# Force to energy proof problem

1. Oct 5, 2013

### 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:
$\delta W=\vec{F}\cdot \vec{\delta r}=I (\vec{dl}\times \vec{B})\cdot \vec{\delta r}=I[ \delta x (dy B_z-dz B_y)+...]=I[(\delta x dy-\delta y dx)B_z+..]$
To complete the proof,I should be able to set $\delta A_z=\delta x dy-\delta y dx$,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

2. Oct 7, 2013

### 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...

3. Oct 17, 2013

### 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.

Last edited: Oct 17, 2013