0
$\begingroup$

To find the emf (voltage) generated for a changing magnetic field around a conducting loop, we use faraday's law with flux defined across the surface of the disk which the loop encloses. This tells you emf between two adjacent points on the conducting loop, right?

The problem I'm facing is that because I do not understand what faraday's law is actually derived from neither from an experimental nor theoretical perspective, I also fail to understand its scope. There are other situations for which I have seen this applied, such as a spinning conducting disk with a constant magnetic field applied parallel to the rotation axis, or a rod passing through a static magnetic field. In these cases, unless a specific intuition is made about choice of surface flux change seems to be 0. For either cases, it is easier to shift the coordinates to a moving frame to instead calculate the lorentz force due to static B field. If the surface is just the area 'swept out' by the conductor, then what does swept out mean and how do we find the two points which produce the corresponding emf?

$\endgroup$
  • $\begingroup$ Faraday's law is derived from experiment. All you need to do is write it down, make a sketch with a conducting loop and some arrows indicting current, magnetic field and EMF. Or google for images of Faraday's law. $\endgroup$ – my2cts Apr 30 '18 at 22:06
  • $\begingroup$ what about when there isn't a conducting loop? $\endgroup$ – BIGFATNIH Apr 30 '18 at 22:07
  • $\begingroup$ See in the proper light Faraday's law is a particular instance of Stoke's theorem, which means that you can fall back on the math to understand that in general you have a lot of freedom to chose the surface and the boundary as you see fit, so long as they agree with one another. How to make best use of that freedom is the next (and harder) question. $\endgroup$ – dmckee --- ex-moderator kitten Apr 30 '18 at 22:29
  • $\begingroup$ You can use any surface, and because div B=0 the result will always be the same. $\endgroup$ – lalala Apr 30 '18 at 22:37
  • $\begingroup$ This is not correct: "Faraday's law is a particular instance of Stoke's theorem". The equivalence of the magnetic flux through a surface and the loop integral of its rotation is mathematical, but the fact that, subsequently, $\vec{\nabla} \times \vec{B} = -\partial\vec{E}/\partial t$ is physics and derived from experiment. $\endgroup$ – my2cts Apr 30 '18 at 22:53
1
$\begingroup$

we use faraday's law with flux defined across the surface of the disk which the loop encloses. This tells you emf between two adjacent points on the conducting loop, right?

EMF mentioned in the Faraday law is meant for a loop (close curve), not for two points. If the physical circuit is disconnected so the wire has two ends that are very close together, then the emf for the closed loop will be approximately equal to voltage across the two ends (exactly equal for perfect conductor), but EMF and voltage are two different concepts.

There are other situations for which I have seen this applied, such as a spinning conducting disk with a constant magnetic field applied parallel to the rotation axis, or a rod passing through a static magnetic field. In these cases, unless a specific intuition is made about choice of surface flux change seems to be 0.

In these cases, the traditional formulation of the Faraday law indeed seems inapplicable, because there is no obvious loop that emf would be defined for. But if some loop in space is chosen nevertheless (it does not have to be inside conducting circuit at all), then the law is applicable. There are two distinct cases.

1) if the loop chosen is not moving or changing in any way, then the Faraday law is valid always without further qualification; EMF for the loop is proportional to rate of change of magnetic flux through that loop;

2) if the loop in space is changing as time passes, then the Faraday law is not always applicable. It is applicable, however, in the special case where the moving part of the loop is fixed to material conductor (for example, a rectangle loop whose one side is always inside the moving rod). If the part of the loop inside conductor was moving with respect to the conductor, the corresponding magnetic force on test charges fixed to the loop would not be in any simple relation to the actual electromotive force acting on the real conducting charges.

So, to apply the Faraday law to loops that move or change in time, one usually chooses loop where the parts in non-conductive medium are stationary with respect to the observer, but the parts in conductor are stationary with respect to the conductor.

The simplest application of this is for a coil of wire that rotates in external magnetic field (as used in simple DC motors): the loop is chosen so that it always passes through the rotating coil, which means the loop rotates in space.

If we wanted to apply this special case of Faraday law to rotating disk, we have to choose a path that is stationary where it is in air and moving along with the conductor where it is in the conductor. One way to define such a loop is this. Imagine a rectangle whose one corner is in the disk center and another corner is on its rim. Then imagine all sides of the rectangle are fixed except the side defined by these two points; this side rotates around the disk center, along with the disk.

This kind of loop is obviously changing its shape in space, and the magnetic flux through this loop will change as well. Now, we apply the Faraday law to this loop.

$\endgroup$
  • $\begingroup$ Right it is the voltage for a path. I've become so accustomed to conservative vector fields that i neglected the actual definition of potential $\endgroup$ – BIGFATNIH May 1 '18 at 16:12
  • $\begingroup$ OK, I'm satisfied with this answer as it at least provides a well defined framework for the law. It's probably not worth trying to justify without deriving third maxwell equation $\endgroup$ – BIGFATNIH May 1 '18 at 16:13

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.