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Induced emf: Faraday's Law and Lenz's Law Question

  1. Feb 5, 2008 #1
    [SOLVED] Induced emf: Faraday's Law and Lenz's Law Question

    The plane of a conductive loop with an area of 0.020 m^2 is perpendicular to a uniform magnetic field of 0.30 T. If the field drops to zero in 0.0045 s, what is the magnitude of the average emf induced in the loop?






    Given that,
    Area (a) = 0.020m2
    Magnetic filed (B) = 0.30 T.
    When the field drops to zero,
    Time taken = 0.0045 s
    I have,
    Induced emf = Change in flux
    Time taken
    = B x a cos 90
    Time taken
    = 0.30 x 0.020 x 1
    0.0045
    = 1.33 V


    Is my work/answer correct?
     
  2. jcsd
  3. Feb 5, 2008 #2

    rock.freak667

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    Homework Helper

    That should be correct.
     
  4. Feb 5, 2008 #3
    Is it really correct...or are you just telling me that?
     
  5. Feb 5, 2008 #4

    rock.freak667

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    Homework Helper

    I calculated it the same way as you and got the same answer

    [tex]E=\frac{BA}{t}[/tex]
     
  6. Feb 5, 2008 #5
    Ok...thanks. I just can't believe that I did it right. :)
     
  7. Feb 5, 2008 #6
    [tex]\Phi = \iint \mathbf{B} \cdot d\mathbf{a}[/tex]

    which in a constant magnetic field perpindicular to a constant area will reduce to

    [tex] \Phi = BA[/tex]

    Then you know that

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

    So the emf would be (assuming a constant change in the magnetic field)

    [tex]\epsilon = \frac{\Delta \Phi}{\Delta t}[/tex]

    and further shows that

    [tex]\epsilon = \frac{A * (B_i - B_f)}{(t_i - t_f)}[/tex]

    where B_f and t_f are zero.

    With numbers

    [tex] \epsilon = \frac{.02*.3}{.0045}\frac{mB}{s^2} = 1.33V[/tex]

    You need some more confidence in yourself.

    Edit:
    Oh right, what about Lenz's law?
     
    Last edited: Feb 5, 2008
  8. Feb 5, 2008 #7
    When it comes to physics I get so confused, that's why I'm on here. lol But thanks for your help.
     
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