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

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SUMMARY

The correct formula for calculating the angular speed of a generator involves using the peak electromotive force (emf) equation E(o) = NABw, where E(o) is the peak voltage, N is the number of loops, A is the area of the coil, and B is the magnetic field strength. In this case, with a coil of 655 loops, an area of 0.03908 m² (19.7 cm side), and a magnetic field of 0.490 T, the angular speed w can be calculated directly. The initial approach incorrectly used sin(2π), which equals zero, leading to an erroneous frequency calculation. The correct angular speed is derived from the equation without the sine function.

PREREQUISITES
  • Understanding of electromagnetism principles, specifically Faraday's law of electromagnetic induction.
  • Familiarity with the formula for peak emf in generators: E(o) = NABw.
  • Basic knowledge of trigonometry, particularly the properties of sine functions.
  • Ability to perform unit conversions, especially between revolutions per second and radians per second.
NEXT STEPS
  • Study the derivation of Faraday's law of electromagnetic induction.
  • Learn about the relationship between frequency and angular speed in rotational systems.
  • Explore the implications of using trigonometric functions in physics equations.
  • Investigate the effects of coil dimensions and magnetic field strength on generator output.
USEFUL FOR

Physics students, electrical engineers, and anyone involved in the design or analysis of electrical generators will benefit from this discussion.

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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