3
$\begingroup$

The title seems confusing, because Eddy currents are induced currents... however, let me explain.

Assume the following closed loop passing a uniform magnetic field like so:

enter image description here

Now, this loop would have an induce EMF to allow current flow to oppose the change that induced it like so: enter image description here

Are there Eddy currents induced as well on the surface?

Vs this diagram(closer look into that portion of the wire):

enter image description here

My confusion lies with Eddy currents on the surface, or just induced current that circulates around the loop opposing the change that induced it... can both coexist?

$\endgroup$
  • $\begingroup$ Both coexist on the same conductor in form of their respective EMFs on the conductor (as described by Kthaxt). However, these give rise to electric fields that would add up and thus for a specific point on the conductor, you'll only have one net value of electric field and thus current, given by adding the two electric fields using principle of superposition $\endgroup$ – Peeyush Kushwaha Mar 21 '16 at 20:21
3
$\begingroup$

The physics creating eddy currents and EMFs in inductors is the same: Faraday's law of induction.

$ \oint_C {E \cdot d\ell = - \frac{d}{{dt}}} \int_S {B_n dA} $

The strength of any induced current and voltage is dependent on:

1) The amount of magnetic flux ($\int_S {B_n dA}$)

2) The rate at which the flux is changing

So for the loop in your first picture, lets assign some dimensions. We'll call the outer radius $ R $ and the inner radius $ r $ so that the thickness of the ring is $R-r$. For a very small thickness, clearly the amount of flux through the ring will be >> than the magnetic field hitting the conductor at any given time. So left side of faraday's law will be much larger for the circuit of the entire loop than for the that of the loop drawn on the conductor.

Moreover, the left side of the equation indicates an induced voltage, so the current generated is proportional the resistance. Eddy currents are primarily on the surface of the conductor (this is why they are often used for non-destructive testing of materials to find cracks on the surface of sheet metal). So the resistance seen by the eddy currents I believe would be much larger than the resistance seen by the currents in the ring as the A in:

$ R = \rho L/A $

Would be much larger for the current induced around the loop.

In conclusion, both can exist (and can oppose each other), and certainly do in your example, typically the induced current in the loop is just dominant for any well designed inductor. In brakes that utilize the drag induced by eddy currents, one would design for the opposite effect.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.