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Simple induction problem

  1. Mar 21, 2004 #1
    Hiyas, I'm having some problems with this problem, its relatively simple, but I just can't get it right for some odd reason..

    The picture (my awesome paint skills), shows two circular regions R1 and R2 with radii r1 = 22.0 cm and r2= 34.0 cm. In R1 there is a uniform magnetic field B1 = 50 mT into the page and in R2 there is a uniform magnetic field B2 = 75 mT out of the page (ignore any fringing of these fields). Both fields are decreasing at the rate of 7.60 mT/s.

    Calculate the integral (E * ds) for each of the three paths.

    I've got the first 2, Path 1 and Path 2 by Faraday's law (-A db/dt), but I don't know how to do Path 3.

    Any help here would be appreciated. Excuse the poor quality of the picture :)

    Attached Files:

  2. jcsd
  3. Mar 22, 2004 #2
    Let [itex]V_B[/itex] be the rate of decrease of the magnetic fields ([itex]\frac{dB}{dt}[/itex]).

    For the 3rd path:
    [tex]\oint E\cdot ds = -\frac{d\phi _B}{dt} = -\frac{{d\phi _B}_1 + {d\phi _B}_2}{dt}[/tex]

    [tex]\phi _B_{(t)} = A_{(t)}B_{(t)}[/tex]
    The area is constant, it's only the magnetic field that's changing:
    [tex]\phi _B_{(t)} = \pi R^2(B_0 - V_Bt)[/tex]

    Since B1 and B2 are in opposite directions, give one of them a minus sign:
    [tex]{\phi _B}_1 + {\phi _B}_2 = \pi R_1^2(B_0 - V_Bt) - \pi R_2^2(B_0 - V_Bt) = \pi (B_0 - V_Bt)(R_1^2 - R_2^2)[/tex]
    Take the derivative of that:
    [tex]\frac{{d\phi _B}_1 + {d\phi _B}_2}{dt} = -\pi V_B(R_1^2 - R_2^2)[/tex]

    And therefore:
    [tex]\oint E\cdot ds = \pi V_B(R_1^2 - R_2^2)[/tex]
    Last edited: Mar 22, 2004
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