Why is induced electric current moving clockwise/counterclockwise?

I know that if I want to identify north pole and south pole of a magnet I can create an electric circuit using a battery (an EMF) and a copper wire. I create a solenoid of N turns, and move a magnet in and out the solenoid. What I know is the following principle:

• if I'm moving magnet and the north pole goes in and out the solenoid, then the induced electric current is moving counterclockwise,

• if I'm moving magnet and the south pole goes in and out the solenoid, then the induced electric current is moving clockwise instead.

I don't know how to prove this principle. I know Faraday-Lenz law is involved, but why is the induced electric current moving counterclockwise/clockwise?

• How can you move only North pole or only South pole? Commented Jun 28, 2022 at 22:06
• @ClaudioSaspinski I'm taking magnet in the south pole and moving north pole in and out the solenoid. It's a famous experiment about Faraday-Lenz law in which you generate an induced electric current by holding a magnet on one side, and the other side generate an induced electric current because you're moving the magnet in and out the solenoid Commented Jun 28, 2022 at 22:09
• @ClaudioSaspinski okay, I fixed the original question, now I think it's better Commented Jun 28, 2022 at 22:26
• Are you mathematically inclined? Are you able to mathematically deduce the reasoning using Faraday-Lenz law? Or are you looking for some other reasoning how this is true? Commented Jun 28, 2022 at 22:39

Consider the Maxwell equation equivalent to Faraday's law:

$$$$\boldsymbol{\nabla} \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$$$

If you perform a surface integral of the terms on both sides you get:

$$$$\begin{split} &\iint _{S} (\boldsymbol{\nabla} \times \mathbf{E}) \cdot \mathbf{dS} = \oint _{\partial S} \mathbf{E} \cdot \mathbf{dl} \\ -&\iint _{S} \frac{\partial \mathbf{B}}{\partial t} \cdot \mathbf{dS} = -\frac{\partial}{\partial t} \iint _{S} \mathbf{B} \cdot \mathbf{dS} = -\frac{\partial \Phi}{\partial t} \end{split}$$$$

The electric field $$\mathbf{E}$$ will have the same direction as the current density $$\mathbf{J}$$ and thus the "direction" of the electric current $$I$$ on a 1-dimensional path (essentially either clockwise or counter-clockwise on the coil).

If you move the north pole of the magnet towards the coil or the south pole away, the direction of the magnetic field is the same as the normal vector of the surface of which the solenoid coils (i.e. your 1-dimensional paths), meaning the inner product is positive and the change in magnetic flux is also positive. Due to Faraday's law, the integral of the electric field on the coil must be negative, meaning the electric field (and thus the current) are in the opposite direction of the one you chose as positive for the coil i.e. clockwise. Hence for north pole approaching or a south pole moving away, the current runs counter-clockwise.

When the north pole is moving further away or the south pole is moving closer, the opposite is true: the change in flux is negative, thus the line integral of the electric field is positive. By the same logic as above, the current then runs clockwise.