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Find the Magnitude of the Magnetic Field in a moving loop

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A conducting square loop placed in the xy plane (oriented with its horizontal and vertical sides parallel to the x and y axes) moves to the right across the y axis at a constant speed of 2.0 m/s, a 0.63-V emf is induced in the loop. If the side length of the loop is 0.30 m, what is the magnitude of the magnetic field?

    2. Relevant equations
    EMF = Blv

    3. The attempt at a solution
    So my EMF is 0.63, V = 2.0m/s, l = .30m and B is unknown

    Thus
    0.63 T*m^2*s^-1/(2.0m*s^-1 * .30m) = B
    0.63 T/.6 = B

    1.05 T = B

    Not sure where I'm going wrong here.
     
  2. jcsd
  3. Apr 8, 2015 #2

    BvU

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    Isn't your equation for a straight section of wire only, not for a square loop ?

    And what is known of the magnetic field? A direction, where are the bounds, is it constant ?
     
  4. Apr 8, 2015 #3
    Magnetic Field is constant, the direction is into the positive z direction (guess I missed that part). I've tried to use ΔΦ/Δt, but not sure I did it right. Tried to take that fact it's traveling at 2m/s and the loop is 0.30 that

    Δt= 0.30m/2m/s giving me 0.15 seconds to go through completely into the magnetic field.
    a = 0.09m2
    emf = 0.63 T*m2

    Thus

    0.63=(B)*(0.09)/0.15s

    0.63*0.15/(0.09) = B

    1.05 T = B

    Gives the same answer.
     
  5. Apr 9, 2015 #4

    collinsmark

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    BvU has a good point here, when asking about the magnetic field's boundaries.

    If the magnetic field and loop have a constant, perpendicular orientation [that is the field is perpendicular to the sides, parallel with the loop's normal], the loop's size is fixed, and the field is not changing with time, then the loop needs to be passing into or out of one of the magnetic field's boundaries in order to get an emf.

    In short, something or other needs to be causing a change to the magnetic flux through the loop, otherwise an emf would not be occurring.

    Is there anything more you can tell us about the magnetic field (boundary edges, etc.), or something about the loop changing size or orientation?

    [Edit: Btw, the 1.05 T figure is correct if the loop is entering or leaving a boundary that is parallel with the loop's side, and spans the entire side. But if that doesn't give the correct answer, is that at least the correct description of the boundary?]
     
    Last edited: Apr 9, 2015
  6. Apr 9, 2015 #5
    The space to the right of the y axis contains a uniform magnetic field of unknown magnitude that points in the positive z direction. As a conducting square loop placed in the xy plane (oriented with its horizontal and vertical sides parallel to the x and y axes) moves to the right across the y axis at a constant speed of
    2.0 m/s, a 0.63-V emf is induced in the loop.
    If the side length of the loop is
    0.30 m, what is the magnitude of the magnetic field?

    Copied and pasted exactly. It's entering the magnetic field. That's all I get.

    It's entering the magnetic field, nothing about magnetic field's boundaries.
     
  7. Apr 9, 2015 #6

    collinsmark

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    The boundary is the y-axis.

    For regions where x < 0, there is no magnetic field at all. For regions where x > 0, there is a magnetic field of unknown magnitude that points in the z direction. That's what was meant by saying, "the space to the right of the y axis contains a uniform magnetic field of unknown [...]" That specifies the boundary, assuming that the space to the left of the y axis contains 0 magnetic field. That's the assumption that I would make anyway.

    That said, your answer of 1.05 T looks right to me. :smile:
     
  8. Apr 10, 2015 #7

    BvU

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    Makes a big difference with post #1, though. I understand you didn't realize that when you posted, but it sure is a reminder that a complete and concise problem statement is extremely important, not only for yourself (to order your thoughts and get lined up for the equation gathering phase), but also to get adequate assistance !
    And putting together the relevant equations is indispensable too - if necessary in more than one iteration.:rolleyes:
     
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