How Does Force Convert to Energy in Electromagnetism?

[ShayanJ]

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

 

[Philip Wood]

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...

 

[Philip Wood]

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

\delta W = \vec{F}.\vec{\delta r} = I \vec{\delta l}\times \vec{B}.\vec{\delta r}

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

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

\delta W = I \vec{\delta r} \times \vec {\delta l}. \vec{B}.

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

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

Thus we have \delta W = I \Phi as required.