Direction of current from electromagnetic induction

Question

I've only scratched the surface of electromagnetism but I figured I might as well ask a question.

When moving a permanent magnet relative to an electric conductor, a current will form. Say you just move a magnet through a coil of wire for this question. I've learned that current will flow when there is a potential difference. However, it doesn't seem to me like there is a potential difference here at all. I figured it could be an electric field that formed; however, this seems incongruous with what I have learned about the formation of electric fields. As of now, I have concluded that the current would form simply by the property of the movement of a magnet (seems too simple). In sum, I am wondering what really causes the current to be able to form.

Answer 1

It isn't true in general that electric field is accompanied by differences in electric potential. This is true only at steady state, when electric and magnetic fields are not changing in time. Your moving permanent magnet violates this condition.

If you replace the coil with a conductive ring for simplicity, it is clear from symmetry considerations that there is no sensible way to define a potential that is not the same everywhere on the ring, i.e. there is no potential difference. Yet, as the magnet approaches, an electric field forms in the ring, which gives rise to a current. One must resort to Faraday's law to explain this:

$$ \oint\limits_{\text{ring}}\vec E\cdot d\vec \ell=-\frac{d\Phi}{dt}. $$

The left-hand side is the electromotive force $\mathcal E$ along the ring, and the right hand side is the rate of change of magnetic flux through the ring. Because of the changing magnetic flux, there is a non-zero electric field in the ring. The current in the ring is $\mathcal E$ divided by the resistance of the ring.

Answer 2

Magnetic induction is an experimental fact discovered by Faraday. About the direction suppose we make several turns of a wire around a permanent magnet in the floor. The N pole is up, so the magnetic field $B$ outside, that crosses our small coil, by convention goes down. The 2 ends of the wire are connected to a voltmeter. The display shows zero initially. If we flip quickly the magnet, $B$ increases upward during the movement. Using our left hand, with the thumb up, the other fingers indicate the sign of the conventional current (positive to negative) in the coil. The voltmeter will show the sign of the measured voltage accordingly.

Matematically, the use of the left hand is expressed by the minus sign in the Maxwell equation: $$\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}$$

The voltmeter here is a handy way to know the direction of the emf. But it is not correct in this situation to say that the electric field is minus the gradient of a scalar quantity called voltage. It is only true when there is no magnetc induction, for example in a DC resistive circuit.

Answer 3

The Faraday-maxwell law, is defined such that the da element points out of the page, if the line integral is measured anticlockwise.

This is to do with the orientation with stokes theorem, and the definition of curl.

If B is increasing out of the page, then negative work is done per unit charge going anticlockwise, aka positive work per unit charge going clockwise, and thus the current is clockwise.

Look up "surface orientation, and stokes theorem"

It is also important to note, faradays law works with any loop, even if a conducting wire isn't there. Faradays law tells us the EMF measured anticlockwise. Which says nothing about the direction of the E field at every point in the wire. However if an actual conducting wire is placed in space, surface charges will rearrange making E point in the direction of the wire, In the direction that positive work would be done per unit charge.