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Induced Current Problem(Electromagnetism)

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A small, 1.60-mm-diameter circular loop with R = 1.40×10−2Ω is at the center of a large 120-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from 1 A to -1 A in 8.00×10−2 s.

    What is the induced current in the inner loop?


    2. Relevant equations
    I = ε/R
    d[itex]\Phi[/itex]/dt = ε
    [itex]\Phi[/itex] = BAcos(0) = BA
    B = (μ_0)(I)/2r

    3. The attempt at a solution

    First, I got (μ_0)(I)([itex]\pi[/itex])(r)(1/2) = [itex]\Phi[/itex], using [itex]\pi[/itex]r^2 as A.

    Then, d[itex]\Phi[/itex]/dt = (μ_0)([itex]\pi[/itex])(r)(1/2)(dI/dt)

    So...I = ((μ_0)([itex]\pi[/itex])(r)(1/2)(dI/dt))/R

    Which is ((4*10^(-7))([itex]\pi[/itex])^2(.03)(2/.08))/.014.

    My answer I got was 2.11 *10^-4 A.


    Please help, thank you. :)
     
    Last edited: May 22, 2013
  2. jcsd
  3. May 22, 2013 #2
    It looks to me like you did not consider the smaller area of the inside loop.
     
  4. May 22, 2013 #3
    I think you need to multiply your answer by the area of the small loop and you will get what I did.
     
  5. May 22, 2013 #4

    rude man

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    Homework Helper
    Gold Member

    Where did this come from?
    You need to compute B in the middle of the large loop. Computing flux thruout the area of the larger loop is both nearly impossible and pointless. Your small loop sees only the B field near the center of the large loop, and fortunately this is easy to calculate using Biot-Savart law.

    EDIT: as barryj pointed out, you already had the correct value for B in your equations list. So use it!
     
    Last edited: May 22, 2013
  6. May 22, 2013 #5
    I think he did compute B in the center . It seems he did not consider the smaller area of the inner loop.
     
    Last edited: May 22, 2013
  7. May 22, 2013 #6
    I think this is incorrect "First, I got (μ_0)(I)(π)(r)(1/2) = Φ, using πr^2 as A"
    You have to consider the R of the big loop and the r of the small loop.

    Big R = .06m and little r = .0008m

    Phi = uI/2R X (pi)r^2
     
  8. May 22, 2013 #7
    Ah, I see. So what I was doing was calculated the flux due to the magnetic field of the big loop, but through the big loop itself. Instead, I need to calculate the flux due to the magnetic field of the big loop, but through the small loop.

    ...Good news, that worked! Thanks guys, I get what I was doing wrong! :)
     
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