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Question on simple static magnetics

  1. Apr 17, 2012 #1
    Suppose a closed loop with constant current around the loop. The loop is in a uniform constant B field. The force is

    [tex]\vec F=I\left (\oint_c d\vec l\right ) \times \vec B.[/tex]

    Being a closed loop [itex]\oint_c d\vec l=0[/itex]. So there is no force acting on the loop.

    BUT at the same time, we know there is torque, so there is force!!!! How is that possible?
     
  2. jcsd
  3. Apr 17, 2012 #2

    mfb

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    You calculate the total force, integrated over the ring. This is 0. But it consists of forces in different directions along the ring, which are not 0.

    I can give you two apples, and then steal two apples from you. Your total number of apples is the same as before. But certainly your number of apples was not the same all the time.
     
  4. Apr 17, 2012 #3
    I understand that, I can see the loop is going to turn as one side being push and the other side being pull and the total force cancel out. But isn't it quite dumb for the book to make a statement like this and the very next page start talking about torque?
     
  5. Apr 17, 2012 #4

    Born2bwire

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    You're comparing two different integrands though. For torque,

    [tex] \mathbf{T} = I \oint \mathbf{r} \times \left( d\boldsymbol{\ell} \times \mathbf{B} \right) [/tex]
     
    Last edited: Apr 17, 2012
  6. Apr 17, 2012 #5
    By the word "force" we typically mean "linear force", as opposed to an "angular force" (torque). Yes, a single loop of current in a static uniform magnetic field experiences zero net linear force and a non-zero net angular force. That is why little electric motors spin instead of shoot out like a bullet.
     
  7. Apr 17, 2012 #6
    I know you use different formulas as torque require an arm to swing. But the circuit loop is the same.
     
  8. Apr 17, 2012 #7
    I understand this, just the idea is very misleading....at least to me. Angular force starts with linear force, difference is it is on a hinge and it is forced to turn.

    Anyway, we all agree on the theory, I just complain about the book. It should say right away about the other possibility. Student read the first time will take it literal.
     
  9. Apr 17, 2012 #8

    Born2bwire

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    No, that is the formula for the torque, they are not the same integrals. The infinitesimal force is on a wire in magnetostatics is

    [tex] dF = I d\boldsymbol{\ell} \times \mathbf{B} [/tex]

    The infinitesimal torque is therefore,

    [tex] d\mathbf{T} = I \mathbf{r}\times d\boldsymbol{\ell} \times \mathbf{B} [/tex]

    Since the position vector changes depending on where you are on your wire, you cannot take the position vector out of your integral. Hence,

    [tex] \mathbf{T} = I \oint \mathbf{r}\times d\boldsymbol{\ell} \times \mathbf{B} [/tex]
     
  10. Apr 17, 2012 #9
    I worded it wrong, I meant they both are of different formulas, one for torque and one for linear. There is not disagreement on the physics here, I just complain about the wording of the book. It's all English. If they want to say that, they should clarify more.
     
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