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Hat is the maximum value of the emf produced when the coil is spun at 1250 rev/min,?

  1. Nov 24, 2005 #1
    Hello everyone, i'm confused on how to do this problem, it says:
    An electric generator contains a coil of 120 turns of wire formed into a rectangular loop 40.0 cm by 20.0 cm, placed entirely in a uniform magnetic field with magnitude B = 3.50 T and with B initially perpendicular to the coil's plane. What is the maximum value of the emf produced when the coil is spun at 1250 rev/min about an axis perpendicular to B?
    ? V

    I tried the following:
    Area = (40)(20) = .08m^2;
    B = 3.50T <-- magnetic field
    Intially
    B Flux = BA = (3.50)(.08) = .28 Tm^2;
    spun = 1250 rev/min = 20.83 rev/sec;


    Emf = N* (change in B flux) / (change in time);
    N is the number of turns;
    So i tried: Emf = (120)*.28*20.83 = 699.888 which was wrong, any ideas? Thanks.
     
  2. jcsd
  3. Nov 24, 2005 #2

    daniel_i_l

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    In one whole rotation the B flux doesn't change at all. It does the maximum change in one fourth of a turn.
     
  4. Nov 24, 2005 #3
    Daniel, how did you figure that? ALso I'm alittle lost on how that helps me, Because i'm not sure how the 16.67 rev/sec i found app;iles to the problem, From the formula, it says number of turns, which is 100 * the change in the Magnetic flux, all over the change in time. I"m very confused on this, and all the other prolbems also tlak about rotation...any help would be great,thanks!!
     
  5. Nov 24, 2005 #4

    mezarashi

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    I'm not sure why it hasn't become obvious if you have the solution, but isn't the EMF suppose to be a sinusoidal?
     
  6. Nov 24, 2005 #5
    I'm sorry i'm having a hard time grasping this concept on how revlutions per second affect the whole value, Like it says the Change in BA, is my intial and final BA calcualted right or am I missing somthing? Also what is sinusoidal mean?
     
  7. Nov 24, 2005 #6

    mezarashi

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    Your application of Faraday's Law is correct. The trick is with the math. The EMF generated is a proportional to the change in flux over time isn't it. Let's look at flux. Flux is a product of both the field B and the area of the coil which is perpendicular to the field. Basically a shadow. I think that last statement may be confusing. But if you see, as the coil turns and it is 90 degrees with the field, then the effective area is zero right?

    So the relationship between the effective area and the original area is:

    [tex]A = A_o sin\theta[/tex]
    where theta is the angle the coils are making with the field.

    If you know that B is constant then surely:

    [tex]\frac{d\Phi_{flux}}{dt} = B\frac{dA}{dt}[/tex]
     
  8. Nov 24, 2005 #7
    Ahhh i understand the A = 0, so that means the final flux will be 0, so all you have is the initial flux, as you wrote: (B*dA)/dt but If i just keep it at that, it isn't right, Because i'm confused on dA, how is the area changing? It isn't changing at all, is it?
     
  9. Nov 24, 2005 #8

    mezarashi

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    [​IMG]

    A diagram I drew to help you visualize, as:

    [tex] \Phi_{flux} = B \times A = BAsin\theta[/tex]

    The B and A fields must be perpendicular for the flux to be there. If you have the A being sideways, not much of the field is going to be able to go through the loop right?

    Also, [tex]\theta = \omega t[/tex]
     
  10. Nov 25, 2005 #9

    daniel_i_l

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    This is what I meant, after a 90deg. turn the B flux will go from 0 (sin(0)=0) to the maximum BA (sin(90)=1).
    The area itself isn't changing, what is changing is the part of the area is in the direction of the B field. So if a is the angle, Asin(a) is the area is the equation.
     
  11. Nov 25, 2005 #10
    Thanks guys, i have no idea why i'm still not getting this problem, i understand what your saying and i also found an example in the book, What i'm trying to figure out is, They tell me "An electric generator contains a coil of 120 turns of wire formed into a rectangular loop 40.0 cm by 20.0 cm" So can i take this as meaning, 120 turns, being N in the equation?
    [tex] \Phi_{flux} = B \times A = BAsin\theta = BA = (3.50T)(.40*.20)*sin(90)= .28 Tm^2[/tex]
    So thats the intial flux, the final flux would be 0, becuase BAsin(0) is 0 as you guys told me. I know the thing is spinning at 20.83 rev/sec So now we have to deal with:
    (Flux Final - Flux INtial)/Change in time;
    (0 - .28)/20.83 = .01346;
    Finally:
    EMF = N*d(Flux)/dt = (120 turns)(.01346) = 1.6153 V, which is of course wrong :(
     
  12. Nov 26, 2005 #11
    Finally got it, my friend told me i had to convert my rev/sec into rad/sec, no wonder it was wrong! Thanks guys.
     
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