Calculating EMF from a Rotating Coil: Max & Avg Values

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SUMMARY

The discussion focuses on calculating the maximum and average electromotive force (emf) produced by a rotating coil in a dynamo with 500 loops and an area of 2.5x10-4 m2 in a uniform magnetic field of 0.400T. The coil rotates at a frequency of 8Hz, resulting in an angular frequency of 16π rad/s. The maximum emf calculated is approximately 2.51V, and the average emf is determined by dividing the peak value by √2, confirming the approach is correct.

PREREQUISITES
  • Understanding of Lenz's Law
  • Knowledge of electromagnetic induction principles
  • Familiarity with the formula for calculating emf: |ε| = dΦ/dt
  • Basic grasp of angular frequency and its calculation
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  • Explore the derivation of the emf formula in electromagnetic induction
  • Learn about the applications of Lenz's Law in real-world scenarios
  • Study the relationship between frequency and emf in rotating systems
  • Investigate methods for measuring emf in laboratory settings
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Physics students, electrical engineers, and anyone interested in the principles of electromagnetism and energy generation through rotating systems.

danago
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On a particular dynamo, the rotating coil has 500 loops, each of area 2.5x10-4 m2. The magnetic field can assumed to uniform and of magnitude 0.400T. When the bicycle is ridden at a particular speed, the coil rotates with a frequency of 8Hz. What is the maximum emf produced? What is the average emf produced?

Ok. According to lenz's law, the emf is equal to the time rate of change of magnetic flux.

<br /> \left| \varepsilon \right| = \frac{{d\Phi }}{{dt}} = \frac{d}{{dt}}(NAB\sin \theta ) = \frac{{d\theta }}{{dt}}NAB\cos \theta <br />

I am told that it rotates 8 times per second, which is means that the angular frequency is given by:

<br /> \frac{{d\theta }}{{dt}} = 16\pi <br />

Therefore, the EMF is given by:
<br /> \left| \varepsilon \right| = 0.8\pi \cos \theta <br />

The maximum value of cos\theta is 1, therefore the maximum EMF is ~2.51V. Is that correct? Also, how should i go about finding the average? Just divide it by \sqrt{2}?

Thanks in advance,
Dan.
 
Last edited:
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Looks right to me. yup divide the peak by sqrt(2).
 
Ah alright. Thanks for the quick reply :smile:
 

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