- #1

Bassalisk

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## Homework Statement

http://pokit.org/get/img/65e8ba92c1d00bf7fc8be2b178757ed8.jpg [Broken]

If a=5b, and I1 and I2 are known, find the force on the triangular loop.

## Homework Equations

## The Attempt at a Solution

For start, the field from the infinitely long wire is :

[itex] \vec B=\large -\frac{\mu _{0} I_{1}}{2 \pi x}\vec a_{z}[/itex]

This is understood and very trivial.

To find the total force on the current loop we will divide the problem into 3 parts, and solve forces on the (1) (2) (3) separately, and then add them all up.

For the force we will use the formula:

[itex] \vec F = I_{2} \int{\vec {dl} × \vec B} [/itex]

From this current view I want first question answered:

Will the only [itex] \vec a_{x} [/itex] component of the force be present?

I got that from the right hand rule, part of the y component vector-crossed with -z will give a positive x component.

The part with the x component crossed with the -z component will give both positive and negative contribution, leading to net 0 y component of the force.

Is this correct and if it is, i have further questions regarding the calculations and solving the line integrals, because I am getting weird answers.

Namely in case (1) and (2).

Is [itex] \vec{dl} =dx \vec a_{x} + dy \vec a_{y} [/itex] in both cases the same or is in second case [itex] \vec{dl} =-dx \vec a_{x} + dy \vec a_{y} [/itex].

If I use the fact that dl is in both cases the same, I get right results.

But keep in mind that when substituting dy in first case i used:

[itex] \vec{dy}=\frac {c}{2b} dx [/itex]

and in second case:

[itex] \vec{dy}=-\frac {c}{2b} dx [/itex]

I should also add that case 3 is not a problem because I got that right.

If i use [itex] \vec{dl} =-dx \vec a_{x} + dy \vec a_{y} [/itex] I get a net y component of the force which is not right in my intuition.

How do you solve these line integrals the right way?

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