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Homework Help: Force on square loop

  1. Jul 15, 2009 #1
    1. The problem statement, all variables and given/known data
    This is Griffiths problem 5.4
    NOTE: this is NOT for class, only for study :) I have the answer, but am unable to reproduce it!
    Problem: Given [tex] B = kz\hat x [/tex] where k is a constant, find the force on a square loop of size a lying in the yz plane and centered at the origin.

    2. Relevant equations

    [tex]F = I\int dl \times B[/tex] is all we need. Clearly then, we get B = IaB.

    3. The attempt at a solution
    Let's assume the current is counterclockwise. According to the answer I have, the force on the left side (towards the left) cancels the right side (towards the right). However, it appears to me that since z is analogous to x in this case, that as we move to the left or right the B-field increases and decreases. Is this not correct? If so, how can you say that the left cancels the right?

    Also, the answer claims that the force on the top points up, and the force on the bottom points up. This is true, only if the force on the bottom is negative, and the answer claims it is, however I don't see why this is so?

    Anyway, I'm clearly missing a few key points, so hope someone can help.
    Last edited: Jul 15, 2009
  2. jcsd
  3. Jul 15, 2009 #2


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    You don't want the spaces in the tags. It should be [noparse][tex] expression [/tex][/noparse]. Just for the sake of being able to read your question, I'll reproduce it here (then maybe I'll be able to think about it):

  4. Jul 15, 2009 #3


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    I would say it's not correct. What do you mean "x is analogous to z?"

    I have visualized my coordinate system so that the yz plane is the plane of the page, i.e positive z means going up, positive y means going to the right, and positive x means coming out of the page.

    There are two separate ideas here:

    1. The B field points in the x-direction.

    2. The B field is a function of z. It increases linearly in magnitude with increasing z. If you move along a left-right line or an in-out line (lines along which z is const) then the magnitude of B will be constant.

    As a result, by symmetry, the varation of B along z on the left edge of the square cancels the effect of the variation of B along z on the right edge of the square.

    The same symmetry is not present on the top and bottom edges, giving rise to a net force.
  5. Jul 15, 2009 #4
    Yes, thank you. I see where my (very silly) error was :)
    Also, just to check: the force on the bottom is negative simply because of z being negative? Thanks! Actually, let me write out the force on the bottom [tex]F = I\int_{x}^{y} dl \times B[/tex] where x = a/2 and y = -a/2 OK yes, it all makes sense now. Easily problem, sometimes it just helps to speak to someone...
    Last edited: Jul 15, 2009
  6. Jul 15, 2009 #5


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    If by "force", you actually mean "B-field", then yes. Since z is negative, kz is negative, and therefore, below the y-axis, the B field changes direction from [itex] +\hat{x} [/itex] to [itex]-\hat{x}[/itex].

    If we assume that the current is counterclockwise, then it moves to the right along the bottom edge. Therefore, the cross product of v and B is given by direction:

    [tex] \hat{y} \times -\hat{x} = \hat{z} [/tex]

    The force points up.

    Along the top edge, the direction of the velocity of charges is opposite, but the B field is also opposite, so the force still points up.
  7. Jul 15, 2009 #6
    Thanks cepheid, I appreciate your help.
    One last question and hopefully everything will be clear :)
    When I integrate my top edge, I have the integral as going from a/2 to -a/2 which means your force for the top edge is -aIB in the positive z-direction. However, the answer should be positive aIB, so where did I go wrong? Thanks in advance!
  8. Jul 15, 2009 #7


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    First of all, I'll point out that you don't have to integrate. B is constant on a horizontal line (B has no y-dependence). Furthermore, we're assuming a steady current. So we could just go with:

    F = qv × B

    The total charge along the top edge is just given by

    q = IΔt ​

    where Δt is the total time it takes for a charge to move from one end to the other. Similarly the velocity is given by:

    v = Δℓ/Δt ​

    where Δℓ is the displacement of a charge. (Is the script l showing up for you?) Of course, the kicker is that the displacement is in the negative y-direction:

    Δℓ = -Δyy

    where of course, Δy = a. So the whole equation for the Lorentz force law reduces to:

    F = (IΔt)(Δℓ/Δt) × B

    F = -Iay × Bx

    F = IaB(-y × x)

    F = IaBz

    I'm sure that if you took this into account in your integral, it would reduce to the same thing, i.e. the infinitesimal displacement is given by:

    dℓ = -dyy

    and so

    Idℓ × B = -I dyy × B

    -I ( dy)B(y × x) ​
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