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.