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Force on triangular current loop: line integrals

  • Thread starter Bassalisk
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  • #1
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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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Answers and Replies

  • #2
vela
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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.
Yes, you're right. By symmetry, the vertical component of the forces will cancel, so any net force has to act in the horizontal direction.

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?
By definition, ##d\vec{l} = dx\,\vec{a}_x + dy\,\vec{a}_y + dz\,\vec{a}_z##. The relative sign dependence of the first two pieces will come in through the relationship between x and y, e.g. y=(2b/c)x or y=-(2b/c)x, and dz=0 for these contours.
 
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  • #3
948
2
Yes, you're right. By symmetry, the vertical component of the forces will cancel, so any net force has to act in the horizontal direction.


By definition, ##d\vec{l} = dx\,\vec{a}_x + dy\,\vec{a}_y + dz\,\vec{a}_z##. The relative sign dependence of the first two pieces will come in through the relationship between x and y, e.g. y=(2b/c)x or y=-(2b/c)x, and dz=0 for these contours.
oh so I dont have to account for it in the differential of l?

It will account for it self when I change the dy into dx through that relationship?

Makes sense. Thank you for the answer. I actually did it this way, but this was purely shot in the dark because I knew the final answer.

Thank you for Your help, You have been very helpful.
 

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