# Force on a Circular Loop Due to an Infinite Wire

1. Oct 27, 2015

### uselesslemma

1. The problem statement, all variables and given/known data
A long straight wire carrying a constant current I1 and a circular wire loop carrying a constant current I2 lie in a plane. The radius of the loop is R, and its center is located at distance D from the straight wire. What is the magnetic force exerted on the loop by the straight wire?

2. Relevant equations

The currents are constant, so $\overrightarrow{F}_{m}=I \int(d\overrightarrow{l}\times\overrightarrow{B})$

For an infinite wire, $\overrightarrow{B}=\frac{\mu_{0}I}{2\pi s}\hat\phi$
(cylindrical coordinates)

3. The attempt at a solution

So I set up the coordinate system like this.

$I=I_{2}$ and $B=B_{1}$, since the magnetic force due to the line's field should be on the current loop.

I thought that the spherical coordinate system would be the easiest to use for this problem. In that case:

$\theta=\frac{\pi}{2}$
$z=0$
$s=y=R sin\phi$ (by symmetry)
$dr=0$ (R is constant for the loop)
$d\overrightarrow{l}=Rd\theta\hat\theta+Rd\phi\hat\phi$

Therefore, above the wire $B_{1}$ is in the $\hat\phi$→$\hat z=-\hat\theta$ direction. In addition, the origin is displaced by length D, so the equations become:

$\overrightarrow{F}_{m}=I_{2}\int(d\overrightarrow{l}\times\overrightarrow{B}_{1})$

$\overrightarrow{B}_{1}=-\frac{\mu_{0}I_{1}}{2\pi(R sin\phi +D)}\hat\theta$

So the cross product says that $\overrightarrow{F}_{m}$ is only in the $\hat r$ direction.

$\overrightarrow{F}_{m}=\frac{\mu_{0}I_{1}I_{2}R}{2\pi}\int_0^{2\pi} \frac{d\phi}{R sin\phi +D}\hat r=\frac{\mu_{0}I_{1}I_{2}R}{\sqrt{D^{2}-R^{2}}}\hat r$

Is this the correct approach and result? I'm not sure if my replacement for s in the equation for $\overrightarrow{B}_{1}$ is valid, or if I set up the coordinate system in the appropriate way. I believe my answer has the correct dimensionality, I just want to verify the approach. Thanks for your help!

2. Oct 28, 2015

### BvU

Your $\hat r$ direction is not a constant of the integration, so I wouldn't take this as the answer: Which way would the ring move if not attached to anything ?

3. Oct 28, 2015

### uselesslemma

Ah yes I see. So $\hat r=cos\phi \hat x + sin\phi \hat y$, and it is therefore not exempt from integration.

$\overrightarrow{F}_{m}=\frac{\mu_{0}I_{1}I_{2}R}{2\pi}(\int_0^{2\pi} \frac{cos\phi d\phi}{R sin\phi +D}\hat x + \int_0^{2\pi} \frac{sin\phi d\phi}{R sin\phi +D}\hat y)$

In this case the $\hat x$ term cancels during integration (as expected), and I'm left with

$\overrightarrow{F}_{m}=\mu_{0}I_{1}I_{2}R (1-\frac{D}{\sqrt{D^{2}-R^{2}}})\hat y$

And since D > R, $\overrightarrow{F}_{m}$ always points down. This makes sense, because the portion of the loop that is attractive is closer to the line than the portion that is repulsive, so the loop should be attracted overall. This makes more sense, thanks!