0
$\begingroup$

The Faraday's induction law states that the temporal variation of the magnetic flux induces an electromotive force, according to the following expression:

$$\varepsilon^i=-\frac{d\theta^m}{dt}$$

I know how this law works on a whorl, coil, but not for a volumic body. How I should write the Faraday's induction law for a solid cylinder? I'm studying induction motors, and so, I need to know how is current inducted on the rotor (cylinder). Should I assume that the current only appears on the surface? Should I assume some "whorl density" like we do for the coil?

I can't find any information on internet about this topic. Do you know some interesting books or sites?

$\endgroup$
  • $\begingroup$ Are you absolutely sure you know how to apply it for a disk? $\endgroup$ – Timaeus Feb 27 '16 at 22:31
  • $\begingroup$ No... I thought that I could assume the disk as a sum of independent whorls, and so we have $\varepsilon^i=\varepsilon^i(\rho)$ where $\rho$ is the distance to the center of the disk. But the disk is a conductor, and maybe this thought is wrong. I will remove that section from the question $\endgroup$ – Élio Pereira Feb 27 '16 at 22:38
  • $\begingroup$ I was trying to find some theory behind the current induction on the rotor of the induction motor, but I didn't have success... $\endgroup$ – Élio Pereira Feb 27 '16 at 22:44
  • $\begingroup$ You need to know the actual electric and magnetic fields and to find the current flowing through a solid. So you need to learn how to find electric and magnetic fields (and the velocities of the charges) so you can find the EMF in a solid. $\endgroup$ – Timaeus Feb 27 '16 at 23:02
0
$\begingroup$

It doesn't hold for a cylinder, instead you have to compute: $$\oint \left(\vec E+\vec v\times \vec B\right)\cdot \mathrm d\vec \ell.$$

Faraday's Law says $$\oint \vec E\cdot\mathrm d \vec\ell=-\iint \frac{\partial \vec B}{\partial t}\cdot \mathrm d\vec A.$$

And the universal flux rule: $$\oint \left(\vec E+\vec v\times \vec B\right)\cdot \mathrm d\vec \ell=-\frac{\mathrm d}{\mathrm d t}\iint \vec B\cdot \mathrm d\vec A$$ only holds for thin wires (where the charges stay inside and there are no magnetic monopoles), not for solid objects.

$\endgroup$

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.