Magnetic force between two halves of a current carrying wire

Question

This is a question from a book "Pathfinder for Physics Olympiad and JEE Advanced"

Consider a long current carrying cylindrical conductor of radius r. Current density j inside the conductor is uniform over it's cross section. Deduce suitable expression for force of interaction per unit length between two halves that are obtained by dividing the conductor by a plane containing the axis of the conductor.

Answer:

$\frac{1}{3}μ_0j^2r^3$

While there certainly exists a way to directly evaluate the force, I wanted to do it using magnetic pressure.

Method 1

We will consider a small ring of width dr at a distance r from the axis of the wire. I am considering the length of the wire as 1 unit so that I don't have to deal with force per unit length later on. The change in magnetic pressure over this dr length would be balancing the magnetic force on this element. So

$$dP × 2πr = idl × B$$

$$dP × 2πr = (2πrdr × j) × \frac{μ_0jr}{2}$$

Integrating this expression and using the fact that pressure must be zero at outer surface of the conductor, we arrive at this expression for pressure as a function of r

$$P = \frac{μ_0j^2(R^2 - r^2)}{4}$$

Calculating force is now easy, we will take small strips at a distance r from axis of width dr and length 1 unit and integrate force on this strip throughout the entire surface. That is

$$dF = PdA$$

$$dF = \frac{μ_0j^2(R^2 - r^2)}{4}dr$$

Integrating this expression from limits 0 to R gives us the desired result.

Method 2

Now I tried writing the magnetic pressure directly using the energy density expression that is $\frac{B^2}{2μ_0}$. My teacher pointed out that this expression for pressure we had derived in the case of a plane sheet of wire carrying current so may not be applicable in this case. So I gave this counterargument that we can take a small element of this wire and write

$$dP = \frac{BdB}{μ_0}$$

This should be applicable since a small element of this wire must behave as a sheet of wire. To find pressure I just need to integrate this back with suitable limits, which gives me the function of pressure as

$$P = \frac{μ_0j^2(R^2 - r^2)}{8}$$

Which is exactly half of the pressure derived earlier.

Afterthoughts

I then discussed this with one of my friends and he skimmed through internet to find that there is a thing named magnetic tension which is also contributing in magnetic pressure. He confirmed that the other half of the pressure is coming from this term only. I am not so acquainted with magnetic tension so I tried searching for some other method.

I then came across this post Is energy density and pressure fundamentally the same thing?

So my final idea was that if I am somehow able to deduce how energy is varying with volume then I should be able to calculate the pressure.

That is if $$ U = kV^n $$

Then $$\frac{dU}{dV} = n\frac{U}{V}$$

This suggested me that in our case energy is proportional to the square of volume. For then only the energy density is half of magnetic pressure. However I immediately faced a difficulty that the position of the volume in space also mattered while calculating energy associated with it. That is a volume $ dV $ at a distance $ r $ and a distance $ 2r $ would have different energies associated with them. Then I further thought that maybe if I took the volume in form of annular discs subtending certain angle at the axis then I might be able to accommodate that. Unfortunately in that case energy varied linearly since magnetic field is not varying in tangential direction.

Questions

1. Why can't we directly use energy density to evaluate pressure? (Just to confirm that magnetic tension is indeed the thing contributing here)
2. Is there a way to avoid using magnetic tension and dwell on the idea of finding proportionality of energy with volume?
3. When will this extra term be coming into our expression of magnetic pressure and is there some similar term in calculation of electrostatic pressure as well?

Answer 1

1. Why can't we directly use energy density to evaluate pressure? (Just to confirm that magnetic tension is indeed the thing contributing here)

Because density of magnetic energy and its variation with position is not related to mechanical pressure in any simple universal way.

It is true that in some cases, we can calculate magnetic force on a current-carrying slab as

$$ \mathbf F_M = \frac{1}{2\mu_0} (B_2^2 - B_1^2) A\mathbf n, $$

where $B_2>B_1$ are magnetic fields on both sides of the slab, $A$ is area of the slab, and $\mathbf n$ is unit vector perpendicular to the slab, pointing in direction from the stronger to the weaker field. Thus it is as if the space with magnetic field acts on the material slab, from both sides, with pressure whose value is the same as magnetic energy density (this is just an imperfect mechanistic interpretation of the magnetic force). And if the slab does not move, there has to be additional force of the same magnitude and opposite direction, acting on the slab, such as the force of gradient of mechanical pressure $P$.

This formula works when the current flows in straight lines (the slab is planar). It also works when the current flows in curved lines (bent slab), provided the difference of magnetic fields on both sides (and thus current in the slab) is significant (non-infinitesimal).

In the case you tried to use this method, where the slab of thickness $dr$ is curved and carries only infinitesimal current, it however fails. This combination of properties breaks one assumption in derivation of the above formula for magnetic force (the assumption that jump in magnetic field between two half spaces equals to $\mu_0 \lambda$, where $\lambda$ is current in between, per unit length). Thus the formula is unjustified and this "magnetic pressure" method is inapplicable. Well, one could derive another formula for this case, where magnetic field varies continuously across the wire, taking into account the fact that jump of magnetic field is $\mu_0 \lambda /2$ here, and the effective pressure of magnetic field is twice the density of magnetic energy, but it is obvious that the formula above and the idea of magnetic pressure equal to energy density is not generally valid.

The correct magnetic force on a material medium element carrying current can be determined, in general, using the "$BIL$" formula for magnetic force on conductor, as you did in the method 1. For infinitesimal element of volume $dV$, this is

$$ d\mathbf F_M = \mathbf j \times \mathbf B_{ext} ~dV $$ where $\mathbf B_{ext}$ is external magnetic field acting on the element. For infinitesmal element of volume $Lrd\varphi dr$, we would have $$ dF_M = B j Lr\,d\varphi\, dr. $$

However, note that in your calculation in Method 1, it makes no sense to add magnetic forces, pointing in different directions, as if they were real numbers; thus the factor $2\pi$ is pointless there. But the result is nevertheless correct.

Then, to find mechanical pressure in the medium, we use the idea that if the current element is at rest, there has to be a counteracting mechanical force on it due to the surrounding material medium. If the element is planar, the counteracting force is due to mechanical pressure $P$, and this allows finding $P$. Thus the method 1 is more universal, and gives the correct answer.