# Magnetic induction: apparent contradiction between local and integral prediction

I'm trying to understand better magnetic induction, and I came across a situation where I find an apparent contradiction between the local form ($$\nabla×E=-\partial B/\partial t$$) and the corresponding integral form ($$e=-d\phi/dt$$, which of course corresponds to the integration of the local form) of induction law.

[EDIT : in order to make the problem easier to visualize in space, I found a simpler situation with a similar problem]

Take these two coaxial coils :

We provide an increasing current $$i_1$$ to coil 1, and want to examine the induced current $$i_2$$ in coil 2.

The question is : in which direction does $$i_2$$ flow?

Integral reasoning

Surface of reference : the cross section of coil 2 (in green).

• The magnetic field $$B_1in$$ in coil 1 is oriented up.
• There's also a downward-pointing magnetic field $$B_1out$$ outside coil 1, in the small annular zone between coil 2 and coil 1.
• But globally the integral $$\phi$$ of the field over S is positive.

So, as $$i_1$$ increases, $$\phi$$ is more and more positive. Therefore $$e=-d\phi/dt <0$$ : $$i_2$$ is such that the induced field points downward, in opposition to the increase of $$\phi$$ ($$i_2$$ < 0)

Local reasoning

the wire of coil 2 is subject to $$B_1out$$ : the wire doesn't "know" what happens inside coil 1. If, instead of coil 1, we had an externally-imposed flux pointing downward everywhere, then coil 2 couldn't tell the difference.

As $$i_1$$ increases, $$B_1out$$ gets stronger, so $$\nabla×E=-\partial B/\partial t$$ points up at each point of the wire loop. If the curl points up, it means that the induced electric field is such that $$i_2$$ is positive (the field induced in coil 2 points up).

This is obviously a contradiction with the integral reasoning. I know that the correct solution is the first one, but I don't see where the mistake is in the local reasoning.

[old description] The situation is simple: a permanent magnet rotates counter-clockwise (the axis of rotation is z) in a wire loop (the loop is in the y-z plan).

I want to examine the induced current at instant t when the magnet "crosses" the plan of the loop (that is to say : the moment seen on the picture above).

If I use the integral form of Faraday's law, I have an integrated flux that is positive at t-dt, and negative at t+dt. So, the variation of the flux is negative, and the induced current is positive. The induced magnetic field points in the +x direction (the induced magnetic field opposes the rotation of the permanent magnet, which is expected).

So far, so good.

Now, let's have a look at the magnetic field that is applied to the wire, to use local form of the equation. If we look from above, the magnetic field at t-dt points slightly to the left. At t+dt, it points slightly to the right (it's crucial to notice that the field lines aren't quite parallel, and that they curve out of the magnet axis faster that the magnet rotates). So, $$\partial B/\partial t$$ has a positive x component, which means that $$\nabla \times E$$ points to the left (negative x component)... which is in total contradiction with the prediction of the integral form.

We have the same problem if we look at the horizontal portions of the loop: the magnetic field points in the opposite direction of that of the magnet, due to the fact that the magnetic lines form a closed loop.

So, when the magnet rotates, the magnetic field variation in the horizontal wire points to the right, and the curl of the induced current points to the left (as well as the induced magnetic field).

You can drive the same analysis at any point of the wire : the induced magnetic field points to the left, because the magnetic field outside the magnet actually isn't aligned with that "inside" the magnet.

We see therefore that the integral form predicts something different from the local examination.

I know that the local analysis I made is wrong, because it predicts that the induced field helps the magnet to rotate (so, any slight pulse at the beginning would lead to a never-ending self-acceleration of the magnet, creating mechanical energy from nowhere). Yet, I don't see what the flaw is in my local analysis. I mean : the wire actually is subject to the local magnetic field, right? And this magnetic field variation points towards the positive x, so...

I don't get what I'm missing.

• Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! Dec 11, 2021 at 10:15
• How is the axis of rotation of the magnet oriented relative to the plane of the loop? Dec 11, 2021 at 14:48
• The two red dots are the cross section of the wire loop in the left hand pictures, the wire loop is a circle centered on the origin in the y-z plane, the magnet's rotational axis is the z axis for it's rotation in the x-y plane. Dec 11, 2021 at 21:07

First of all, you seem to have put some effort in your pictures, but it looks like the purpose was to make reading the question as hard as possible. For example, the magnet and the wire are drawn in different planes in the first two pictures. This is already an excellent source of confusion, but it is also unclear what is the x coordinate of the loop and what the following phrase means: "instant t when the magnet "crosses" the plan of the loop".

Second, the induced current in the second picture looks out of the plane of the wire. Is this an error?

Third, the induced magnetic field in the first picture is directed in the x direction. It is not clear for which point this is true and how this is relevant to the induced field in the loop.

So I would say what you are missing is clarity.

Haven't read all of this, But you confusion seems to stem from the orientation of the surface.

If there is a surface that the orientation is positive on the top.

If there is a magnetic flux through this that the field points upwards

Then da and B are in the same direction, so there is a positive flux. If this flux increases then dphi/dt is positive , so so the line integral of E .da would be negative ( as -positive) , the line integral in this manner is defined anticlockwise. meaning negative work is done going anticlockwise. aka the field then goes clockwise

The differential form. if db/dt is positive then the curl is negative, ( as -db/dt) if the curl is negative then the field goes clockwise relative to my orientation