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Deriving the Force on an Infinitessimal current-loop in a Magnetic Field

  1. Apr 5, 2012 #1
    We're asked to derive the following equation:

    [tex] F=\triangledown(\vec{m}\cdot \vec{B}) [/tex]

    by evaluating [tex] F=\int I(d\vec{l}\times \vec{B})[/tex] along a sqaure loop with sides of length ε, parralel to the yz plane. The square's bottom left corner is situated at the origin.

    so far I have,
    [itex] d\vec{l} \times \vec{B}= -dz \times \hat{z} \times \vec{B}(0,0,z) + dy \hat{y} \times \vec{B}(0,y,0)+dz\hat{z} \times \vec{B}(0,\epsilon ,z)-dy \hat{y} \times \vec{B}(0,y,\epsilon) [/itex]

    then using taylor expansion to turn B(0,y,ε) and B(0,ε,z) into expressions of B(0,y,0) and B(0,0,z) respectively and combining like terms i've got:

    [itex]
    - \hat{y} \times \left \epsilon dy\frac{\partial \vec{B}} {\partial z} \right |_{0,y,0} + \hat {z} \times \left \epsilon dz \frac{\partial \vec{B}} {\partial y} \right |_{0,0,z}
    [/itex]

    the next step would be to evaluate ∫dF and the solutions manual says that,

    [itex]
    \left \int dz\frac{\partial\vec{B}}{\partial y} \right |_{0,0,z}\approx \left \epsilon \frac{\partial\vec{B}}{\partial y} \right |_{0,0,0}
    [/itex]

    from this point the solution comes pretty easily, but I still don't understand why the above is true. Can anyone explain this to me?
     
  2. jcsd
  3. Apr 5, 2012 #2
    Does anyone know what is wrong with the LaTex coding here?
     
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