Force on triangular current loop: line integrals

In summary, the conversation discusses finding the force on a triangular loop using known variables and equations. The conversation also includes a question about the presence of the vertical component of the force and how to solve line integrals correctly. After clarifying the relationship between x and y and the definition of dl, the correct method of solving the line integrals is determined.
  • #1
Bassalisk
947
2

Homework Statement



http://pokit.org/get/img/65e8ba92c1d00bf7fc8be2b178757ed8.jpg

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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  • #2
Bassalisk said:

Homework Statement



http://pokit.org/get/img/65e8ba92c1d00bf7fc8be2b178757ed8.jpg

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
vela said:
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 don't 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.
 

1. What is a triangular current loop?

A triangular current loop is a closed, triangular-shaped path through which an electric current flows. It is a common type of current loop used in physics experiments and calculations.

2. How is force calculated on a triangular current loop?

The force on a triangular current loop can be calculated using line integrals, which take into account the direction and magnitude of the current, as well as the shape and orientation of the loop.

3. What factors affect the force on a triangular current loop?

The force on a triangular current loop is affected by the strength of the current, the size and shape of the loop, and the orientation of the loop relative to an external magnetic field.

4. Can the force on a triangular current loop be negative?

Yes, the force on a triangular current loop can be negative if the loop is oriented in the opposite direction to an external magnetic field. In this case, the force acts in the opposite direction to the current flow.

5. How is the direction of the force on a triangular current loop determined?

The direction of the force on a triangular current loop is determined by the right hand rule, which states that if the fingers of your right hand curl in the direction of the current, your thumb will point in the direction of the force on the loop.

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