Tension in a flexible circular loop

In summary, the conversation discusses a problem involving a small circular flexible loop of wire carrying a current placed in a uniform magnetic field. The question is which factor will result in a doubled tension in the loop. The relevant equation for finding tension is F=I(lxB). The conversation goes on to discuss the correct approach to solving the problem and concludes that the correct answer is A or C.
  • #36
The tensions will change direction, if that's what you are thinking. The endpoints of the circuit are assumed fixed so that one can analyze the forces. When the current changes direction, the magnetic force will change direction and you will have the same force diagram as in post #11 except that will be reflected 180o about the horizontal axis passing through the endpoints (dashed line in post ##11). In other words, the tensions will reverse their perpendicular components to the dashed line, but not their parallel components. In the end, the entire loop will still be under tension except "upside down" in a sense.
 
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  • #37
kuruman said:
but not their parallel components.

Please reconsider this .

I think when the current is reversed , both the perpendicular and parallel components should be reversed .

If parallel component is not reversed (as in post#11 ) then the force at the end points of the tiny element will not be tangential .
 
  • #38
kuruman said:
... you will have the same force diagram as in post #11 except that will be reflected 180o about the horizontal axis passing through the endpoints ...
Perhaps I wasn't very clear with this statement. The entire diagram will be reflected. This means that the arc under consideration will curve down instead of up.

LoopInField_2.png
 

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  • #39
Jahnavi said:
@haruspex , @kuruman If the current flows in anticlockwise direction (instead of clockwise as in the OP ) and magnetic field is perpendicular to the plane of the page , the direction of force on any infinitesimal portion of the wire will be radially inwards .

Would the force on this tiny element from the neighbouring wire element be exactly opposite to the one shown in post#11 ?

But this force doesn't look like some tension force . Is it some type of compressive force ?

But magnitude of this force will still be BIR just as in the OP .Right ?
I believe @Jahnavi has it correct. If ## \vec{B} ## is upward and ## I ## is counterclockwise the force points outward from each part of the loop. ## \\ ## If the current is clockwise the force from each part of the loop will be of a compressive form, and it will be towards the center of the loop for each part of the loop.
 
  • #40
Charles Link said:
If ## \vec{B} ## is upward and ## I ## is counterclockwise the force points outward from each part of the loop.

No .

I think if the magnetic field is perpendicular to the plane of the page (going in) and current is anticlockwise then force due to magnetic field will be towards the center .The net compressive force on an infinitesimal element should be radially outward .
 
  • #41
Charles Link said:
I believe @Jahnavi has it correct. If ## \vec{B} ## is upward and ## I ## is counterclockwise the force points outward from each part of the loop. ## \\ ## If the current is clockwise the force from each part of the loop will be of a compressive form, and it will be towards the center of the loop for each part of the loop.
Minor correction, the magnetic field is perpendicular to the screen. Anyway, yes the force will be compressive, but consider this: imagine a supple segment of a wire tied at both ends such that the distance between the anchor points is less than the length of the wire. Place it in a uniform magnetic field and run a current through it. When the current runs "left to right" the wire will flex and be taut in one direction and if you run the current "right to left" it will be taut and flex in the opposite direction. If you have a closed loop of supple wire all crumpled up and somehow manage to run a current through it, it will stretch and be under tension as long as the current is running. If you reverse the current, yes it will crumple up but it will not stay crumpled up. It will stretch and be under tension again.
 
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  • #42
Jahnavi said:
No .

I think if the magnetic field is perpendicular to the plane of the page (going in) and current is anticlockwise then force due to magnetic field will be towards the center .The net compressive force on an infinitesimal element should be radially outward .
The description of my coordinate system needs a little more detail. By "up" (for the magnetic field) I mean positive ## z ##, and counterclockwise refers to the direction of the current in the x-y plane as viewed from above.
 
  • #43
kuruman said:
Perhaps I wasn't very clear with this statement. The entire diagram will be reflected. This means that the arc under consideration will curve down instead of up.

View attachment 224767

I feel this diagram is same as the one in post#11 .

Please see this .
LoopInField1.PNG

Don't you think if the current is in anticlockwise direction , direction of force due to neighbouring current wire element will be as shown by the red vectors ?
 

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  • #45
Yes . I have seen post 41 .

Please clarify , is the direction of force (red vectors ) in the above picture correct ?
 
  • #46
Jahnavi said:
Yes . I have seen post 41 .

Please clarify , is the direction of force (red vectors ) in the above picture correct ?
Yes. I agree with it. We need to see if @kuruman agrees. ## \\ ## Edit: If this is a loose loop of wire resting on a surface, it would have a tendency to flip over, i.e. compress itself and then expand again with the current running the other way, so that the magnetic moment from the loop of wire ## \vec{\mu}=I \vec{A} ## was aligned with the magnetic field, because energy ## E=-\vec{\mu} \cdot \vec{B} ##. The system would tend to go to the state of lowest energy.
 
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  • #47
Charles Link said:
Yes. I agree with it. We need to see if @kuruman agrees. ## \\ ## Edit: If this is a loose loop of wire resting on a surface, it would have a tendency to flip over, i.e. compress itself and then expand again with the current running the other way, so that the magnetic moment from the loop of wire ## \vec{\mu}=I \vec{A} ## was aligned with the magnetic field, because energy ## E=-\vec{\mu} \cdot \vec{B} ##. The system would tend to go to the state of lowest energy.
I wholly agree with the "Edit" statements by @Charles Link. My arguments rely on the wire being supple as state in post #41. By "supple" I mean "extremely flexible". Consider what will happen to two identical circuits placed in a uniform magnetic field perpendicular to the screen (see below) with currents running in opposite directions. To minimize the energy, one of the two subloops in each circuit will flip over and the the circuit will be under tension and form a circle with the current running in the same direction.

TwoLoops_1.png


If the circuits are made of very stiff wire but free to rotate, then the subloop with highest magnitude of magnetic moment (here the bottom subloop) will determine the final configuration.

TwoLoops_2.png


And if the stiff wire circuits are sandwiched between two pieces of stiff plastic, then they will stay as they are with one subloop under tension and the other under compression in each.
 

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