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Describe the construction and working theory of cyclotron.

Answer
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Hint- Here, we will proceed by telling what is the function of a cyclotron. Then, we will discuss the principle on which this device is based. Also, the construction of this device along with its working theory will be explained.

Complete step-by-step solution:
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Cyclotron is a device used to accelerate charged particles to high energies.
Principle
Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic Lorentz force due to which the particle moves in a circular path.
Construction
It consists of a hollow metal cylinder divided into two sections D1 and D2 called Dees, enclosed in an evacuated chamber. The Dees are kept separated and a source of ions is placed at the centre in the gap between the Dees. They are placed between the pole pieces of a strong electromagnet. The magnetic field acts perpendicular to the plane of the Dees. The Dees are connected to a high frequency oscillator.
Working Theory
When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a negative potential at that instant of time. Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in a circular path. By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed. Hence the particle is once again accelerated and moves into the other Dee with a greater velocity along a circle of greater radius. Thus the particle moves in a spiral path of increasing radius and when it comes near the edge, it is taken out with the help of a deflector plate (D.P). The particle with high energy is now allowed to hit the target T.

Note: According to the concept of magnetic lorentz force, when the particle moves along a circle of radius r with a velocity v then the magnetic lorentz force provides the necessary centripetal force which means that here the magnetic lorentz force will be equal to the centripetal force i.e., Bqv = $\dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}$.