Compression of a superconducting spring

[etotheipi]

Homework Statement

What is the change in length of a spiral superconducting spring, which has $N$ turns, radius $R$, natural length $x_0$ and spring constant $k$, when an initial current $I_0$ flows through it?

Relevant Equations

N/A

Let the length of the spring be $x$, so that the extension in any given configuration is $\delta = |x_0 - x|$. The magnetic flux through such a coil is $$\Phi = \frac{\mu_0 I N^2 A}{x}$$The fact that the coil is superconducting means that the flux linked will always remain constant even with changes of current, which means that the current must satisfy$$I(x) = \frac{I_0 x}{x_0}$$Since $\Phi = LI$ we have that the inductance of the coil is$$L(x) = \frac{\mu_0 N^2 A}{x}$$and consequently the magnetic energy of the coil, $E_m = \frac{1}{2}LI^2$, is$$E_m = \frac{\mu_0 N^2 I_0^2 A}{2x_0^2}x$$Now when a current flows, the spring compresses (adjacent coils attract), so I thought to try and equate the decrease in magnetic energy to the increase in spring energy,$$\frac{1}{2}k(x_0-x)^2 = \frac{\mu_0 N^2 I_0^2 A}{2x_0^2}(x_0 - x)$$However this yields $$\delta = x_0 - x = \frac{\mu_0 N^2 I_0^2 A}{x_0^2}$$which is apparently incorrect. The solution manual uses a force approach which I am struggling to understand. They say that from the magnetic energy, we can define a "magnetic force of contraction" $F_0$ where$$F_0 = \frac{E_m}{x} = \frac{\mu N^2 I_0^2 A}{2x_0^2}$$and say that equilibrium occurs when the elastic force $k(x_0 - x)$ "balances" this magnetic force of contraction, which gives an answer that is half of mine.

I wondered what the flaw in my method is, but would also like to understand how their method works. Specifically, on what sub-system are the magnetic force of contraction and spring force acting, and are balanced? Thanks!

 

[haruspex]

Energy won't do it. As with any spring, suddenly applying a force leads to overshoot, then a damped oscillation, ending at the new equilibrium. And the shift in equilibrium is the amplitude of undamped oscillation, which is half the total range.

 

[hutchphd]

You also get the same 1/2 when you suddenly supply a fixed battery potential to a capacitor. Very fundamental effect. (and as I mentioned once before maybe related to fluctuation dissipation theorem ?)

 

[etotheipi]

Energy won't do it. As with any spring, suddenly applying a force leads to overshoot, then a damped oscillation, ending at the new equilibrium. And the shift in equilibrium is the amplitude of undamped oscillation, which is half the total range.

I see! In that case, what system are we taking to be in equilibrium in order to equate the two forces? Is it, for instance, the final turn of the coil?

 

[haruspex]

I see! In that case, what system are we taking to be in equilibrium in order to equate the two forces? Is it, for instance, the final turn of the coil?

How about differentiating your expression for the magnetic energy wrt x?

 

[etotheipi]

How about differentiating your expression for the magnetic energy wrt x?

That would be the quantity $F_0$, which I'm still struggling to interpret (specifically I'm not sure where it acts). The tension in the spring would be $k(x_0 - x)$.

 

[haruspex]

That would be the quantity $F_0$, which I'm still struggling to interpret (specifically I'm not sure where it acts). The tension in the spring would be $k(x_0 - x)$.

You don't need to worry about what the force represents. What you know is that at equilibrium a small change dx will transfer a little energy between the spring and the magnetic field. Virtual work, and all that. So just differentiate your energy balance equation and solve for x.

 

[etotheipi]

You don't need to worry about what the force represents. What you know is that at equilibrium a small change dx will transfer a little energy between the spring and the magnetic field. Virtual work, and all that. So just differentiate your energy balance equation and solve for x.

That's interesting, so we would say that for our equilibrium length $x$, the energy of the configuration after the system comes to rest (having been damped) is$$E = \frac{1}{2}k(x_0-x)^2 + \frac{\mu_0 N^2 I_0^2 A}{2x_0^2}x$$If the system is in stable equilibrium, then an infinitesimal change in length $\delta x$ in either direction will not change the energy $(\frac{dE}{dx} = 0)$, so$$\delta E = -k(x_0 - x) \delta x + \frac{\mu_0 N^2 I_0^2 A}{2x_0^2} \delta x = 0$$and that let's us solve for $x$. It works!

As a side-note, I think it is slightly misleading that the book calls that term a force, even though it would be dimensionally consistent. Without knowing what system a force acts on I'm very hesitant to do any analysis involving them.

Thanks, again, for the help!

 

[haruspex]

it is slightly misleading that the book calls that term a force, even though it would be dimensionally consistent. Without knowing what system a force acts on I'm very hesitant to do any analysis involving them.

We could set k=0 and instead apply some external force to hold the coils at a certain total length. That is balancing the force from the magnetic field, so the force n question is that net force on the coil from that field.

 

[etotheipi]

We could set k=0 and instead apply some external force to hold the coils at a certain total length. That is balancing the force from the magnetic field, so the force n question is that net force on the coil from that field.

Right but it's still not too obvious what system is in equilibrium.

If we return to the original scenario and consider one coil somewhere in the middle, then that's acted upon, vaguely, by two tensions on either side (which will be equal if the spring is light), and two magnetic ($BIL$)-type forces from adjacent coils. It seems fair to assume that we only need consider the magnetic force from the two adjacent coils and not the rest behind them, as a first approximation. But for most such systems we choose, it doesn't seem like we can extract anything useful from the equilibrium equations, because in each we have two equal and opposite tensions and two equal and opposite magnetic forces, which is a bit like getting 0 = 0.

So I thought maybe we'd need to consider the final coil, which would only be acted upon by a tension on one end and an opposing magnetic force from that same end, which we could equate for the equilibrium condition. But it's not obvious that our expression $F_0$ is this required magnetic force...