Calculating the accurate emf of an electric generator

Let's consider the following simple model of an electric generator: Many textbook say the following:

The magnetic flux through the coil is $$\Phi=BA\sin \theta$$, where $$\theta=\omega t+\phi$$ is the angle the plane on which the coil lies makes with the upward vertical. $$A$$ denotes the area bounded by the coil, and $$B$$ the magnetic field strength. The emf induced, according to Farady's Law, is therefore $$\mathcal E=\frac{d\Phi}{dt}=\omega BA \cos (\omega t+\phi)$$.

This is, or course, under the assumption that the magnetic field is uniform and constant. However, the current in the coil produces some magnetic field, according to Ampere's Law, so the magnetic field should not be constant in general.

Question: Is the emf $$\mathcal E$$ still a sine wave if the change in magnetic field caused by current in the coil is taken into account? Of course, the answer depends on the type of load the generator is connected to, but now let's assume the load is a resistor. How will the waveform be different form $$\mathcal E=\omega BA \cos (\omega t+\phi)$$? Can I find an analytical expression for $$\mathcal E$$ in this case?

• Consider the limit where you short the two output wires together (giving the maximum loop current). Now what's $\mathcal E$? – The Photon Aug 6 at 15:34
• Epsilon here is electromotive force for a circuit, it is not appropriate to call it potential. – Ján Lalinský Aug 6 at 22:37
• @ThePhoton Is it zero in that case? – Ma Joad Aug 6 at 22:41

The effect that you are worried about is self-induction. Given circuit (including the coil and the rest of wires) can be assigned quantity called self-inductance, denoted usually as $$L$$. The meaning of self-inductance is that magnetic flux through the circuit due to its own current is
$$\Phi_{self} = LI.$$
$$emf = -\frac{\Phi_{external}+\Phi_{self}}{dt},$$ $$emf = - \frac{d}{dt}\left(BA\sin (\omega t +\phi) + LI\right),$$ $$emf = -\omega BA\cos (\omega t +\phi) - L\frac{dI}{dt}.$$