A cyclotron designed to accelerate protons has an outer radius of 0.350 m. The p
ID: 2075436 • Letter: A
Question
A cyclotron designed to accelerate protons has an outer radius of 0.350 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.800 T. (a) Find the cyclotron frequency. (b) Find the speed at which protons exit the cyclotron and (c) their maximum kinetic energy. (d) How many revolutions does a proton make in the cyclotron (e) For what time interval does one proton accelerate? a. omega = b. V = K_max d. N = e. T =Explanation / Answer
Given Data
outer radius of the proton r = 0.35 m
potential difference V = 600 V
magnitude of magnetic field B = 0.8 T
Let
mass of the proton m =1.67*10-27 kg
charge of the proton q =1.67*10-19 C
Solution : -
(a) assume that speed of proton is v
we have
B*v*q = m*v2/r
=> r = mv / Bq
and period T = 2?m/qB
or frequency, f = 1/T = qB/2?m
= (1.67*10-19 )*0.8 / 2?*(1.67*10-27)
= 12.732*106Hz = 12.732 MHz
(b) the speed at which protons exit the cyclotron:
v{max} = r{max}*B*q/m
= 0.35*0.8*(1.67*10-19 ) / (1.67*10-27)
= 28*106 m/s = 28 Mm/s
(c) maximum kinetic energy.
K = 1/2*m*v{max}2
= 1/2*(1.67*10-27)*{28*106}2
= 6.55*10-13 J
d)
(d)revolutions does a proton make in the cyclotron
each cycle gives the proton an amout of energy, which is E = 2*?V*q where ?V=600 V
=> N turns = K/E
= K/2*?V*q
= 6.55*10-13 / 2*600* (1.67*10-19 )
= 3410 revolutions
(e)what time interval does one proton accelerate
the period is constant
=> time required t = N*T = N / f
= 3410 / 12.732*106
= 2.68 *10-4s = 268 micro sec
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