How is this the graph of induced emf against angle for a rotating coil in a magnetic field?

Question

So a coil is rotating in a magnetic field, and at $\theta=0$ the coil is perpendicular to the field lines. At $\theta=90$ the coil is parallel to the field lines.

The the angle, theta, is the angle between the normal to the coil and the direction of the magnetic field lines.

This implies that the magnetic flux induced in the coil will be at a maximum when $\theta=0$ and 0 when $\theta=90$.

The question I was asked was to sketch a graph of induced emf, $\epsilon$, against $\theta$. Since $\epsilon=\dfrac{\Delta\phi}{\Delta t}$, where $\phi$ is the magnetic flux linkage, the induced emf should be at a maximum when the magnetic flux is a maximum. So at $\theta=0$, $\epsilon$=maximum and at $\theta=90$, $\epsilon=0$.

This means that the curve sketched should look like a cosine graph, starting at the maximum value for emf induced.

However, in the mark scheme for this exam question the correct answer was a sine graph. It says that at $\theta=0$, the emf induced is zero. How can this make sense? Is my logic wrong?

Answer 1

As you have written, $$ \epsilon = {\Delta \phi \over \Delta t}$$ This equation says that $ \epsilon $ is greatest when the change in flux with respect to time is greatest, not when the flux itself is greatest. In order to find when the change in flux is the greatest, you need to come up with an equation for the flux, then take the derivative with respect to time. In this case the flux equation looks like: $$ \phi (\theta)=BAcos(\theta)$$ So: $$ \phi'(\theta)=-BAsin(\theta){d\theta \over dt }$$ In this case, I assume that $ {d\theta \over dt}$, namely the angular velocity, is constant, though it isn't specifically stated in your question. Under this assumption, you can find from this equation where $ \epsilon$ is maximized.