Force derived from magnetic energy of a current carrying spring

[etotheipi]

The magnetic energy of a current carrying spring, with $N$ turns, length $x$ and cross sectional area $A$, is $$E_m = \frac{\mu_0 N^2 I^2 A}{2x}$$The (negated) spatial derivative of this yields a quantity with dimensions of force,$$F = - \frac{dE_m}{dx} = \frac{\mu_0 N^2 I^2 A}{2x^2}$$How can we interpret this quantity, and where specifically does it act? Thanks!

 

[Dale]

How can we interpret this quantity, and where specifically does it act?

It doesn’t act anywhere. It has units of force but it isn’t force. It is the energy per unit length required to make the configuration longer.

 

[etotheipi]

Thanks, I see! The term is only applicable in a 'd'Alemberts principle' sense.

I wonder if it's possible to parameterise the spring with something like $\vec{r}(\lambda) = (a\cos{\omega \lambda}, a\sin{\omega \lambda}, b\lambda$) and calculate explicitly the magnetic force on any given segment of spring with Biot-Savart. Presumably in the equilibrium case that would be balanced by axial stresses in the adjacent pieces of spring. It would be interesting to see if that bears any resemblance to the equation above (though I would suspect not!).

Maybe the behaviour of a current carrying spring would be easier to analyse using a magnetic pressure approach. I'll have to do a bit more reading!

 

[Dale]

You probably want to look into the Maxwell stress tensor. It can be used to get forces from the fields.

 

[etotheipi]

You probably want to look into the Maxwell stress tensor. It can be used to get forces from the fields.

Cool, I'll give it a go and see how far I get. I've found a few references treating similar problems, although I haven't look at their maths in too much detail yet:

https://ocw.mit.edu/courses/nuclear...spring-2003/lecture-notes/forces_stresses.pdf

https://nvlpubs.nist.gov/nistpubs/jres/69C/jresv69Cn4p287_A1b.pdf

https://dqmp.unige.ch/senatore/wp-content/uploads/2018/12/Senatore_Lecture10_v2018.pdf

 

[kuruman]

It doesn’t act anywhere. It has units of force but it isn’t force. It is the energy per unit length required to make the configuration longer.

Wouldn't an alternate interpretation be the force that must be added to $kdx$ in order to stretch the spring by the same $dx$ when the current is turned on?

 

[Dale]

Wouldn't an alternate interpretation be the force that must be added to $kdx$ in order to stretch the spring by the same $dx$ when the current is turned on?

Possibly, but I am not certain because that would change the geometry of the spring. I can’t recall if that formula was based on some important assumptions about the geometry.

 

[etotheipi]

Wouldn't an alternate interpretation be the force that must be added to $kdx$ in order to stretch the spring by the same $dx$ when the current is turned on?

That could work; the total energy of the system is$$E = \frac{\mu_0 N^2 I^2 A}{2x} + \frac{1}{2}k(x-x_0)^2$$the work done by an external force stretching the spring over an extra incremental extension $dx$ is, if the current is assumed constant,$$F dx = dE = -\frac{\mu_0 N^2 I^2 A}{2x^2} dx + k(x-x_0)dx$$so the external force holding the spring in place is$$F = -\frac{\mu_0 N^2 I^2 A}{2x^2} + k(x-x_0)$$The assumption that the current is constant is slightly off, though, not least because this result seems to show that the external force required is actually less than without the current (when we know full well that the coils will attract each other and try to compress the coil axially).

We can improve this if we make some assumptions about the coil. For instance if it is superconducting, then $\Phi$ is constant always and we must have that $I = \frac{x}{x_0}I_0$, which gives us the relation$$E = \frac{\mu_0 N^2 I_0^2 A}{2x_0^2}x + \frac{1}{2}k(x-x_0)^2$$With the same method, we can now determine the external force required to hold the spring as$$F = \frac{\mu_0 N^2 I_0^2 A}{x_0^2} + k(x-x_0)$$which is much more like we expect!

 

[etotheipi]

Possibly, but I am not certain because that would change the geometry of the spring. I can’t recall if that formula was based on some important assumptions about the geometry.

You might be right that in a more advanced treatment we need to worry about the geometry, I just took the simplest approach and worked in the $x\gg r$ regime, so that we can take the magnetic field as uniform along the one edge of a rectangular closed curve that passes through the solenoid$$\oint_C \vec{B} \cdot d\vec{r} = Bx = \mu_0 N I \implies B = \frac{\mu_0 N I}{x}$$ $$\Phi = NBA = \frac{\mu_0 N^2 I A}{x} = LI \implies L = \frac{\mu_0 N^2 A}{x}$$ $$E_m = \frac{1}{2}LI^2 = \frac{\mu_0 N^2 I^2 A}{2x}$$