0
$\begingroup$

Whenever we have a changing magnetic flux through a conductor an induced electric field is generated. It is often said that these induced electric field lines are circular in shape.But why is this so?Why are these field lines not elliptical for example? How can this shape be mathematically derived?

Also does this have anything to do with equipotential surfaces?Do such non-conservative electric fields even produce a potential field?If not why?

$\endgroup$
  • 1
    $\begingroup$ They can be elliptical. The shape of the induced electric field depends on the shape of the changing magnetic field. $\endgroup$ – S. McGrew Oct 19 at 14:33
  • $\begingroup$ How can this shape be mathematically derived? $\endgroup$ – Schwarz Kugelblitz Oct 19 at 14:44
  • $\begingroup$ What can be done to attract answers $\endgroup$ – Schwarz Kugelblitz Oct 20 at 18:34
  • $\begingroup$ "How can this shape be mathematically derived?" It is pretty straightforward if you know vector calculus. Just use Maxwell's equations to describe a changing current density, and solve for the induced electric field. $\endgroup$ – S. McGrew Oct 26 at 20:38
  • $\begingroup$ I don't know vector calculus $\endgroup$ – Schwarz Kugelblitz Oct 26 at 20:56
2
$\begingroup$

The induced fields can be many shapes other than circular, such as elliptical. It can easily be shown using Maxwell's equations that unless the changing current density is axially symmetric, the induced electric field lines will not be circular.

If you don't know calculus, you won't understand the derivation. But think about it this way: A uniform bundle of straight magnetic field lines (e.g., formed by an infinitely long solenoid coil) that is steadily increasing in strength will indeed generate a circular electric field around the bundle, because the bundle and fields will be axially symmetric. But two such bundles, separated by some distance, will each produce their own circular field. When the two fields are added together (as vectors), the sum field will not be circular because there will not be axial symmetry (due to the two separated bundles of magnetic field lines).

$\endgroup$
  • $\begingroup$ Can u still post the mathematical derivation anyway?I basic differential and integral calculus and understand vector calculus somewhat $\endgroup$ – Schwarz Kugelblitz Oct 27 at 8:25
  • $\begingroup$ No, I won't post the derivation but I'm willing to guide you to derive it yourself. We may need to do it in Chat. But first step: can you derive the magnetic field around a wire carrying a constant current? $\endgroup$ – S. McGrew Oct 27 at 13:17
  • $\begingroup$ Yes,it's UI(sin(alpha)+sin(beta))/4πz $\endgroup$ – Schwarz Kugelblitz Oct 27 at 13:55
  • $\begingroup$ You need to define your terms. $\endgroup$ – S. McGrew Oct 27 at 13:59
  • $\begingroup$ Sure, I - current..alpha ,beta are angles From the point where the field is to be found to the end of the wire,..z is perpendicular distance from point to wire....U is permeability of vacuum $\endgroup$ – Schwarz Kugelblitz Oct 27 at 14:01

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.