# Magnetic force on triangular loop of wire

1. Apr 2, 2017

### rmiller70015

1. The problem statement, all variables and given/known data
Find the force on the triangular loop in the figure:
I am trying to do this in cylindrical polar coordinates, because it builds character.

2. Relevant equations
$$\vec{B}(\vec{r}) = \frac{\mu _0}{4\pi} \int \frac{I(d\vec{l} \times \hat{\alpha})}{|\vec{\alpha}|^2}$$
Where alpha is the script r vector (I couldn't figure out how to do script r in mathjax).
and

$$\vec{F} = I\int (d\vec{I} \times \vec{B})$$

3. The attempt at a solution
I've found the magnetic field due to the lower wire by using the Biot-Savart law and an example in the book. It is $$\frac{\mu _0 I}{2s\pi} \hat{\phi}$$

The solution uses Cartesian coordinates to integrate along the two wires. I figure that the (x,y) coordinates would be the (z,s) coordinates from the way I set up my cylindrical system (+z is along the wire and +s is up).

The cross product is:
$$d\vec{l_A} \times \vec{B} = \begin{vmatrix} s & \phi & z \\ 0 & 0 & -dz \\ 0 & \frac{\mu _0 I}{2s\pi} & 0 \end{vmatrix} = \frac{\mu _0 I}{2s\pi}\hat{s}$$

Then the force is:
$$\vec{F_A} = I\int^{a}_{0}\frac{\mu _0 I}{2\pi}dz\hat{s} = \frac{\mu _0 I^2a}{2\pi}\hat{s}$$

Then, finding the cross product for the B and C lengths become more difficult because I'm not quite sure if I am using the correct directions and I'm unsure of my limits of integration.
$$d\vec{l_B} \times \vec{B} = \begin{vmatrix} s & \phi & z \\ ds & 0 & dz \\ 0 & \frac{\mu _0 I}{2s\pi} & 0 \end{vmatrix} = -\frac{\mu _0 I}{2s\pi}dz\hat{s} + \frac{\mu _0 I}{2s\pi}ds\hat{z}$$

And the Force integral is:
$$\vec{F_B} = I\int-\frac{\mu _0 I}{2s\pi}dz\hat{s} + I\int \frac{\mu _0 I}{2s\pi}ds\hat{z}$$

Again I am uncertain if these integrals are correct and I'm unsure of the limits. I believe for dz differential I should be integrating between 0 and a. For the ds differential I should be integrating between s and √¾a. The integrals will give me a root3/2 and a natural log expression which are in the answer key, but I am not confident that what I am doing is correct.

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2. Apr 2, 2017

### kuruman

Do yourself a favor. Don't use "s" as a variable because the symbol has been reserved as the fixed distance between the long wire and the base of the triangular loop. Call it something else, r is good if you insist on cylindrical coordinates.

3. Apr 2, 2017

### rmiller70015

The book likes to call it s, so I went with that, but I think I see what you're saying. If I do that integral with respect to s, I might accidentally integrate it and the s should show up in the limits.

4. Apr 2, 2017

### kuruman

If the book calls it "s" and you like it, then relabel in the figure the distance between the long wire and the base of the triangle something other than "s".