Conversation started Feb 19, 2020 at 10:09.
Feb 19, 2020 10:09
@JohnRennie: What is meant by "force of compression" on a circular wire if each element is under a magnetic force which it towards the centre?
Is that tension?
@GuruVishnu yes
Ok sir. Thank you.
But from the above diagram, I could only infer that the wire breaks. Or in other words, I think there could be no equilibrium as net force is not equal to zero, @JohnRennie sir.
@GuruVishnu I would do it using virtual work ...
@JohnRennie Cool! "virtual work" = I haven't learnt about that so far. Is that "real" work we're used to?
Could you explain how to use virtual work? It seems like a good tool.
Suppose the radius of the wire decreases by $dr$.
Feb 19, 2020 10:23
Ok sir.
Then the work done is $dr$ times the total inwards force on the wire $BI\ell$. OK so far?
@JohnRennie Yes sir.
I see, now you'd say the work done by the compression force is opposite and finally I get $F=iBl$
Now suppose the tension in the wire is $T$. As the wire moves inwards its length changes by $dL = 2\pi dr$ i.e. it decreases from $2\pi r$ to $2\pi (r + dr)$.
@JohnRennie Ok sir.
So the work done is $dW = T2\pi dr$
And the two works must be the same.
So equate the two works and solve for $T$.
Feb 19, 2020 10:29
@JohnRennie: Yes sir. This method is really great and I obtained $F=iBl$ when you were explaining itself (but using some hypothetical assumptions :)). And I got the correct answer. But could you tell whether there is any mistake in the above force diagram? Am I missing any other forces?
Your diagram looks fine.
@JohnRennie But it seems every single particle will not be in equilibrium as net force is not zero, sir.
Each particle in the wire exerts an equal and opposite force on every other particle. The forces cancel so the ring doesn't move.
I'm wrong the current answer is $T=iBr$
@JohnRennie Still I don't understand sir. I think we need to include only one of the action-reaction pair of forces to find the equilibrium.
This is how the free body diagram looks like.
The virtual work method gives me the correct answer. Do you want to go through the virtual work calculation?
Feb 19, 2020 10:39
@JohnRennie If it's fine for you, I'd say "yes" sir.
OK. Step 1: the length of the wire is $\ell = 2\pi r$ so the inwards force is $F = B i 2\pi r$.
@JohnRennie Yes sir,
So the work done by the inwards force as the radius decreases by $dr$ is $dW = Bi2\pi r dr$.
OK so far?
@JohnRennie Yes sir.
As the wire moves in its circumference dereases from $2\pi r$ to $2\pi (r - dr)$ so the circumference changes by $2\pi dr$.
If the tension in the wire is $T$ then the work done by the tension as the length of the wire changes is $dW = T 2\pi dr$.
@GuruVishnu OK so far?
Feb 19, 2020 10:44
@JohnRennie Does the tension as seen in the previous diagram have a direction opposite to that in the diagram?
The tension is the force in the wire i.e. it is tangential to the wire.
@JohnRennie Fine then we shall consider the tension vectors to be nearly vertical but not tangential. If they were exactly tangential, then the net force will be towards the centre and the tension can't counteract the magnetic force, sir.
Let's get this virtual work calculation done and worry about the stability in a moment.
@JohnRennie Ok sir.
8 mins ago, by John Rennie
So the work done by the inwards force as the radius decreases by $dr$ is $dW = Bi2\pi r dr$.
6 mins ago, by John Rennie
If the tension in the wire is $T$ then the work done by the tension as the length of the wire changes is $dW = T 2\pi dr$.
Do you agree with my working so far?
Feb 19, 2020 10:49
@JohnRennie Understood this but
@JohnRennie not completely this one sir.
Because of the direction argument. So I think it's safe to proceed.
OK. It's the same work we are calculating, just calculated in two different ways, and that means the two $dW$s must be the same.
So we equate the two expressions, and solve for $T$.
@JohnRennie Ok sir. And on equating we get $T=iBr$
And it gives the correct answer. I understood this point. If possible, shall we discuss about that direction, sir?:
Let me draw a diagram
Ok sir. I arrived at the following after some tangential modifications:
Here are some metal atoms in a circle.
When we try and shrink the circle the atoms exert forces on each other and don't move.
Feb 19, 2020 10:56
Ok sir.
The forces are indeed not tangential, but metal atoms are very small so if we use a macroscopic radius, i.e. mm or cm, the forces are so nearly tangential that in practice the deviation from the tangential direction is undetectably small.
@JohnRennie Ok sir.
More precisely, if we consider an element of the circle $d\ell$ then the force becomes tangential in the limit of $d\ell \to 0$.
Which is fine for us, because we are doing the calculation of the virtual work in the limit $dr \to 0$.
@JohnRennie Ok sir. Then how does it stay in equilibrium. There is a force exerted by the magnetic field towards the centre.
Because In the real world $d\ell$ never goes to zero since atoms have a finite size.
Feb 19, 2020 11:02
@JohnRennie Ok sir. If so is the following nearly accurate as it answers the equilibrium:
The tension is massive enough so that even it small horizontal component matches the inwards magnetic force.
@JohnRennie Thank you sir :-)
 
Conversation ended Feb 19, 2020 at 11:03.