When studying electromagnetism, I came across this equation, and made some mathematical manipulations

Faraday's Law Of Electromagnetic Induction: $ \varepsilon = -N\frac{\mathrm{d}\phi }{\mathrm{d} t} $

Here, $Φ$ should be the dot product of the area of coil with the magnetic field: $\phi = ab \cos\theta $

In my model, area of coil will not be changing, and neither will the strength of the magnetic field. Therefore, the equation I have come up with is: $$\varepsilon = -NAB \frac{\mathrm{d} \cos\theta }{\mathrm{d} t}.$$ Is this equation correct? I also wanted to ask if $\frac{\mathrm{d} \cos\theta }{\mathrm{d} t}$ can be written as $$\frac{\mathrm{d}\cos\theta }{\mathrm{d} \theta }\frac{\mathrm{d} \theta }{\mathrm{d} t},$$ and since $\frac{\mathrm{d}\cos\theta }{\mathrm{d} \theta }$ is $-\sin \theta$, the final equation be:

$$\varepsilon = NAB \sin\theta\frac{\mathrm{d} \theta }{\mathrm{d} t}.$$

Now, I am confused how to translate this into average emf produced, as in suppose, 1 second. I saw these two stack exchange answers, and couldn't quite understand what they meant. They also seem to contradict each other

Average induced emf in a rotating coil after rotating by 180 degrees

Will there be any induced EMF?

So how would one go on about translating this above equation, which has derivatives, and assumes instantaneous rate, to an equation that gave average emf produced, per rotation, ie $\theta$ goes from 0 to 2$\pi$.

  • $\begingroup$ That's completely correct. We usually denote $\frac{d\theta}{dt}$ by $\omega$. It is the so-called angular speed (or angular velocity) of the coil. [Indeed, we might put $\theta=\omega t$ right at the start, and if $\omega$ is constant your final result emerges from the differentiation wrt $t$. But there's nothing at all wrong with your method!] $\endgroup$ Commented Nov 3, 2023 at 16:54
  • $\begingroup$ I see no reasoning errors here. Yes, differentiation variable replacement is perfectly valid step. Note, that last term is angular speed of magnetic field rotation, so $\varepsilon = Nab \omega \sin \theta$. If you want to experiment with EM induction for a very small angles (which probably results in weak EM too),- like $|\sin \theta| \ll 1$, then you can even reduce formula to $\varepsilon = Nab \omega \theta$, because for tiny angles $\sin \theta \approx \theta$. $\endgroup$ Commented Nov 3, 2023 at 16:57
  • $\begingroup$ Thank you, that helps a lot. I have heard about sin theta = theta approximations before $\endgroup$ Commented Nov 3, 2023 at 16:59
  • $\begingroup$ re your supplementary questions... Did you spot that the two Stack Exchange questions you cited dealt with different cases of a 180° turn: in one case, (a), from the coil axis being parallel to the field to its pointing in the opposite direction and (b) from its axis being at right angles to the field to its being at right angles to the field but pointing in the opposite direction. In each case the mean emf over the half turn is $(-)N \Delta \phi/\Delta t$ in which $\Delta t =T/2$. In (a) $\Delta \Phi=BA-(-BA)=2BA$. In (b) $\Delta \Phi=0-0=0$. Closure of question prevents a fuller answer. $\endgroup$ Commented Nov 4, 2023 at 17:56