Magnetic force on triangular loop of wire

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The discussion focuses on calculating the magnetic force on a triangular loop of wire using cylindrical polar coordinates. The user has successfully derived the magnetic field from a lower wire using the Biot-Savart law and is attempting to compute the force on the loop. There is uncertainty regarding the integration limits and the use of the variable "s," which is reserved for a fixed distance in the problem. Suggestions are made to rename the variable to avoid confusion and ensure accurate integration. The conversation emphasizes the importance of clear variable definitions in solving the problem correctly.
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".
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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