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Homework Help: Expanding Loop: Induced EMF?

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Imagine a pliable round metal loop that can expand or contract. In a region with a constant magnetic field B0 that is oriented perpendicular to the plane of the loop, suppose that the loop expands, with its radius growing with time as r = r0(1+at2). As the loop expands and grows thinner, its resistance per unit length changes according to R = R0(1+bt2). Find an expression for the current induced in the loop as a function of time. To check your answer, suppose that B0 = 3.30 mT, r0 = 16.5 cm, R0 = 7.55Ohm/m, a = 0.245 x 10-4 s2, and b = 0.590 x 10-2 s2. What is the value of the induced current at t = 27.5 s? (Note: Give the direction of the current where when viewed from above a positive current will move counterclockwise.)

    2. Relevant equations

    V = IR
    EMF = d(flux)/dt
    flux = Sb*dA (integral of dot product of B and dA, or |B||dA|cos(90) in this case --> BdA)

    3. The attempt at a solution

    First, find EMF induced and then use V = IR to solve for I. This is the equation I came up with:

    I(t) = B∏(ro(1+at^2))^2/(Ro(1+bt^2) = 7E-6 A

    This answer wasn't correct in my online homework program. The negative value was also incorrect. Then, I reread the question and saw the part about the equation for resistance being per unit length, so I got this equation (multiply above by circumference):

    I(t) = 2B∏^2(ro(1+at^2))^3/(Ro(1+bt^2) = 7E-6 A

    It didn't change my answer. Which equation is right (if either)?
    Last edited: Apr 15, 2012
  2. jcsd
  3. Apr 15, 2012 #2


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    Neither one is correct. What is the induced EMF in the loop?
  4. Apr 15, 2012 #3
    Should be -d(flux)/dt = integral (B*dA)
    the derivative of an integral of an equation with respect to t is just B*dA

    So my mistake was to forget to differentiate at all. Whoops.

    A = ∏r^2 but r= ro(1+at^2)
    so dA = derivative of ∏(ro(1+at^2)^2 = 4∏*a*t*ro(1+at^2)

    EMF = B*4∏*a*t*ro(1+at^2)

    Does my derivative look ok? And, was I right in thinking that in order to get the resistance in the equation V = IR you have to multiply by the circumference because the equation gives resistance/unit length?
  5. Apr 16, 2012 #4


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    OK, you're on the right track now. Your derivative looks OK, except it should be r0^2. And yes, you are right that since the resistance is pre unit length, you need to multiply by the circumference.
  6. Apr 16, 2012 #5
    Thanks, got it!
  7. Apr 16, 2012 #6
    The right side of this equation is the flux:

    [tex]\Phi = \int \vec{B} \cdot d\vec{A}[/tex]

    So, after evaluating [itex]\Phi(t)[/itex], you would then take its time derivative to get the emf. But to get this flux, you only need to multiply the magnetic field by the area, since

    [tex]\Phi = \int \vec{B} \cdot d\vec{A} = BA[/tex]

    if the magnetic field is perpendicular to the area and is uniform throughout the area. Then you do

    [tex]\text{emf} = -\frac{d\Phi}{dt}.[/tex]

    In particular, none of this needs to involve taking the derivative of the area.
  8. Apr 17, 2012 #7


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    Um....what? If B is constant, and phi = BA, then isn't d(phi)/dt = B * dA/dt? Isn't this taking the derivative of the area? What are you trying to say?
  9. Apr 17, 2012 #8
    I'm correcting the OP's terminology. It looks like he has the correct expression for the magnitude of the emf but seems confused on the proper derivative and naming conventions.
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