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Force to energy proof problem

  1. Oct 5, 2013 #1


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    One of the problems in my textbook of electromagnetism is about proving that the work done by the force [itex] \vec{F}=I \vec{dl} \times \vec{B} [/itex],is [itex] \delta W=I \delta \phi [/itex] where the circuit isn't rigid and the displacement vector of the element of interest is [itex] \vec{\delta r} [/itex] with a constant current and [itex] \delta \phi [/itex] is the change in magnetic flux.My calculation is as follows:
    \delta W=\vec{F}\cdot \vec{\delta r}=I (\vec{dl}\times \vec{B})\cdot \vec{\delta r}=I[ \delta x (dy B_z-dz B_y)+...]=I[(\delta x dy-\delta y dx)B_z+..]
    To complete the proof,I should be able to set [itex] \delta A_z=\delta x dy-\delta y dx [/itex],etc.([itex]\delta A_z[/itex] being the change in area caused by [itex] B_z [/itex]).My problem is,I don't know how to justify it!
    Any ideas?
  2. jcsd
  3. Oct 7, 2013 #2

    Philip Wood

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    What you are (correctly) trying to evaluate is a scalar triple product. You can cyclically shift the terms round (look it up, say, in Wiki) so that the cross-product is of the dl and dr vectors, giving you a directed area... Then use the definition of phi...
  4. Oct 17, 2013 #3

    Philip Wood

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    Did you succeed? This is what I had in mind...

    [itex]\delta W = \vec{F}.\vec{\delta r} = I \vec{\delta l}\times \vec{B}.\vec{\delta r}[/itex]

    [This is unambiguous without brackets. The cross product has to be executed first, because if we tried to do the dot product, [itex]\vec{B}.\vec{\delta r}[/itex], first, it would yield a scalar, rendering the cross product meaningless.]

    We are allowed to re-arrange this scalar triple product cyclically...

    [itex]\delta W = I \vec{\delta r} \times \vec {\delta l}. \vec{B}.[/itex]

    Now, [itex] \vec{\delta r} \times \vec {\delta l}[/itex] is easily shown to be a vector [itex]\vec{\delta S}[/itex] of magnitude equal to the area swept our by [itex]\delta l[/itex], and directed normally to this area..

    But, by definition of flux, [itex]\delta \Phi = \vec{\delta S}.\vec B[/itex]

    Thus we have [itex]\delta W = I \Phi[/itex] as required.
    Last edited: Oct 17, 2013
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