Transformer with wire running through as secondary

[stephen163]

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?

 

[Bob S]

Yes, you are correct. Review the physics of a particle (electron) accelerator called a betatron. A sinusoidal magnetic field B_A(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 B_v(t) field on the vacuum chamber itself is half the average field B_A(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 B_v(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.

 

[charng]

Yes, you are thinking about this in the right way. The wire in the secondary of the transformer can be considered as part of a closed loop, with one end connecting to the primary coil and the other end connecting to the rest of the secondary coil. This means that an electric field will be induced in the wire due to the changing magnetic field in the primary coil. This is in accordance with Faraday's law, which states that the induced electromotive force (EMF) in a closed loop is directly proportional to the rate of change of the magnetic flux through that loop. In this case, the wire acts as part of the closed loop and the induced EMF will result in a voltage across the two ends of the wire.