1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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! :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Induced Current Problem Date
Problem on induced EMF Jan 24, 2015
Induced Current problem from review Dec 11, 2014
Induced Current GRE Problem Sep 5, 2013
Magnetic Field Induced Current Problem Mar 9, 2012
Induced Current Problem Oct 28, 2008