magnetic field B at the electron path r path of electrons evacuated doughnut o m v F B magnetic field B directed up E o B directed up and increasing E Figure 15 When the strong central field B 0 in the betatron is rapidly increased, it produces a circular electric field that is used to accelerate the electrons. The electric field E is related to the flux Φ B of the central field B 0 by Faraday's law E ⋅ d = – d Φ B / dt . Figure 14b Top view of the betatron showing the evacuated doughnut, the path of the electrons, and the magnetic fields B 0 in the center and B r out at the electron path. In order to keep the electrons moving on a circular path inside the doughnut, the magnetic force F B = – e v × B r must have a magnitude F B = m v 2 r where r is the radius of the evacuated doughnut.
30-18 Faraday's Law Let us consider an explicit example to get a feeling for the kind of numbers involved. In the 100 MeV betatron built by General Electric, the electron orbital radius is 84 cm, and the magnetic field B 0 is cycled from 0 to .8 tesla in about 4 milliseconds. (The field B 0 is then dropped back to 0 and a new batch of electrons are accelerated. The cycle is repeated 60 times a second.) The maximum flux Φ m through the orbit is Φ m = B 0 max π r 2 = .8 tesla × π × (.84m) 2 Φ m = 1.8 teslam 2 If this amount of flux is created in 4 milliseconds, then the average value of the rate of change of magnetic flux Φ B is d Φ B dt = Φ m .004 sec = 1.8 .004 = 450 volts Thus each electron gains 450 electron volts of kinetic energy each time it goes once around its orbit. Exercise 3 (a) How many times must the electron go around to reach its final voltage of 100 MeV advertised by the manufacturer? (b) For a short while, until the electron’s kinetic energy gets up to about the electron’s rest energy m 0 c 2 , the electron is traveling at speeds noticeably less than c. After that the electron’s speed remains very close to c. How many orbits does the electron have to make before its kinetic energy equals its rest energy? What fraction of the total is this? (c) How long does it take the electron to go from the point that its kinetic energy equals its rest energy, up to the maximum of 100 MeV? Does this time fit within the 4 milliseconds that the magnetic flux is being increased? TWO KINDS OF FIELDS At the beginning of the chapter we showed that the line integral E ⋅ d around a closed path was zero for any electric field produced by static charges. Now we see that the line integral is not zero for the electric field produced by a changing magnetic flux. Instead it is given by Faraday ’ s law E ⋅ d = – d Φ B /dt . These results are shown schematically in Figure (16) where we are looking at the electric field of a charged rod in (16a) and a betatron in (16b). In Figure (17), we have sketched a wire loop with a voltmeter, the arrangement we used in Figure (12) to measure the E ⋅ d . We will call this device an “ E ⋅ d meter ” . If you put the E ⋅ d meter over the changing magnetic flux in Figure (16b), the voltme- ter will show a reading of magnitude V = d Φ B /dt . If we put the E ⋅ d meter over the charged rod in
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- Fall '19
- Magnetic Field, Faraday