Faraday's Law

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Homework Statement



A square coil (25 x 25 cm) that consists of 50 turns of wire rotates about a vertical axis at 1000 revolutions per minute. The horizontal component of the Earth's magnetic field at the location of the coils is 2 x 10-5. Calculate the maximum voltage induced in the coil by this field.

Homework Equations



[tex]\epsilon = - \frac{d \Phi_B}{dt}[/tex]

The Attempt at a Solution



I know that I should use Faraday's law

[tex]| \epsilon | = N \frac{d \Phi_B}{dt} = N \frac{BA}{dt} = 50 \frac{(2 \times 10^{-5}) \times 0.0625}{dt}[/tex]

How do I continue from here? How do I obtain "t" from 1000 revolutions per minute? The correct answer must be 6.54 mV... :confused:
 

Answers and Replies

  • #2
Doc Al
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Hint: The flux is not simply BA. (That would be the flux at the instant that the field is exactly perpendicular to the coil.) How does the flux depend on the angle between the field and the coil? How does that angle depend on the time?
 
  • #3
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Hint: The flux is not simply BA. (That would be the flux at the instant that the field is exactly perpendicular to the coil.) How does the flux depend on the angle between the field and the coil? How does that angle depend on the time?
I don't know the relationship between the angle and time. I think at an angle the flux would be [tex]\Phi_B = BA cos \theta[/tex]. So [tex]\epsilon = -\frac{d}{dt} (BA cos \theta)[/tex]? But we don't know what the angle is exactly and if we assume it is 90, then cos 90=0. :frown:
 
  • #4
Doc Al
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I don't know the relationship between the angle and time.
Assume that the rotation rate--which is given--is constant. ω = dθ/dt.

I think at an angle the flux would be [tex]\Phi_B = BA cos \theta[/tex].
Good.

So [tex]\epsilon = -\frac{d}{dt} (BA cos \theta)[/tex]? But we don't know what the angle is exactly and if we assume it is 90, then cos 90=0.
First find the derivative, then find its maximum.
 
  • #5
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First find the derivative, then find its maximum.
The derivative is [tex]-BA \sin \theta[/tex]. And the maximum for sine is 1, so [tex]-(2 \times 10^{-5})\times(0.0625) \sin (1) = 2.18 \times 10^{-8}[/tex]. But why this is still not right?
 
  • #6
Doc Al
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The derivative is [tex]-BA \sin \theta[/tex].
Almost, but not quite. You're taking the derivative with respect to time, not just θ.
 

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