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Consider the $yz$-plane. There is a wire in which a current of magnitude $I$ runs from $(y=a/2, z=a/2)$ to $(y=-a/2, z=a/2)$. There is a uniform magnetic field $\mathbf{B}$ along the length of the wire. I want to find a formula for the magnetic force on the wire.

$$\mathbf{F}_{mag}=I\int(d\mathbf{l}\times\mathbf{B})$$

I understand that there may be many ways of doing this problem. I want to find $\mathbf{l}$, differentiate it to find $d\mathbf{l}$ and then integrate $ I\int(d\mathbf{l}\times\mathbf{B})$ between the appropriate limits. So, what is $\mathbf{l}$?

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$\mathbf{l}$ is the line or curve (generally expressed parametrically) in space describing the shape of the wire. Its differential $d\mathbf{l}$ represents an infinitesimally-long element of the wire.

In your case, the wire has a very simple shape and, if you just visualize it, it should be clear how to express $d\mathbf{l}$ using coordinates.

The idea is that each little piece of wire feels a force from the magnetic field at the location of that piece. (In general, the field might vary in space, although in this problem it is uniform.) The integral adds up the forces on all the pieces to get the total force on the whole wire.

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You needn't have to find $\textbf{l}$ and differentiate it and then integrate. It is just a matter of notation.

$d\textbf{l}$ is simply the infinitesimal line element at a location on the wire. More generally, $Id\textbf{l}$ is the current element at that location such that this current over the infinitesimal length of the wire would be responsible for the magnetic force on that length element. Then you can integrate it to find the magnetic force on the whole wire.

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  • $\begingroup$ Thanks. The reason I wanted to find $\mathbf l$ rather than just write $d\mathbf l$ as $-dy\ \hat{\mathbf y}$ was because I wasn't sure if $y$ ran from $-a/2$ to $ a/2$ or $a/2$ to $-a/2$. $\endgroup$ – Thomas May 28 at 10:24
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    $\begingroup$ In the above formula you have to consider the current element $Id\textbf{l}$ as an entity. $d\textbf{l}$ alone is just a length element whose direction is determined by the direction of the current. $\endgroup$ – Richard May 28 at 11:40

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