Tension in a flexible circular loop

[Jahnavi]

@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 ?

 

[haruspex]

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 ?

Yes.

 

[Jahnavi]

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

@haruspex ,please reply this .Is the direction of force from the neighbouring elements exactly opposite to the one shown in the diagram in post#11 ?

 

[kuruman]

Yes, it will be opposite. If $d\vec F = I d\vec l \times \vec B$, then $I d (-\vec l) \times \vec B = -d \vec F$.

 

[Jahnavi]

I am not asking about the direction of force on the tiny wire element by the magnetic field

I am asking about the direction of force on the infinitesimal length element from the neighbouring elements . Just like in post#11 you showed the two forces by $T$ .

My question is that if an anticlockwise current flows then the direction of force would be exactly opposite to the one showed by two $T$'s in post#11 ?

 

[kuruman]

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 180^o 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.

 

[Jahnavi]

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 .

 

[kuruman]

... you will have the same force diagram as in post #11 except that will be reflected 180^o 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.

 

[Charles Link]

@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.

 

[Jahnavi]

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 .

 

[kuruman]

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.

 

[Charles Link]

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.

 

[Jahnavi]

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 .

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 ?

 

[Charles Link]

@Jahnavi Did you see post 41 by @kuruman ? I think he has concurred that it is compressive for the other case.

 

[Jahnavi]

Yes . I have seen post 41 .

Please clarify , is the direction of force (red vectors ) in the above picture correct ?

 

[Charles Link]

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.

 

[kuruman]

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.

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.

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.