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Homework Help: Induced voltage in rectangular loop

  1. May 27, 2014 #1
    1. The problem statement, all variables and given/known data
    There's a magnetic field B in +[itex]\hat{z}[/itex]. A rectangular loop is lying in the xy-plane. Three sides are static, the 4th one is moving with velocity v along the direction of +[itex]\hat{y}[/itex], making the rectangular larger and larger. The length of this moving side of the rectangle is L.

    Determine an expression for the magnitude of the electromotoric voltage induced in the loop using the flux through the loop.

    2. Relevant equations
    1. [itex]\Phi[/itex]m=[itex]\int[/itex]BdS
    2. ε=-[itex]\frac{∂\Phi}{∂t}[/itex]
    3. s = v*t (s = displacement, v = velocity, t=time)

    3. The attempt at a solution
    Using 1) and 3), with dS=dxdy[itex]\hat{z}[/itex] I get
    [itex]\Phi[/itex]m=[itex]\int[/itex]BdS = B[itex]\int[/itex][itex]\int[/itex]dxdy (with limits x:0→L and y:0→y0+v*t) = ... = BL(y0+v*t) = BLy0 + BLvt

    Then using 2)
    ε=-[itex]\frac{∂\Phi}{∂t}[/itex] = -[itex]\frac{∂}{∂t}[/itex](BLy0 + BLvt) = -Bvl

    Everything here is correct, except the sign in the end. It should, apparently, not be negative. It should be positive, i.e. just Bvl, not -Bvl. I don't understand why. In the answers, the only thing that differs from my solution is their definition of dS. They define it as dS=dxdy(-[itex]\hat{z}[/itex]) instead of dS=dxdy[itex]\hat{z}[/itex]. I don't understand why they define it in (-[itex]\hat{z}[/itex]). Can anyone shed some light on this, please?
  2. jcsd
  3. May 27, 2014 #2


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    Since the problem says "Determine an expression for the magnitude of...", it seems that the answer should be positive by definition. To say whether it is positive or negative, you need to define a convention for which direction represents positive current flow in the loop. The easiest way to determine the sign in these problems is to use Lenz's law, which says that the induced current will oppose the motion. In this problem, that means that the induced current will flow in a clockwise direction when viewed from above. Is this a positive or negative induced voltage? It depends on your convention. This is probably why the problem only asked for the magnitude of the induced voltage.
  4. May 28, 2014 #3

    rude man

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    Yes, magnitude is by definition positive.
  5. May 28, 2014 #4
    Ok guys. Thanks for your replies. This was driving me nuts. From your answer, phyzguy, I take it there's no correct way of determining the direction of dS? Whatever floats your boat?
  6. May 28, 2014 #5

    rude man

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    Oh yes, the polarity of the voltage is definitely determinable.

    The algebra is well defined, based on Maxwell's equations and Stokes' theorem, but you don't need to worry about it if you follow Lenz's law: the induced mag field will always oppose a CHANGE in the existing flux. So, in your case, flux is in the +z direction and increasing as the moving side moves in the +y direction, so the induced field will be in the -z direction to oppose the buildup of flux within the loop. So, if you draw your figure with x along the right , y up and z out of the page, the current must flow clockwise to generate a B field in the -z direction. So the right side of the moving side is + and the left is -.

    Physically, you can think of it as follows: force F on a free charge will be qv x B, In your case v = v j and B = B k so F = qv j x B k = qvB i so free charge is going to be forced to the right so that end will have a pile-up of + charge. (I use bold for vectors).
  7. May 29, 2014 #6
    @rude man, thanks for your explanation.

    Also thanks for pointing me in the direction of Lenz's law. I should've revisited that one before even posting the question here.
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