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EMF and magnetic flux

  1. Apr 1, 2010 #1
    Here is how the question is stated exactly:

    A circular conducting loop is held fixed in a uniform magnetic field thatt varies in time according to B(t) = B0exp(-at) where t is in s, a is in s-1 and B is the field strength in T at t=0. At t=0, the emf induced in the loop is 0.0758 V. At t= 1.76 s, the emf is 0.0192 V. Find a. Answer in units of s-1



    2. Relevant equations
    emf= [tex]\frac{d\Phi}{dt}[/tex]


    3. The attempt at a solution

    I'm quite lost on this problem! I tried plugging in 0.0192 = B(t)/ 1.76 and solve for a, but obviously that did not produce the correct answer. Any help would be appreciated!
     
  2. jcsd
  3. Apr 1, 2010 #2

    gabbagabbahey

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    Hint: How is magnetic flux related to magnetic field?
     
  4. Apr 1, 2010 #3
    The magnetic flux is a way of counting the total magnetic field lines. A changing magnetic field (given by the B(t)) will produce the emf's given.
     
  5. Apr 1, 2010 #4

    gabbagabbahey

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    No it isn't (thinking of it in this way allows you to visualize things, but the total number of field lines is infinite for all non-zero fields)

    There is an equation that tells you how to determine the magnetic flux through a surface from the magnetic field (it involves an integral)...what is that equation?
     
  6. Apr 1, 2010 #5
    [tex]\int[/tex]BdA.......B(dot)dA = BAcos[tex]\theta[/tex]
     
  7. Apr 1, 2010 #6

    gabbagabbahey

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    Right, when the magnetic field is uniform (such as in this case), the flux is given by [itex]\Phi=\textbf{B}\cdot\textbf{A}=BA\cos\theta[/itex]

    What is the surface area [itex]A[/itex] in this case? What does that tell you the EMF is?
     
  8. Apr 1, 2010 #7
    I am not sure I understand because I don't see any values given in the problem for the dimensions of this loop. Do you mean A = [tex]\frac{\phi}{B}[/tex]?
     
  9. Apr 1, 2010 #8

    gabbagabbahey

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    Sorry, you aren't given [itex]A[/itex], and you don't need to find it.

    You have two equations, [itex]\text{EMF}=-\frac{d\Phi}{dt}[/itex] and [itex]\Phi=BA\cos\theta[/itex]...put them together...what do you get?
     
  10. Apr 1, 2010 #9

    gabbagabbahey

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    Sorry, you aren't given [itex]A[/itex], and you don't need to find it.

    You have two equations, [itex]\text{EMF}=-\frac{d\Phi}{dt}[/itex] and [itex]\Phi=BA\cos\theta[/itex]...put them together...what do you get?
     
  11. Apr 1, 2010 #10
    EMF= -[tex]\frac{B_{0}exp(-at)A}{dt}[/tex] ? Or take the derivative of BAcos[tex]\phi[/tex] with respect to time and set that equal to EMF? (Thank you for your help thus far though)
     
  12. Apr 2, 2010 #11
    I'm thinking I have to use the emf at time t=0 to find the initial B at time t=0.
     
  13. Apr 2, 2010 #12

    gabbagabbahey

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    When you substitute your equation for [itex]\Phi[/itex] into the emf equation, you get

    [tex]\text{EMF}=-\frac{d}{dt}\left(B_0e^{-at}A\cos\theta\right)[/tex]

    Do [itex]A[/itex], [itex]\cos\theta[/itex] or [itex]B_0[/itex] depend on time? Why or why not?

    What does that make the derivative?
     
  14. Apr 4, 2010 #13
    The only one that would be dependent on time would be B[tex]_{0}[/tex] because the loop is held fixed and the angle does not change.
     
  15. Apr 4, 2010 #14

    gabbagabbahey

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    [itex]B_0[/itex] is a constant. The only thing that varies with time is the exponential, so

    [tex]\text{EMF}=-\frac{d}{dt}\left(B_0e^{-at}A\cos\theta\right)=-B_0A\cos\theta\frac{d}{dt}\left(e^{-at}\right)[/tex]

    right?

    Carry out the derivative and then plug in the information you are given.
     
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