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Faraday's Law

  1. Jan 14, 2010 #1
    Hi All,

    Please take a very long tightly wound solenoid of any radius with another coil wound around the outside of it. Now, if a current is applied to the solenoid, the changing flux of the magnetic field causes an induced emf in the outside coil which is perfectly consistent with the math of the situation.

    Although the math is perfectly obvious to me, I don't understand how such a thing could happen. The wire in which the emf is generated is in a part of space where absolutely nothing happens. Put the whole set up in a vacuum and the wire is then in a part of space where there is absolutely nothing!

    How could such a thing happen?

    Thank you,

    Bob
     
  2. jcsd
  3. Jan 14, 2010 #2
    A changing magnetic flux induces an electric field that propagates outward so that it may induce a current through the outer coil.
     
  4. Jan 14, 2010 #3
    Thank you. That makes perfect sense.

    Bob
     
  5. Jan 14, 2010 #4
    To add to what Wannabeagenius says, a changing current in the solenoid causes changing electric and magnetic fields external to the solenoid that propagate radially at velocity c.
     
  6. Jan 15, 2010 #5
    It's a good question, particularly if the solenoid is infinitely long (so that B = 0 at all external points). The key here is that the vector potential is NOT zero at points external to the solenoid. It circulates around the solenoid's axis, and the circulation equals B inside the solenoid. The existence of a time-varying vector potential circulation is equivalent to a time varying magnetic field inside the solenoid. An emf is induced in the outer coil, even though B=0 at all times outside of the solenoid!
     
  7. Jan 15, 2010 #6
    An emf is generated in the coil because there is a net

    EMF = -N∫(dB/dt)·n dA inside the coil (and inside the solenoid). (Faraday's Law).

    This is identical to a conventional transformer, where the dB/dt is confined to the iron core, and the coil represents the windings outside the iron.

    Bob S
     
  8. Jan 15, 2010 #7
    I've had enough of reposting blown-out posts from a bad internet connection, containing all the supporting equations. So simply put, there is no required action at a distance influences from magnetic fields interior to both an idealized infinite solenoid and interior to the iron of a transformer that produced a current flow in exterior windings.

    A changing current in either one results in exterior electric and magnetic fields that obey the harmonic equations of traveling and attenuated waves in a vaccum with appropriate boundry conditions at the surface of the solenoid or core material.
     
  9. Jan 15, 2010 #8
    If im lucky, and short winded, this will post the fields surrounding an ideal solenoid.

    [tex]\partial_{r}{}^{2}E_{\theta}(r) = \frac{1}{c^2} \partial_{t}{}^{2} E_{\theta}(r)[/tex]
    [tex]\partial_{r}{}^{2}B_{z}(r) = \frac{1}{c^2} \partial_{t}{}^{2} B_{z}(r)[/tex]
     
  10. Jan 15, 2010 #9
    An example (because it was the easiest to solve.), is a constantly increasing current in an ideal solenoid.

    [tex]E_{\theta}(r)= E_{r_0} \exp[(-r+ct)/R][/tex]

    [tex]B_{z}(r) = B_{r_0} \exp[(-r+ct)/R][/tex]
     
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