# Homework Help: Tension in a flexible circular loop

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1. Mar 31, 2017

### cheapstrike

1. The problem statement, all variables and given/known data

A small circular flexible loop of wire of radius R carries a current I. It is placed in a uniform magnetic field B. The tension in the loop will be double if

A) I is doubled B) B is halved C) r is doubled D) both B and I are doubled.

2. Relevant equations

3. The attempt at a solution

(Assuming B to be inside plane of paper and I to be clockwise)
The way I am looking at the problem is that because the loop is flexible its area will increase and because of the increase in area that total flux passing through the loop will also increase. This change in flux will induce an EMF which leads to an anticlockwise current, thereby reducing net current.
I think we then have to find a relation between I and r but I can't understand how to do it.
I don't even know if this is correct way to approach.

2. Mar 31, 2017

### kuruman

What is the relevant equation showing where the tension comes from?

3. Mar 31, 2017

### cheapstrike

I think F=I(lxB) where l will be 2*pi*r?

4. Mar 31, 2017

### kuruman

No. Consider first an element dl = r dθ and the force acting on it at each of its ends.

5. Mar 31, 2017

### cheapstrike

Will the force acting on the dl element not be radially outwards?

6. Mar 31, 2017

### kuruman

It will indeed be radially outwards, but the rest of the loop will keep it in place by exerting tension at its ends. Draw a FBD.

7. Mar 31, 2017

### cheapstrike

Ok.. so in the FBD, there will be an outward force and a force on both sides of dl due to the other two ends. But can we get any relation from here?
I mean the loop is elastic so won't it expand?

8. Mar 31, 2017

### kuruman

According to the problem, the wire is flexible not elastic and its radius is fixed at R. I don't interpret this as meaning that the loop can stretch. All it means is that it wll assume a circular shape of radius R.

9. Mar 31, 2017

### cheapstrike

Ah.. I got mixed up with that. Ok. So the element dl will remain at it's place meaning net force on it is zero? So, the net outward force = I(dlxB) will be balanced by the two forces at it's ends. Btw how will we do it?

Last edited: Mar 31, 2017
10. Mar 31, 2017

### cheapstrike

I think the force acting radially outwards on the element dl will be I(dlxB) which is IdlB as dl is perpendicular to B. This force will be balanced by sin of both the tension forces. Also, their cosines will cancel each other.
So we will have an equation IdlB=2Tsin(dx), where T is the tensional force and dx is the angle which each tension vector makes with the tangent at dl (dx is d(theta)). As dx is small, sin(dx)=dx.
So equation becomes IdlB=2Tdx
Giving T=IdlB/2dx.
dl/dx can be written as r where r is the radius.
So, T=IBr/2.
But this seems incorrect to me as, in this case, T will be doubled on doubling I as well as r.
But I believe the question is single correct (I don't know the answer but I think it's single correct).
Can you please tell what is wrong with this and also the correct equations.
Thanks.

11. Mar 31, 2017

### kuruman

It looks like your tension is half of what it ought to be. Here is the FBD. You need to balance forces in the y-direction. Remember to integrate the y-component of dF using integration variable φ.

12. Mar 31, 2017

### cheapstrike

Ok.. so I took the angle that T vector makes incorrectly. It should be dx/2.

13. Mar 31, 2017

### kuruman

Can you finish now?

14. Mar 31, 2017

### cheapstrike

dFcosφ=Tθ (i wrote sinθ=θ) and then wrote dF as IdlB where dl is rdφ and integrated it from -θ/2 to θ/2 and wrote sinθ/2=θ/2 and finally got
IBr=T.
Is this correct?
If this is, then shouldn't the answer be A) and C)?

15. Mar 31, 2017

### kuruman

That's what I got.
I would say the answer is A or C. Are you sure this is not a multiple answer question?

16. Mar 31, 2017

### cheapstrike

I am not sure, i don't even know the answer. But the method seems correct.. :D
Thanks a lot

17. Apr 1, 2017

### haruspex

English is not great for expressing logical predicates.
The tension will double if A or C is done; the tension will double if A is done and the tension will double if C is done.

18. Apr 1, 2017

### kuruman

Exactly why this should be put forth as a "multiple answer" question where A and C are the two correct answers. It cannot be a "multiple choice" question which guarantees that only one answer is correct and which can be answered either by proving/deriving the correct answer or by showing that each of the n - 1 answers is incorrect.

Here, the statement "the tension will double" is true if either A is true or C is true, but not if both are true. Perhaps I should have written, "I would say the answer is A xor C" in an effort to remedy the shortcomings of the English language with logic.

19. Apr 3, 2017

### cheapstrike

There is no point in writing this but...
I think, saying A and C are correct is fine coz these two are two different cases. In A) only I is being doubled and in C) only r is being doubled. That's why option D) is there which shows that if the asker of the question wants to change two parameters in 1 case together, he must mention them together under a single option.

As far as this question is considered, either there is a printing mistake or it's a multiple correct. I think it's former coz question claimed to be single correct.

20. Apr 5, 2017

### Piyushagg

You all r putting a lot of brain...it has now formed a closed loop so F=mxB
m=(pi*r*r)*i
This implies on increasing current i or magnetic field B...Force will double.
In ur case it is "on incresing current". Hope it was correct.

21. Apr 5, 2017

### kuruman

You r incorrect. Torque, not force is mxB. We are concerned with force here.

22. Apr 5, 2017

### Piyushagg

I was thinking that net force is suppose to be zero in a closed loop in uniform magnetic field but torque is mostly non zero so if torque increases then force increases that's what i thought, sorry for putting it in that way, get me corrected if i am wrong again. any kind of help is appreciated.

23. Apr 5, 2017

### kuruman

You are correct in saying that the net force is zero in this case, but you misunderstood the question. If you place a flexible limp loop in a uniform magnetic field and run a current through it, equilibrium will be reached when the loop forms a circle with its plane perpendicular to the magnetic field. In this configuration the net torque and the net force on the loop will be zero. However, each arc segment ds on the loop will experience a magnetic force I ds B directed radially out. This will put the wire loop under tension T which is what this question is about.

24. Apr 5, 2017

### haruspex

Right, but I note this in the OP:
which seems to be saying the field is perpendicular to the axis of the loop. I assume the OP is wrong on that.

25. Apr 5, 2017

### kuruman

I would assume so too. OP mentions next
If OP is considering flux through the loop then we have a contradiction.