How Does a Single-Phase Generator Calculate EMF?

[danago]

A simple single-phase generator has coils of 200 turns. The coil is 14cm long and 9cm wide. The magnetic field in the generator is 0.15T. The generator coil is turned at a rate of 3000rpm.

Calculate the emf produced by this generator.

$$\begin{array}{c} \\<br /> \frac{{d\theta }}{{dt}} = 3000rpm = 100\pi {\rm{ rad/sec}} \\<br /> \varepsilon = - N\frac{d}{{dt}}(AB\cos \theta ) \\<br /> = \frac{{d\theta }}{{dt}}NAB\sin \theta \\<br /> = (100\pi )(200)(0.09)(0.14)(0.15)\sin (100\pi t) \\<br /> = 37.8\pi \sin (100\pi t) \\<br /> \end{array}$$

Now it asks for 'the emf', which is a bit hard to give as a numerical answer, since the emf varies with time. I looked at the answer, and they give an average emf.

So i proceeded as follows:

$$\varepsilon _{{\rm{av}}} = \frac{{\varepsilon _{\max } }}{{\sqrt 2 }} = \frac{{37.8\pi }}{{\sqrt 2 }} = 83.97V$$

Now the book does it differently. They find the change in flux over a 90 degree turn (and i think are assuming that the rate of change of flux is constant), and then find the change in time.

$$\begin{array}{c}<br /> \Phi _B = AB\cos \theta \\ <br /> \Phi _i = 0.00189\cos \frac{\pi }{2} = 0 \\ <br /> \Phi _f = 0.00189\cos 0 = 0.00189 \\ <br /> \Delta \Phi _B = 0.00189 \\ <br /> \Delta t = 0.005 \\ <br /> \varepsilon = - N\frac{{\Delta \Phi _B }}{{\Delta t}} = 75.6V \\ <br /> \end{array}$$


Which way is the correct way?

Thanks in advance,
Dan.

 

[danago]

Ahh everything seems to be contradicting everything else.

My teacher gave us the formula:
$$\begin{array}{l}<br /> \varepsilon(t) = 2\pi fNBA\sin (2\pi ft) \\ <br /> \therefore \varepsilon _{\max } = 2\pi fNBA \\ <br /> \end{array}$$

For a coil of N turns with cross sectional area A rotating through a magnetic field of flux density B, with a rotational frequency of f. Now i completely agree with this formula, and am able to derive it using some basic calculus and faradays law. However, another textbook uses another method, which gives a different answer.