Force on Loop Due to Infinite Wire?

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Homework Help Overview

The discussion revolves around the force exerted by an infinitely long current-carrying wire on a circular loop placed in the X-Y plane. The loop is positioned at a distance 'r' from the wire and carries a current in an anticlockwise direction. Participants are exploring the magnetic field generated by the wire and its interaction with the loop.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the equation dF=i {dl x B} to find the force on the loop. There are attempts to express the magnetic field B in terms of the integration angle, with some questioning how to integrate B along with the angle. The importance of unit vectors and the dependence of distances on the integration angle are also highlighted.

Discussion Status

There are multiple interpretations being explored regarding the integration limits and the dependence of the magnetic field on the angle. Some participants have provided guidance on resolving forces and integrating over specific intervals, while others express confusion about the setup and calculations.

Contextual Notes

Participants note the potential confusion arising from using the same symbol 'r' for different distances in the problem. There is also mention of the symmetry in the problem, which influences the approach to integration.

zd1899
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Q)An infinitely long current carrying wire , carrying current I moving towards positive Y is placed along Y-axis. In the X-Y plane , a circular loop of radius 'r' is placed at a distance 'r' from the wire carrying current 'i' in anticlockwise circulation. Find the force exerted by the infinite wire on the loop.



Homework Equations



Acc. to me the only equation required should be
dF=i {dlxB} where B is the magnetic field due to the infinitely long wire at the length element dl which is to be integrated .

The Attempt at a Solution



dF=i{dlxB}
dF= (i*U*I/2pi) * 1/r *dl (assuming dl to be a small length element on the loop.
where r is the distance of dl.

i replaced dl by rd$ where $ is the small angle and now wished to integrate along $ from 0 to 2pi but B has a dependence on 'r' only as it decreases along X axis. so i am stuck on how to integrate B along with d$.
 
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zd1899 said:
Q)An infinitely long current carrying wire , carrying current I moving towards positive Y is placed along Y-axis. In the X-Y plane , a circular loop of radius 'r' is placed at a distance 'r' from the wire carrying current 'i' in anticlockwise circulation. Find the force exerted by the infinite wire on the loop.



Homework Equations



Acc. to me the only equation required should be
dF=i {dlxB} where B is the magnetic field due to the infinitely long wire at the length element dl which is to be integrated .

The Attempt at a Solution



dF=i{dlxB}
dF= (i*U*I/2pi) * 1/r *dl (assuming dl to be a small length element on the loop.
where r is the distance of dl.

i replaced dl by rd$

You are missing your unit vectors here for dl and B. It is important here to keep track of the direction or you will miss some important cancellations. (Notice that dl points in different directions as you move along the circle.)

where $ is the small angle and now wished to integrate along $ from 0 to 2pi but B has a dependence on 'r' only as it decreases along X axis.

That's true, but this r is the distance from the wire to the current element dl, and that distance will depend on your integration angle. (That is, replace the distance with an expression that involves theta.)

(By the way, you are using the symbol r for two different things in this problem: the given distances, and the distance from wire to current element. I would suggest changing one of them so there's no confusion.)
 
alphysicist said:
You are missing your unit vectors here for dl and B. It is important here to keep track of the direction or you will miss some important cancellations. (Notice that dl points in different directions as you move along the circle.)



That's true, but this r is the distance from the wire to the current element dl, and that distance will depend on your integration angle. (That is, replace the distance with an expression that involves theta.)

(By the way, you are using the symbol r for two different things in this problem: the given distances, and the distance from wire to current element. I would suggest changing one of them so there's no confusion.)

Thats the main trouble , i am not able to find a way to get the dependence of B on $ because B is just dependent on distance from the wire. And i have already judged the directions , its the magnitude that bothers me.
 
zd1899 said:
Thats the main trouble , i am not able to find a way to get the dependence of B on $ because B is just dependent on distance from the wire. And i have already judged the directions , its the magnitude that bothers me.

So we're talking about the same thing, look at this picture:

http://img11.imageshack.us/img11/1123/wireandloop.jpg

So when the integration angle is theta, what is the distance between the wire and the current element dl (the bold arrow on the circle)?
 
Last edited by a moderator:
r + r(1-cos$) ?
 
zd1899 said:
Q)An infinitely long current carrying wire , carrying current I moving towards positive Y is placed along Y-axis. In the X-Y plane , a circular loop of radius 'r' is placed at a distance 'r' from the wire carrying current 'i' in anticlockwise circulation. Find the force exerted by the infinite wire on the loop.

dF=i{dlxB}
dF= (i*U*I/2pi) * 1/r *dl (assuming dl to be a small length element on the loop.
where r is the distance of dl.

i replaced dl by rd$ where $ is the small angle and now wished to integrate along $ from 0 to 2pi but B has a dependence on 'r' only as it decreases along X axis. so i am stuck on how to integrate B along with d$.
The force on a small element dl on the circular loop is given by
dF = μο/2π*I*i*1/x*dl.
dF is perpendicular to dl. Its vertical components dF*sinθ from upper and lower semicircular coils get canceled out. The horizontal components 2dFcosθ add up.
Substitute dl = r*dθ and x = 2r + r*cosθ.
To find F, take the integration form 0 to π.
 
Last edited:
zd1899 said:
r + r(1-cos$) ?

If you have theta as in my picture, it would be r + r(1+cos theta) = r (2+cos theta)

If you are starting your integration on the left side (so that theta=0 corresponds to the leftmost point of the circle), then your expression would be correct.
 
@ rl.bhat why 0 to pi?? why not o to 2pi?

@alphysicist
yes i took it from the left actually and after pi/2 cos which change the sign as it will be pi-theta.
 
can some1 tell me the final answer so that i can confirm it? i do not have the answer.
 
  • #10
zd1899 said:
@ rl.bhat why 0 to pi?? why not o to 2pi?
.
Because I have taken two symmetric elements from upper and lower semicircles and I have resolved the forces acting on them. Horizontal components add up and vertical components cancel each other.
So F =2* μo/2π*I*i*Intg[ r*cosθ*dθ/r(2 + cosθ)] form 0 to π
 
Last edited:

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