I'm a little bit confused with this problem: Find the induced current in a rotating coil with resistance R and inductance L. What I tried is this (assuming N=1): $$\epsilon =-\frac{\text{d}\phi}{\text{d}t}=-L\frac{\text{d}I}{\text{d}t} \tag{1}$$ Since $\phi =AB\cos\omega t$: $$\epsilon=-AB\omega \sin\omega t$$ $$I=\frac{AB\omega \sin\omega t}{R} \tag{2}$$ So far, so good. But I don't understand why if we work with the second relation of equation (1), we get: $$-\frac{\text{d}\phi}{\text{d}t}=-L\frac{\text{d}I}{\text{d}t}\Rightarrow \frac{\text{d}}{\text{d}t}(LI-\phi)=0\Rightarrow LI=\phi $$ $$I=\frac{AB\cos\omega t}{L} \tag{3}$$ Isn't a contradiction there? A maximum in equation (2) implies a minimum in equation (3). I know that the correct answer is the equation (2), but I can't see anything wrong in the derivation of equation (3).
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The derivation is not wrong, but there is a little issue with your premises: When using $\phi = L I$ you assume, that $I$ is a current through a conductor loop which creates the flux $\phi$. This is not the case here, because you have an outer magnetic field $B$. The magnetic field created by the induced current, on the other hand, is not of interest for your problem, but it will indeed be phase shifted in comparison with the current, like your equations (2) and (3) suggest.
