Working Principle of a Cyclotrons and Some Q&A

1.   describe the working principle of a conventional cyclotron

2.   show that the time required by a particle to travel through  a semi-circle , within Dee, is independent of the radius of the circle.

3.   deduce the kinetic energy of an emerging particle

4.   a cyclotron has a frequency of 14 X 10⁶HZ and a Dee radius of 53cm.

·       what value of magnetic field is needed to accelerate a deuteron?

·       what deuteron energy results?

answers

1)    a cyclotron is a device that can accelerate charge particles to very high speed,  the magnetic field produced are used to bombard atomic nuclei and thereby produce nuclear reactions of interest to researchers.

the cyclotron shown above consists of two hollow metal containers held between the pole faces of a large and powerful magnetic field. the two hollow metal metals are known as dees, because they are more or less like the letter D.  a positively charged is introduced into the cyclotron at the center, the edge of the left and Dee is negatively charged at this instant, as a result the particle is attracted to the left Dee.

 however , once the particle is inside the hollow metal, it does not experience any further force because there is no electric field  inside the hollow metal, but there is a magnetic field in the hollow metal. But there is a magnetic field in the hollow metal which is perpendicular and maintains the charged particles in a circular path and returns it to the edge of the Dee. By the time the particle is ready to cross the gap , the polarity of the voltage is reversed and the particle is accelerated  across the gap to the right Dee.

in other word , if the emerging deuteron from the ion source find the Dee that is facing it to be negative it will accelerate towards this Dee and will enter it. once inside , it is screened from electric fields by  the metal walls of the dee. the magnetic field is not screened by the dee so that the ion bends in a circular path of radius r=mv/qB. where m is the mass of the particle, v is the velocity, q is the charge and B is the magnetic field.

After time t₀ the ion emerges from the dee on the other side of the ion source, assume that the accelerating potentials has how changed sign (that is to negative), thus the ion again faces a negative dee and is further accelerated and again describes a semi-circle, of larger radius in the dee. The time of passage through this dee is however still t₀, this process goes on until the ion reaches the other edges of one dee where it is pulled out by a negatively charged deflector plate.

 The key to the operation of the cyclotron is that he characteristic frequency at which the ion circulates in the field must be equal to the fixed frequency of the electric oscillator.

2)   The centripetal force is produced by the magnetic force,  therefore F= qvB=mv²/r .

solving for r we obtain r=mv/qB

the time required by a particle to travel through  a semi-circle , within Dee,  which is a semi-circle is given by the period, which is T=πr/v

Substituting the value of r we obtain

T= πm/qB

3)    we can also deduce the kinetic energy of the emerging charged particles from the above equations, since r=mv/qB

making v=qBr/m

the kinetic energy K.E= ½ mv² = ½ [q²B²/m]r².

4)    solution

·       Frequency is given by 14x10⁶HZ and r=53cm=0.053m

F=qB/2πm,

14 x10⁶=(1.6x10^-19 x B)/[2X3.142x 3.343 x 10^-27 kg]

B=2.941X10^-19/1.6x10^-19 =1.838T

·       the energy of the deuteron must be a kinetic energy.

he kinetic energy K.E= ½ mv² = ½ [q²B²/m]r².

½ [q²B²/m]r²

½ x {[(1.6x10^-19)²x 1.838²]/ 3.343 x 10^-27 kg} x0.53²

= 3.633x 10^-12j,

=22.71MeV