Direction of force on circular loop.

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Homework Help Overview

The discussion revolves around determining the direction of the force acting on a circular loop carrying current, particularly in relation to the effects of changing current direction in a straight conductor. Participants are exploring concepts related to electromagnetism and the forces between current-carrying conductors.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the right-hand rule to determine force direction, while expressing uncertainty about its application. Questions arise regarding the influence of current direction in straight conductors and how this affects the circular loop. Some participants suggest considering symmetry and the forces between small segments of the loop.

Discussion Status

The discussion is ongoing, with participants offering insights and guidance on how to approach the problem. There is an exploration of different interpretations of the forces involved, particularly in relation to the configuration of the loop and the current direction.

Contextual Notes

Participants mention previous experiences with similar tasks involving straight conductors and metal frames, indicating a potential gap in understanding how these principles apply to the circular loop scenario. There is also mention of using unit vectors, suggesting a mathematical approach that may be constrained by the problem's complexity.

diinnoo
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Homework Statement



Find direction of force on circular loop

http://pokit.org/get/img/b886e52efb703d47d09fa6e781c74461.jpg

What would hapend if direction of the current in a straight conductor were different ?

Homework Equations



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The Attempt at a Solution



http://pokit.org/get/img/7b5d76e176716f45edb5dbf331ff9e3e.jpg

I try to do this by using right hand rule, but this is just assumptions, I am not sure is this correct or can it be done like this. When direction of current change i don't know what would happened, or how straith conductor influence? I hope you would help me ;)
 
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welcome to pf!

hi diinnoo! welcome to pf! :smile:

you're only asked for the direction, so can't you just use general principles, and symmetry?

the force between two straight currents in the same direction is … ?

in the opposite direction is … ?

now apply that to tiny bits of the circle, in pairs at opposite ends of the same diameter :wink:
 
thanks for welcome tiny-tim.

Yes only direction was asked, i think two straight current same direction force would attract each other, when they are opposit they wil reject each other, two forces.

I tried this using unit vectors i tried something like this

http://pokit.org/get/img/96569110fd7b5c1228052db376c5fecd.jpg

but i can do this onle wher direction of current is same as unit vector i or j, i don't know how to do it on whole circut. but in this examle above i again don see influence of straight current.
 
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if the loop is is placed parallel to the field there will be a couple acting on it, when it's perpendicular the two forces will be equal opposite and have the same line of action
 
I really didnt understand you, sorry, or you just can simply draw what you thought. In school we did task like this, with straight conuctor and metal frame, in same plane. Direction we find using units vector, as my example up.

http://pokit.org/get/img/23dd764ee7a54aef8c120e0932b3eb97.jpg

And this task should be on same principes but much harder :(
 
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hi diinnoo! :smile:
diinnoo said:
… two straight current same direction force would attract each other, when they are opposit they wil reject each other, two forces.

yes … so first consider two very small parts of the circle, at 12 o'clock and 6 o'clock …

which one will have the larger force, and so what will be the direction of the resultant force on the pair?

(this is similar to the question in your diagram, but your "hollow" arrows are wrong …)
diinnoo said:
http://pokit.org/get/img/23dd764ee7a54aef8c120e0932b3eb97.jpg

then consider the resultant force on four very small parts, at 1 5 7 and 11 o'clock :wink:
 
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