Force on small wire segment in ring

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Homework Statement



Part (a): Describe what happens to current when ring contracts
Part(b): Find how much energy is stored, changing per unit time and rate of mechanical work being done
Part (c): Show the pressure is given by:
Part (d): Why is the force on a small wire segment not BIL?

Typed exam question 7 on a long straight solenoid with n turns per unit length carrying current I, covering magnetic field, force direction, stored energy and pressure, and a final part on the force on a small wire segment.


Homework Equations


The Attempt at a Solution



Part(d)

I tried a geometric reasoning, saying the horizontal components cancel out. So the net force is [tex]2Fcos(\frac{\theta}{2}) = 2F\sqrt{1-(\frac{L}{2r})^2} = BIL\sqrt{4 - (\frac{L}{r})^2}[/tex]

Hand-drawn diagram of a small arc of a circular wire with two tension forces F drawn tangent to the ends, a dashed radius to the centre, and angle theta marked at the centre.


Somehow I don't think that's the answer they are looking for..
 
If you take L << r for a segment of wire, then you can consider it as essentially straight when calculating the force on the segment.

I think the purpose of the problem is to show that the magnetic force on the segment is not BIL, where B is the field inside the solenoid. This can be surprising since BIL is the usual expression for the force.

As the solenoid expands in radius, the energy supplied by the current source goes somewhere. If you can identify where it goes, then you can use energy concepts to deduce the amount of magnetic force on a segment of wire.