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The induced emf in a coil in AC generator is given as:

$$\mathbb E = NAB\omega \sin \theta $$

$\omega = d\theta/dt$

Now, when the angle between the normal of plane and magnetic field is zero degrees, the induced emf is zero i.e.

$$\Bbb E = NAB\omega \sin 0 = 0$$

But we also define emf as the time rate of change of magnetic flux so, why do we get zero emf in the above case, magnetic flux is still changing with time?

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3 Answers 3

The magnetic flux as a function of time is $$ \Phi(t) = \mathbf B \cdot \mathbf A(t) = BA\cos(\omega t) $$ where $\mathbf B$ is the magnetic field and $\mathbf A(t)$ is the area vector as a function of time and $\omega t$ is the angle between the field and the area vector as a function of time. Then the rate of change of the flux as a function of time is $$ \Phi'(t) = -BA\omega\sin(\omega t) $$ Notice that if the angle between the area vector vector and the magnetic field is zero, then the flux is nonzero and equal to $AB$, its maximum, but the rate of change of the flux vanishes because $\sin(0) = 0$.

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You said "why do we get zero emf in the above case, magnetic flux is still changing with time." I showed that when the angle between the area vector and the magnetic field is zero, the magnetic flux is not changing with time. –  joshphysics Mar 6 '13 at 22:42
    
But the coil inside the generator is continuously rotating so shouldn't there be a change in magnetic flux all the time which will result in some induced emf all the time? –  Rafique Mar 6 '13 at 22:47
    
No. If you look at the calculation, you'll see that at the very moment when the magnetic field vector and area vector are aligned, the rate of change of the magnetic flux is zero. The derivative of the flux with respect to time vanishes at that particular instant. –  joshphysics Mar 6 '13 at 22:50
    
Yes, i can see that mathematically but i am asking physicaly. –  Rafique Mar 6 '13 at 22:52
    
How about this, the flux is a periodic function with maxima and minima. At every point where the flux is either a maximum of minimum, its rate of change is, by definition (since local maxima and minima are defined as points with vanishing derivative), zero. So by virtue of the fact that the flux is maximized when the field and area vector are aligned, we can see that its rate of change must vanish. –  joshphysics Mar 6 '13 at 22:57
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Muhammad Rafique the induced EMF in a AC generator also depend's on the rate change of magnetic flux linked with the coil, that is $\Delta\phi =\phi(f)$ - $\phi(i)$ and rate change of magnetic flux $\frac{\Delta\phi}{\Delta(t)}$ = $\frac{BASin(\theta(f)- \theta(i))}{\Delta(t)}$ so it can go zero for some values of $\theta$ but we are taking an average value in this case but the induced $EMF$ in an AC generator which generate alternating current which mean's it change's it's value from positive to negative in which it also become's zero we cannot take an average value for a complete cycle because it will result in zero EMF so we take the average for a half cycle. Hope it cleared your doubt if not you can notify me

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Not sure if this helps any future readers or not, but it helped me a lot. http://macao.communications.museum/eng/exhibition/secondfloor/moreinfo/2_4_1_ACGenerator.html

enter image description here

The graph shows how the area of the coil changed over time with respect to the angle of rotation. When the area is at 0, the gradient is steepest. Emf is directly proportional to the rate if change in flux, the field is constant, therefore the flux is proportional to the change in area. So at the moment when the coil's velocity is perpendicular to the field, the rate of change in flux is at a maximum and when the velocity is parallel the derivative is at a minimum.

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