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Transformer with wire running through as secondary

  1. Apr 24, 2009 #1
    I want to confirm that i'm thinking about this in the right way...


    Imagine a transformer. The primary has an integer number of turns. Now imagine the secondary has less than 1 turn, i.e., just a piece of wire passing through, forming essentially half a turn.

    I know voltage is induced in this piece of wire. It doesn't form a closed loop, but faraday's law relates the induced electric field around a CLOSED loop to the time varying magnetic flux passing through the loop.

    Can I assume that the wire itself forms PART of a closed loop and hence an electric field is indiced across the two ends?
     
  2. jcsd
  3. Apr 24, 2009 #2
    Yes, you are correct. Review the physics of a particle (electron) accelerator called a betatron. A sinusoidal magnetic field BA(t) is applied to the area (A = pi R2) inside a toroidal vacuum chamber of major radius R, inducing an azimuthal electric field E(t) inside the vacuum chamber of minor radius r << R, per Faraday's law. If the Bv(t) field on the vacuum chamber itself is half the average field BA(t) in the area A, the particle will be simultaneously smoothly accelerated by Faraday's law and deflected by the Lorentz Force F(t) = q[v(t) x Bv(t)] , and wlll remain inside the vacuum chamber as it is accelerated. This demonstrates that the accelerating electric field is everywhere inside the vacuum chamber.
     
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