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Problem : A frequency of an electric oscillator is 10 MHz. What should be the magnitude of magnetic field for accelerating doubly ionized alpha particle? Assume mass of alpha particle 4 times that of proton.

Solution : The frequency of cyclotron is :

ν = q B 2 π m B = 2 π m ν q

Putting values,

B = 2 X 3.14 X 4 X 1.66 X 10 - 27 X 10 X 10 6 2 X 1.6 X 10 - 19 B = 1.3. T

Energy of charged particle

The energy of the finally accelerated particle corresponds to the speed when it travels in the outermost semicircular path having radius equal to that of Dees.

R = m v max q B v max = q B R m K max = 1 2 m v max 2 = q 2 B 2 R 2 2 m

Problem : Compare the final velocities of a proton particle and ionized deuteron when accelerated by a cyclotron. It is given that radius of cyclotron is 0.3 m and magnetic field is 2 T. Assume mass of deuteron twice that of the proton.

Solution : Let subscripts 1 and 2 correspond to proton and deuteron respectively. Note that deuteron is an isotope of hydrogen comprising of 1 proton and 1 neutron in the nucleus. The ionized deuteron thus carries one electronic positive charge same as proton. Now, final velocity of the charged particle accelerated by cyclotron is given as :

v max = q B R m


v max1 v max2 = q 1 B 1 R 1 m 2 q 2 B 2 R 2 m 1

But R 1 = R 2 , q 1 = q 2 , B 1 = B 2 and m 2 = 2 m 1 . Thus,

v max1 v max2 = 2

Numbers of revolutions

The kinetic energy of the charged particle is increased every time it comes in the gap between the Dees. The energy is imparted to the charged particle in “lumps”. By design of the equipment of cyclotron, it is also evident that the amount of energy imparted to the particle is equal at every instance it crosses the gap between Dees.

Since particle is imparted energy twice in a revolution, the increase in energy corresponding to one revolution is :

Δ E = 2 q V

Let there be N completed revolutions. Then total energy,

E = N Δ E = 2 q N V

Equating this with the expression obtained earlier for energy, we have :

2 q N V = q 2 B 2 R 2 2 m N = q 2 B 2 R 2 4 m q V

Magnetic field and energy

From the expression of kinetic energy of the accelerated particle, it is clear that kinetic energy of the charged particle increases with the magnitude of magnetic field – even though magnetic field is incapable to bring about change in speed of the particle being always perpendicular to the motion. It is so because increasing magnetic field reduces the radius of curved motion inside Dees. Therefore, there are greater numbers of revolutions before reaching to the periphery. See that numbers of completed revolutions are directly proportional to the square of magnetic field.

N = q 2 B 2 R 2 4 m q V

Greater numbers of revolutions result in greater numbers of times electrons are subjected to electrical potential difference in the gap between Dees. The maximum kinetic energy of the particle is :

K max = 2 q N V

As such, energy of the emerging particle increases for a given construction of cyclotron when magnetic field increases.

Potential difference and energy

Again, it is clear from the expression of kinetic energy of the accelerated particle that the energy of emerging particle from the cyclotron is independent of potential applied in the gap. It appears to contradict the fact that it is the electric force which accelerates the charged particle in the gap. No doubt, the greater potential difference results in greater electric force on the charged particle. This, in turn, results in greater acceleration of the particle and hence velocity. But then, particle begins to rotate in greater semicircle. This results in lesser numbers of rotations possible within the fixed extent of Dees. In other words, the greater potential difference results in greater acceleration but lesser numbers of opportunities for acceleration. Now,

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Source:  OpenStax, Electricity and magnetism. OpenStax CNX. Oct 20, 2009 Download for free at http://cnx.org/content/col10909/1.13
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