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Faraday's Law and coil EMF

  1. Nov 13, 2004 #1
    A closely wound, retangular coil of 80 turns has dimensions of .25m x .4 m. The plane of the coil is rotated, in .06 sec from a position where it makes an angle of 37 degrees with a magnetic field of 1.1T to a position perpendicular to the field. What is the average emf induced in the coil?

    Here's what I came up with:

    In .06 sec, the coil moves 53 degrees(90 - 37). So, I converted this to raidians, which is .925025 rad. It moves this far in .06 sec, so it's rotational speed is 15.42 rad/sec.

    [tex] |E| = NwBA|Sin Wt|[/tex]

    My book states that to find the average emf, replace Sin(wt) with the average value. They did this and came up with [tex]2/\pi[/tex]

    So, the average value of [tex] E = 2NwBA/\pi[/tex]

    Plugging into the equation, I get 86.39Volts, far from the 58.4 Volts answer. Can anyone tell me what I'm doing wrong? Thanks.
  2. jcsd
  3. Nov 13, 2004 #2

    Doc Al

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    Staff: Mentor

    How are they defining the angle of the coil with respect to the field? By the normal to the plane or by the "plane of the coil"? I suspect that when they say the "plane of the coil" makes an angle of 37 degrees to the field, then the normal makes an angle of -53 degrees. And it's the angle the normal makes that you need to use in applying Faraday's law. (Otherwise you'll mix up sines and cosines.)
    No problem.

    The book is giving the average value from 0 to pi radians; but your problem is over a different range of angles, so the average is different.

    You need to find the average value of [itex]sin(\omega t)[/itex] over the range of angles given in this problem. (Hint: integrate.)
  4. Nov 13, 2004 #3
    I had a feeling that I had to integrate, but for some reason I just blew it off.

    Ok, So, I need to integrate over the range of angles, which is 37 to 90, or [tex]\frac{37*\pi}{180*w}[/tex] to [tex]\frac{\pi}{2*w}[/tex]. I divided them by 'w' to get the proper period for each. Is this right? I saw what the book did and I'm just thinking that's how they got [tex]\frac{\pi}{w}[/tex] for their ending range.

    So, I integrated that and then divided it by the time interval, [tex]\frac{\pi}{2*w}[/tex] to get the average. Finally, I plugged it back into the original equation and got 68.99 Volts. It's closer, but not it. The only thing I can think of is if I did the limits of integration wrong, but I don't see what I'm doing wrong.
  5. Nov 14, 2004 #4

    Doc Al

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    To find the average value of [itex]sin(\omega t)[/itex], you need to integrate [itex]sin(\omega t)[/itex] over the proper range of angles ([itex]\omega t[/itex] varies from -53 degrees to 0) and then divide by the range of angles (53 degrees measured in radians).
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