Betatron ppt
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A betatron is a device that accelerates electrons using an expanding magnetic field within a doughnut-shaped vacuum chamber. Electrons are injected into the chamber and accelerated as the magnetic field strength increases over time. This increasing magnetic flux induces an electric field that increases the electrons' energy, allowing them to gain extremely high speeds. The betatron condition requires that the rate of change of magnetic flux through the circular orbit equals 2π times the radius squared times the rate of change of the magnetic field, in order to maintain the electrons' constant orbital radius as they accelerate.
Betatron ppt
- 2. BETATRON Betatron is a device for speeding up electron to extremely high energies with the help of expending magnetic field. It was constructed in 1941 by D.W.Kerst.
- 3. K.W. Kerst with BETATRON
- 4. Betatron Differs from cyclotron The electrons are accelerated by expending magnetic field. The circular orbit has a constant radius.
- 5. Different Betatrons According to their Generations
- 8. Constant Radius of Betatron
- 9. Construction Betatron consists of highly evacuated angular tube D known as doughnut chamber. The chamber is placed between the poles of an electromagnet excited by an alternating current (frequency of 60 or 180 Hz) Electrons are produced by electron gun and are injected into doughnut at the beginning of each cycle of alternate current. The increasing magnetic flux gives rise to a voltage gradient(electric field) round the doughnut which accelerates the orbiting electrons
- 11. PRINCIPLE The principle of the betatron is the same as that of a transformer in which an Alternating current applied to the primary coil induces an alternating current In the secondary. In betatron secondary coil is replaced by a doughnut shaped vaccum chamber. When the electron is injected in doughnut, the alternating magnetic field has two effects : An electromotive force is produced in the electron orbit by changing magnetic flux that gives an additional energy to the electrons. A radial force is produced by the reaction of magnetic field whose direction is perpendicular to the electron velocity which keeps the electrons moving in the circular part.
- 13. OPERATION Electrons from the electron gun are injected into doughnut shaped vacuum chamber when the magnetic field is just rising from its zero value in the first quarter cycle. The electrons now make several thousand revolution and gain energy. When the magnetic field has reached its maximum value, the electrons are pulled out from their orbit. Either they strike a target and produce X-rays or emerge from the apparatus through a window
- 15. BETATRON CONDITION Consider an electron is moving in a circular orbit of radius ’r’ in the magnetic field. Let at any instant, B be the magnetic field at this orbit and the total magnetic flux through the orbit is ΦB. The flux ΦB increases at the rate of d/dt (ΦB) and the induced e.m.f. In the orbit is given by Induced e.m.f. = d /dt (ΦB) .....(i) work done on the electron in one revolution = induced e.m.f. X Charge = - d/dt (ΦB) x e Thus work done must be equal to the tangential force F acting on the electron multiplied by the length of the orbit path i.e., work done = Force x Distance =Fx2πr Therefore, F x 2 π r = - d/dt (ΦB) x e F =- e/ 2 π r x {d/dt (ΦB)} ....(ii)
- 16. The force F will increase the electron energy and which in turn would tend to increase the orbit of large radius. In order to maintain the radius of the orbit, The force experienced by the electron must be counteracted. Suppose the velocity Of the electron is v and its mass is m. When the electron moves in an orbit of Radius r under the action of field of magnetic induction B, the inward radial force B e v is to be equal to the upwards centrifugal force mv2/r . Therefore, B e v = m v2 / r m v=Ber ....(iii) According to Newton’s law, the force is defined at the rate of change of momentum (p=m v) i.e., F = d/dt (B e r) = er dB/dt ....(iv)
- 17. To maintain the radius constant, the value of F given in equation (ii) and Equation (iv), should numerically, hence e/ 2 π r x {d/dt (ΦB)} = e r dB/dt d/dt (ΦB) = 2 π r2 dB/dt Integrating, we get ΦB = 2 π r2 B This is known as Betatron condition