If you have drop a bar magnet through a coil so that it goes all the way through I was told the graph of emf induced in the coil vs time looks something like this:
(emf induced is on y axis, time is on x axis)
The area of the pink part and the blue part are equal (sorry that they don't look equal in my diagram)
I understand why the blue part looks more stretched than the pink part -- the magnet is accelerating. What I don't understand is the direction of the blue part -- it is opposite (negative) to the pink part (positive).
Why does the s-pole induce an emf in the opposite direction to the one induced by the n-pole when a bar magnet falls through a coil? The field lines at the s-pole and n-pole are pointing in the same direction, so the induced emf should continue increasing in the pink part (rather than turning and going below zero) because the field from the n-pole is complemented by the field from the s-pole which is cutting the flux in the same direction as the one from the n-pole.
Please explain in layman terms.
Thanks!
Note: Layman terms means you can use stuff from high school physics (left/right hand rules, lenz's law, etc.), but not e.g. calculus.
Answer
One of Maxwell's equations is $$ \nabla \times \vec{E} = - \frac{d\vec{B}}{dt} \, .$$ Consider an imaginary disk whose normal vector is parallel to the axis of the coil and which is inside the coil. If you integrate this equation over the area of that disk you get $^{[a]}$ $$\mathcal{E} = - \dot{\Phi}$$ where $\Phi$ is the flux threading the disk and $\mathcal{E}$ is the EMF drop around the loop. This is called Faraday's law.
So, for each imaginary disk inside your coil we get some EMF as the flux through that disk changes in time.
Now think about the bar magnet's descent. Suppose we drop it starting way above the entrance to the coil. It's far away, so there's no flux and no EMF. As it descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF. This is the initial rise in the red part of the diagram. As the magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves the time rate of change of total flux increases, so the EMF goes up. Note that the field lines above and below the bar magnet point in the same direction.
At some point, the bar reaches the middle of the coil. At this point, the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. This is the midopint of the diagram where the EMF crosses the horizontal axis. The falling part of the red section is just the approach to the mid section of the coil.
As the bar magnet exits the coil, more flux is leaving than is entering, so the EMF versus time in the blue section is just the opposite (except for the stretching which you already understand) of the red section.
[a]: On the right hand side, the area integral of the magnetic field is the flux $\Phi$ by definition, and the time derivative just goes along for the ride. On the left hand side you are doing an area integral of a curl of a vector, which by Stokes's theorem is equivalent to the line integral of the vector itself around the boundary of the area. The line integral of the electric field vector is the EMF by definition.
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