# Why (physically not mathematically) is the induced EMF of a rotating coil the negative derivative of the graph for flux?

The graphs for a rotating coil of the flux and then the induced EMF looks like this.

So I understand mathematically that the emf will be at a maximum when the gradient of the flux is at its biggest. Because of the equation:

However, I don't understand visually why this is the case. Why for instance (visually) is the change in flux so little when it is the coil is perpendicular to the field, why is it different to any other point in the rotation?

Then I don't understand why the current graph is just the negative of the flux graph? Surely the EMF causes the current so it must be tied to the emf graph not the flux graph?

I am just trying to make sense of the graphs intuitively not mathematically, so sorry if the question seems a bit weird. Any help much appreciated.

It comes from the fact that when the coil is perpendicular to the field, small movements of it lead to small changes in the flux. You can see visually that where the black curve ($$\Phi$$) has a maximum it is also very "flat", i.e. counter-intuitevely you have a big flux, but it changes very little in time. Basically, the coil is very "flat" in that position and moving it a bit does not change the flux a lot. On the other hand, when it is parallel, a small change in the coil changes the flux a lot.

I know you did not ask for this, but the best way to see it is math!

Assume the coil rotates with constant angular speed $$\omega$$ so that $$\theta(t)=\omega t$$ and thus, according to your table $$\Phi = ABcos(\theta(t))=ABcos(\omega t)$$

Actually, the way your graph is drawn, implies there also is a phase so that $$\Phi(0)=0$$. We are going to ignore it. Imagine you have the same graph but just shifted in $$t$$ so that the first time were the flux is maximum is $$t=0$$.

From the expression of the flux we cain extract the emf by a time derivative (i.e. $$\Delta\Phi / \Delta t$$ with $$\Delta t \sim 0$$) [if you don't know about derivatives go down for the visual explanation] $$\epsilon = - Nd\Phi / dt = NAB\omega sin(\omega t)$$ And from the properties of sine and cosine you get very easy that where the cosine is max (very high flux) the cosine is 0 (no emf, $$\epsilon=0$$ and vice versa)!

You can also see it visually on the shape of the curve: where the cosine (i.e. the flux) is max its curve is very flat, whereas where it crosses 0 the curve is very steep i.e. its change with time [i.e. the derivative] ($$\epsilon$$) is very high.

So physically, the reason you see that shift between $$\Phi$$ and $$\epsilon$$ is the fact that according to the position of the coil you get different instantaneous variations of the flux in time, i.e. $$\Delta \Phi / \Delta t$$ is not constant. This is true if the coil rotates with a constant angular velocity! If the coil rotates in another way you might get different graphs.

As far as the current $$I$$ is concerned, I am afraid your graph is wrong, as it should be in phase with the emf unless the coil contains some elements which are slowing it down.