Energetic argument for the tension in a current loop

[pop_ianosd]

TL;DR Summary

Considering the energy of a current loop as a function of current intensity and inductance, it seems that the energy decreases with decreasing the size of the loop. However, we would expect an outward force on the current elements of the loop.

The energy stored in a current loop equals $\frac{LI^2}{2}$. From a dimensional argument, it follows that the inductance grows with the size of the loop. This would mean that, if we assume the current stays constant, enlarging the loop would require external positive work, so, the force associated with the magnetic energy - the electromagnetic force, I guess - should point inwards towards the loop. This however does not really correspond with the consideration that diametrically opposed current elements repel each other, since they are antiparallel currents.
I think the issue with this thought experiment is the assumption of constant current. Somehow, keeping the current constant while increasing the size of the loop is supposed to feed some energy into some other place than the field-energy of the current loop, but I can't identify where.

 

[jartsa]

I think the issue with this thought experiment is the assumption of constant current. Somehow, keeping the current constant while increasing the size of the loop is supposed to feed some energy into some other place than the field-energy of the current loop, but I can't identify where.

Well, I would say that keeping the current constant while increasing the size of the loop feeds some energy into the field-energy of the current loop. Why do you think the energy can not go there?We could generate electricity from the motion of the expanding loop and feed that electricity to the loop. This would keep the field-energy of the loop constant. Current would decrease, inductance would increase.

 

[pop_ianosd]

Sorry, I had the wrong picture in mind, and the statement that energy must be fed into some other place is misleading. What I see is happening if you increase the loop while somehow maintaining constant current is, indeed, as you are saying, that you have an increase in the field-stored energy. However you are also extracting positive work, because of the outward electromagnetic force. So there must be energy coming in from an external source. Is this what you have in mind as well? Now what I would want to work out is how come the energy increase in the magnetic field equals the work output.

For the case where there is no external energy source, I can see now that the current should decrease. I'm curious how you reasoned that the current should decrease though. My reasoning is that the angular momentum, which is proportional to current times radius, is conserved. I would disagree that the energy in the field remains constant. I think it decreases, corresponding to the mechanical work produced by the expansion.

 

[jartsa]

For the case where there is no external energy source, I can see now that the current should decrease. I'm curious how you reasoned that the current should decrease though. My reasoning is that the angular momentum, which is proportional to current times radius, is conserved. I would disagree that the energy in the field remains constant. I think it decreases, corresponding to the mechanical work produced by the expansion.

Well, I said that we put any energy that leaves the loop back into the loop. That is why the energy of the loop was unchanged, and as the inductance of the loop did increase, and energy is proportional to inductance, the current had to decrease to compensate for the increased inductance.

(First the loop underwent an accelerating expansion, then the loop met some strong permanent magnets arranged on a circle, at that point induction caused all of the kinetic energy of the loop to be converted to electric energy. That is one way how it could have happened. )

 

[pop_ianosd]

Well, I said that we put any energy that leaves the loop back into the loop.

Ah, now I get your point. Yes, it makes sense now, thank you!

 

[alan123hk]

Since the inductance is proportional to the loop area, suppose that the inductance $L=cA$ , where c is a constant and A is the loop area.

If the inductor current is constant $I$ , then when the loop area changes, the energy difference $δE=(\frac 12 c I^2) δA$

On the other hand, the voltage of the inductor $V=\frac {dθ} {dt}$ , since $θ=cAI$ , $V=cA\frac {dI} {dt} +cI \frac {dA} {dt} =cI \frac {dA} {dt}$, the energy difference $δE= \int VI \, dt =∫cI^2 \frac {dA} {dt} \,dt=(cI^2) δA$

The actual energy difference is $(\frac 12 c I^2) δA$, but the electric energy provided from the outside is $(cI^2) δA$. If I am not mistaken, then why dose this happen? Why does the outside provide twice as much electric energy?

I'm a bit confused. So far, I can't think of any other explanations other than thinking that part of the energy may be lost through mechanical energy during the edge movement of the loop area.

 

[pop_ianosd]

So far, I can't think of any other explanations other than thinking that part of the energy may be lost through mechanical energy during the edge movement of the loop area.

That is what I understand to be happening as well. It must be the case since, as mentioned in the original question, there is an outward electromagnetic force. Half the energy provided by the external power source gets stored in the magnetic field, and half goes into mechanical work. Your computation gives indeed the mathematical derivation of this fact. To me it is a rather surprising fact that it works out this way - that the energy gets split into two.

I have the impression that the inductance is proportional not to the area, but to a linear dimension, like the radius of the loop, if we consider it circular. The intuition behind this is that you do increase the flux, but you decrease the field in the central regions of your area. The dependence can be argued for by dimensional analysis:

\begin{align*}
\nabla \times \mathbf{B} &= \mu_0 \mathbf{j} \Rightarrow [ \mathbf B] = [ \mu_0] \times [ I] \times m^{-1}\\
[ L] &= \frac{[ \mathbf B]\times m^2}{[ I]} = [ \mu_0] \times m
\end{align*}

But even with this, your argument holds, just by replacing $A$ with, say, a radius $R$.

It is nice to see that you get the same mechanical energy dissipation for a current loop with no external energy source, upon expansion of the radius. Assuming 0 resistance:
$$-\frac{d\theta}{dt} = \oint{\mathbf{E}\cdot d\mathbf{r}} = \oint{\rho \mathbf{j}\cdot d\mathbf{r}} = 0 $$

$\theta = LI = cRI \Rightarrow \frac{d}{dt} (RI) = 0 .$ This is what I previously argued for, in a rather awkward manner, by conservation of angular momentum.
The mechanical power is then $$ P_\text{mech} = -\frac{dE_\text{mag}}{dt} = -\frac{1}{2} \frac{d}{dt} (cRI^2) = -\frac{1}{2} c RI \frac{dI}{dt}$$
Use that $I = \frac{RI}{R}$ so $\frac{dI}{dt} = -\frac{RI}{R^2} \frac{dR}{dt}$ and finally get
$$P_\text{mech} = \frac{1}{2} cI^2 \frac{d}{dt} R$$
Integrate this over time and you get the same result for the produced work.

This is of course quite expected, since the ratio between the work and the displacement corresponds to the force, whose value should not be different if we consider a process where we keep the current constant by means of an external energy source, or if we let the current drop. The force is there without any process.

 

[alan123hk]

Very good, if we agree with this conclusion, that is, when we keep the inductor current constant, and then change the geometry of the inductor to change the inductance value, so the magnetic field energy stored in the inductor changes accordingly, assuming the rate of change magnetic field energy is dE/dt, then there will inevitably be another mechanical energy loss rate dE/dt of equal value.

This makes people think about the practical application of this conclusion, and I guess it may be useful in the relatively simple cases, such as a cylindrical inductor, we seem to be able to use this concept to find its approximate magnetic self radial force and circumferential force. If we apply Lorentz force law and Biot-Savart law to solve this problem, it must be much more complicated and difficult.

 

[alan123hk]

The following is an interesting attempt to find the radial force of two different inductors.

Since $E = \frac 1 2 L I^2~$, $~\frac {dE} {dt} = LI\frac {dI}{dt} + \frac {I^2}{2} \frac {dL}{dt} = \frac {I^2}{2} \frac {dL}{dt} ~~(I=constant)$Circular loop inductor $~~L_1 \approx u_0r\ln \left( \frac {8r} {a} -2 \right)~~$ where r=loop radius, a=wire radius
$\frac {dE} {dt} = F_1 \frac {dr}{dt} = \left( \frac {2u_0r I^2} {4r-a} \right) \left ( \frac {dr} {dt} \right) ~~~\Rightarrow~~~ F_1 = \frac {2u_0r I^2} {4r-a}$
Let I=1A, r=5mm, a=0.5mm, F₁=6.4e-7N, which is so small that it can be ignored.Cylindrical ferrite inductor $~~ L_2 \approx \frac {u_0u_r N^2 \pi r^2} {\lambda} ~~$ where u_r= relative permeability, r=radius, $~\lambda$ = length, N=number of turns
$\frac {dE} {dt} = F_2 \frac {dr}{dt} = \frac {u_0 u_r \pi r (NI)^2} {\lambda} \frac {dr} {dt} ~~~\Rightarrow~~~ F_2 = \frac {u_0 u_r \pi r (NI)^2} {\lambda}$
Assuming $~~$ I=1A $~~$ u_r=1000 $~~$ N=100 $~~$ r=5mm $~~$ $\lambda$ = 10mm, then F₂ = 19.7N
The radial force per turn is 19.7/100 = 0.197N (Based on an assumption that the radial force is evenly distributed along the axis of the inductor).
Tangential force along the wire = $\frac {0.197N} {2\pi}$ = 0.031N = 3.2 gram-force
I believe this value is still much smaller than the tensile force that a copper wire with a diameter of 0.3mm can withstand.