = 9.268 x 10^-13 J  

KE (in MeV) = 5.79

(C) 2 pi f = (q B / m)

2 pi f = (1.6 x 10^-19 x 1.40) / (1.67 x 10^-27)

f = 21.4 x 10^6 Hz or 21.4 MHz

(D) 21.4 x 10^6 = v / (2 x pi x 0.505)

v = 3.33 x 10^7 m/s

KE = m v^2 / 2 = 3.84 x 10^-12 J  

KE (in MeV) = 24

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