What is the Correct Formula for Calculating Speed in a Basic Generator?

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To calculate the speed of a generator with a 655-loop square coil measuring 19.7 cm on each side in a 0.490 T magnetic field to achieve a 110V peak output, the correct formula is E(o) = NABw, where E(o) is the peak electromotive force (emf). The initial approach incorrectly used sin(6.28), which equals zero, leading to an erroneous result. The area A of the coil should be calculated as A = (0.197 m)^2. By substituting the values into the corrected formula, the angular speed w can be accurately determined. The discussion emphasizes the importance of using the correct peak emf notation and avoiding mistakes in trigonometric calculations.
mrbling
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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
 
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Well, sin(2*pi) is equal to zero using trig circles. There is one problem.
 
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.
 
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