Magnetic force on triangular loop of wire

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Homework Help Overview

The discussion revolves around calculating the magnetic force on a triangular loop of wire using cylindrical polar coordinates. The original poster attempts to apply the Biot-Savart law and integrate to find the force, while also expressing concerns about the coordinate system and limits of integration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of cylindrical coordinates and the implications of variable naming, particularly the use of "s" versus "r". There are questions about the correctness of the integrals and limits of integration for different segments of the triangular loop.

Discussion Status

There is ongoing exploration of the coordinate system and variable naming conventions. Some participants suggest clarifying the variable used for the distance between the wire and the loop's base to avoid confusion. The original poster expresses uncertainty about the integration process and the limits involved.

Contextual Notes

Participants are navigating the constraints of the problem as defined by the textbook, which influences their variable choices and integration limits. There is an acknowledgment of potential pitfalls in variable naming that could affect the setup of the integrals.

rmiller70015
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Homework Statement


Find the force on the triangular loop in the figure:
I am trying to do this in cylindrical polar coordinates, because it builds character.
triangle_loop.png

Homework 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

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

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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rmiller70015 said:
I figure that the (x,y) coordinates would be the (z,s)
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.
 
kuruman said:
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.

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.
 
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".
 

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