A closed loop of super-conducting wire has a self-inductance of L1,
and a carries a current of I1. Mechanical work W is performed on the
wire to deform it to a new physical configuration. The inductance and
current at the new configuration are respectively L2 and I2.

Questions:

1. Does the value of I2 depend on the manner in which the wire is deformed? Why?
2. Can I2 be written in terms of L1, L2, and I1? If so, what is the
expression? Explain how you got the result.

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Clarification of Question by gerlach-ga on 30 Aug 2004 21:31 PDT

A closed loop of super-conducting wire has a self-inductance of L1,
and a carries a current of I1. Mechanical work W is performed on the
wire to deform it to a new physical configuration. The inductance and
current at the new configuration are respectively L2 and I2.
 
Questions (clarified):

1. Given a fixed starting and fixed ending physical configuration of
the wire (and hence a fixed L1 and L2), and given the starting current
I1, does the value of I2 depend on the manner in which the wire is
deformed from the staring to the final configuration? Why?
2. Can I2 be written solely in terms of L1, L2, and I1? If so, what is the
expression, and explain how you got the result. If not, explain why.

Let W2 represent the new energy condition after work W has been added
to the initial condition.  The initial condition is just I1^2*L1 so
therefore:

W2 = I1^2*L1 + W

Since we know the total work in the superconductor is W2, and that
since it is a superconductor it does not lose energy (R = 0 ohms) we
can say that the energy doesn't change after the wire is deformed.  So
therefore:

W2 = I2^2*L2

Solving for I2 we get:

I2 = sqrt(W2/L2)

and substituting the original equation for W2:

I2 = sqrt((I1^2*L1 + W)/L2)

Does that answer the question? or did I miss something subtle in the problem?

The answer to part 1 is sort of.  Since the manner in which the wire
is deformed will determine the new inductance.  If the area remains
constant, then the flux density will too.  Total flux must remain
constant because any change would be counteracted by a change in
current (this is accounted for in the change in the inductance value).

If the wire is deformed in such a way that the area changes, then the
inductance will also change, and therefore, so will the current.  But
if the area doesn't change, then no net energy will be required to
deform it.


I said "Sort of" above because you don't need to know how the wire was
deformed if you can measure the change in inductance.  But you could
presumably develop an expression for the change in area (and therefore
inductance) by looking at the total work required to deform the wire. 
The *manner* would then be unimportant as long as the total work was
known.

I'll leave the derivation of that expression to whoever wants to
answer this question for real...

Dear upstartaudio-ga,

Thanks for the comments. Your expression for I2 does not answer the
question, because I'm looking either for an expression for I2 which
depends solely on L1, L2, I1 (see the clarification to the question),
or a proof that no such expression exists.

There is one little nit. The expression for energy stored in an
inductor is .5*L*I^2, not L*I^2.

With this nit corrected, the expression you wrote for I2 becomes:

I2 = sqrt( (I1^2*L1 + 2*W ) / L2 )	Eq1

Quantities L1 and L2 are determined solely by the geometry of the wire
configuration at the start and finish. There are integral expressions
for the inductance given the wire geometry.

It is my conjecture that I2 is also determined solely by the start and
finish geometry, and the starting current I1. One reason I suspect
this is because it checks out for the case I1=0. If I1=0, then it
takes no work to deform the wire (W=0), and therefore I2=0, regardless
of how the wire was deformed in going from start to finish.

However if I1~=0, then it will take work to deform the wire, and Eq1
states whatever work W we do will show up in the magnetic energy of
the system. An equivalent way of stating question #2 is ?How much work
W needs to be done to go from start to finish?? Once you have W in
terms of L1, L2, and I1, you can plug it into Eq1, and the problem is
solved.

Is this reasonable?