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Calculate the net force on square loop (Magnetism)

  1. Jun 16, 2013 #1
    1. The problem statement, all variables and given/known data

    A square loop of wire of side length L lies in the xy-plane, with its center at the origin and its sides parallel to the x- and y- axes. It carries a current i, in a counterclockwise direction, as viewed looking down the z-axis from the positive direction. The loop is in a magnetic field given by B = (B0/a)(zx-hat + xz-hat), where B0 is a constant field strength, a is a constant with the dimension of length, and x-hat and z-hat are unit vectors in the positive x-direction and positive z-direction. Calculate the net force on the loop.

    2. Relevant equations

    F = iL x B

    3. The attempt at a solution

    What's throwing me off is the zx-hat + xz-hat. I'm not sure how to figure this out. When I tried to solve the problem, though, everything cancelled out because the current on opposite sides of the square is flowing in opposite directions, so I got the force to be 0. However, I know this answer is not right. Any help would be appreciated.
     
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  3. Jun 16, 2013 #2

    Simon Bridge

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    You mean: $$\vec{B}=\frac{B_0}{a}\left ( z\hat{x} + x\hat{z} \right )$$ i.e. ##\hat{x}## is a unit vector pointing in the positive x direction ... which would more commonly be written: ##\hat{\imath}##.

    The current flowing in opposite directions may produce the same force in opposite directions - but the forces act at different places, so they do not cancel. Hint: "torque".

    The equation you gave for the magnetic field, though, suggests that the field strength, and thus the force for the same current, is different in different places ... so I'd expect the force to be different along different bits of the loop.
     
  4. Jun 16, 2013 #3
    This is what I have so far now. For torque, I have the equation [itex]\tau[/itex] = NiABsin[itex]\Theta[/itex]. Since there is only one loop, N would be 1. And since [itex]\tau[/itex] = F x r, I can equate the two expressions and solve for F.

    But I'm still confused with what to do. Do I have to find the torque for each side of the square and then add them to get the net torque?
     
  5. Jun 16, 2013 #4

    Simon Bridge

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    This expression is for a specific set of circumstances - do they apply in this situation?

    If I were you, I'd start by understanding the first equation you wrote down (it's telling you about the net torque on the loop in terms of the flux enclosed by it) ... or you can go back to first principles with ##\vec{F}=q\vec{v}\times\vec{B}## and work out the force on each bit of the wire.
    The fact that it is a square loop makes the geometry easier.

    What I need you to do is confirm that what I wrote for the ##\vec{B}## expression in post #2 is correct. You have to actually say - don't make me guess.
     
  6. Jun 17, 2013 #5
    Oh sorry, that was the right equation. I'm bad at formatting so that's why I said x-hat and z-hat.

    And it's the magnetic field equation that's really throwing me off. I don't understand what xz-hat and zx-hat (again, sorry with the formatting but I can't seem to find the proper unit vector symbols), are supposed to represent. Are they cross products or planes or something else?

    And I'm trying to figure out what to do with torque, but besides the equation I mentioned above, which was derived in my textbook for square loops, I don't have anything else about relating torques to magnetism.
     
  7. Jun 17, 2013 #6

    Simon Bridge

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    The "hat" notation indicates a unit vector ... so ##\hat{x}=\hat{\imath}##, ##\hat{y}=\hat{\jmath}##, and ##\hat{z}=\hat{k}## ... you know, like i,j,k notation?
    The numbers right in front are the components of the vector.

    So: $$\vec{B}=\frac{B_0}{a}\begin{pmatrix} z\\0\\x \end{pmatrix}$$

    Presumably you have some relation telling you the force on a wire carrying a current?
    You also have ##\vec{F}=q\vec{v}\times \vec{B}##

    You have the book's derivation for the formula you wrote too - that should tell you how to derive an equation for this particular situation.
     
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