Potential difference of an inductor equal to minus the induced emf?

[BomboshMan]

Hi,

Two questions I'm finding it hard to get my head around with inductors...

Say we have an increasing current, this is going to cause an increase in flux in one direction, which will induce an emf to drive a current which produces a magnetic field to oppose the change in flux. Than doing the calculation to find induced emf,

ε = |\frac{d\Phi}{dt}| so my first question is, is \Phi the magnetic flux just from the inductor current, or is it the flux from the net magnetic field, including the induced one?

Then my book says that the potential difference of the inductor is minus the induced emf. I get the minus, but wouldn't the inductor have an 'original' potential difference, as in shouldn't the overall potential difference of the inductor be the 'original' pd plus the induced emf?

Hope that made sense!

Thanks,
Matt

 

[vanhees71]

First of all one cannot emphasize enough that an electromotoric force, \Phi, is not a potential difference, because precisely if you have a time-varying magnetic flux, the electric field is not a potential field, according to Faraday's Law
\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.
When reduced to quasistationary circuits in general the total flux has to be taken into account. It's included in the self-inductance of the circuit. The sign of the emf in the Kirchhoff Laws is precisely given by the above given local form of Faraday's Law.

Take a circuit at rest. Then the integral Faraday law is obtained from its local form by integrating over an appropriate surface and using Stokes's integral theorem. The sign is given by the relative orientation of the surface normal vectors with the orientation of the boundary curve according to the right-hand rule.

For details, see my (handwritten) writeup for a freshman lecture, I've given some years ago:

http://fias.uni-frankfurt.de/~hees/physics208.html

See the Lecture Notes, part III, and the worked examples on AC circuits.