Question

15.2) Cyclotron

A cyclotron consists of two dee magnets each of radius 30 cm. They create a 0.75 T magnetic field

perpendicular to their faces. The potential difference across the gap between the dee magnets is 500 V. The

cyclotron is accelerating an alpha particle (helium-4 nucleus).

(a) How much time does it take the alpha particle to travel one full revolution?

(b) What is the alpha particle’s maximum kinetic energy achievable in this cyclotron?

(c) How many complete revolutions must the particle make to achieve this maximum energy? (Note: It passes

the gap twice every revolution.)

Answer 1

we know,
mass of He-nucleus, m = 4*1.67*10^-27 kg
charge of He-nucleus, q = 2*1.6*10^-19 C

a) Time taken to complete one revolution, T = 2*pi*m/(B*q)

= 2*pi*4*1.67*10^-27/(0.75*2*1.6*10^-19)

= 1.75*10^-7 s <<<<<<<<<<----------------Answer

b) let v_max is the maximum speed of the alfa particle

we know, r_max = m*v_max/(B*q)

v_max = B*q*r_max/m

= 0.75*2*1.6*10^-19*0.3/(4*1.67*10^-27)

= 1.08*10^7 m/s

KE_max = (1/2)*m*v_max^2

= (1/2)*4*1.67*10^-27*(1.08*10^7)^2

= 3.90*10^-13 J <<<<<<<<<<----------------Answer

c) let N is the number of revolutions made.

Work done by both Ds = gain in kinetic energy

2*N*q*delta_V = KE_max

N = KE_max/(2*q*delta_V)

= 3.90*10^-13/(2*2*1.6*10^-19*500)

= 1219 <<<<<<<<<<----------------Answer