Basic Generator problem

1. Mar 10, 2004

mrbling

The question is:
A simple generator has a 655 loop square coil 19.7cm on a side. How fast must it turn in a .490 T field to produce a 110V peak output? give your answer in radians/second

I used V=NBAwsin(wt)
where V = 110, N=655, B=.490, A=.197*.197, w simplifies to 2*pi*f, wt simplifies to 2*pi
so plugging in, 110=655*.490*.197*.197*2*3.14*f*sin(6.28)

i get f = 12.85 rev/s * 360/57.3 = 80.76 rad/s

but this is wrong..

anyone know where i went wrong?
places i suspect are: wrong formula, and i am using the sin(6.28) incorrectly..

any hints?
thanks

2. Mar 10, 2004

AshleyF708

Well, sin(2*pi) is equal to zero using trig circles. There is one problem.

3. Mar 10, 2004

AshleyF708

Also the peak emf is E(o) not simply E so you can use the following equation:

E(o)* sin(wt) = NAB w sin(wt)= E

This simplifies to:

E(o) = NAB*w

Solve for w.