Magnetic energy density, and pressure due to magnetic force

[Steve Spence]

Homework Statement

What is the pressure exerted on the outer conductor of a coaxial cable due to the current in the inner conductor?

Relevant Equations

$$\mu_{B}=\frac{B^{2}}{2\mu_0}=159\ Pa$$

Hi,
The problem I am working on requires me to work out the the pressure on the outer conductor of a coaxial cable due to the current on the inner one.
This cable carries a dc current of 5000 Amps on the inner wire of radius 2 cm. The outer cylindrical wire of radius 5cm carries the return current of 5000 amps in the opposite direction.
I had this idea that the value for the magnetic energy density at the outer cylinder is the same as the pressure on it from the current on the inner wire. I'm not quite sure why, but this coincidence seems to occur in other problems involving pressure. I decided to try and work it out this way to see how my answer compares to the official one :

The magnetic field strength at the outer cylinder is
$$B=\frac{\mu _{0}I}{2\pi r}=\frac {4\pi\times 10^{-7}\times 5000\ amps}{2\pi\times 5\ cm}=20\ mT$$
Plugging this value into the energy density formula I get :
$$\mu_{B}=\frac{B^{2}}{2\mu_0}=159\ Pa$$
This value is wrong, but it is half the value of the correct answer. It's a bit frustrating as this seemed to be a quicker and more convenient way of calculating the pressure, but I don't understand why I am only half right ! Of course, I don't fully understand the relationship between energy density and pressure other than that they share units. The official solution (which I understand) is as follows :

The magnetic field created by the inner conductor exerts a force of repulsion on the current in the outer sheath. The strength of this field is 20.0 mT. Consider a small rectangular section of the outer cylinder of length L and width W. It carries a current of
$$(5000\ amps)\left (\frac{w}{2\pi(5\ cm)} \right )$$
and experiences an outward force
$$F=ILBsin\theta = \frac{(5000\ amps)w}{2\pi(5\ cm)}L(20\times 10^{-3}T)sin \ 90$$
The pressure is then
$$P=\frac{F}{A}=\frac{F}{wL}=\frac{(5000\ amps)(20\times 10^{-3}\ T)\cancel{wL}}{2\pi(5\ cm)\cancel{wL}}=318\ Pa$$

My question is therefore why is the energy density at the outer conductor only half the value of the pressure ?

B=μ0I2πr=4π×10−7×5000 amps2π×5 cm=20 mT

 

[haruspex]

I don't know if this matches an actual explanation, but it is common for an energy to have a factor 1/2. Consider e.g. $\frac 12mv^2$, $\frac 12kx^2$. The corresponding force in each case has no such factor.
I note in this case you have $\frac 1{2\mu_0}B^2$. Tracking how that factor 1/2 arises might answer your question.

 

[Gordianus]

Recently we faced a similar problem: the force between platas of charged capacitor. Let me repeat (badly) Feynman's reasoning. You ha assumed the outer wire is completely immersed in the inner wire's field. But, just outside the outer wire, the field drops to zero. Feyman said "lets use the average value of those fields" and so he explained the 1/2 factor.

 

[Steve Spence]

Thanks Gordianus, this kind of makes some sense, but sorry I'm still a bit confused. Why don't both methods give the same answer, as they both assume the same value of the field (20mT) ?

 

[Gordianus]

The field Is 20 mT just inside the outer wire and drops to zero just outside. What value would you choose?

 

[Steve Spence]

I would choose 20mT. I am happy with this value for the field just inside the outer wire, and therefore for the pressure on it : 318 Pa. But I thought the energy density (perhaps clumsily) was the same as pressure. Why is that only 318/2=159 Pa ?

 

[Gordianus]

What Is the field AT the outer wire?
We're facing the same problem we had with the Parallel plate capacitor.

 

[Steve Spence]

OK, the magnetic field at the outer wire is 10mT (20+0)/2. The energy density at the wire is now 79.6 Pa, but I want it to be 318 Pa ! The energy density at any radius between the the outer wire and inner wire is never more than 159 Pa, so my idea that energy density = pressure must be inaccurate ?

 

[Gordianus]

The idea is: go back to the líne where you computed the force F on the sector of length L and width w. You replaced B= 20 mT. Try with the averaged value we've discussing.

 

[Steve Spence]

ok :

$$F=ILBsin\theta = \frac{(5000\ amps)w}{2\pi(5\ cm)}L(10\times 10^{-3}T)sin \ 90$$
The pressure is then
$$P=\frac{F}{A}=\frac{F}{wL}=\frac{(5000\ amps)(10\times 10^{-3}\ T)\cancel{wL}}{2\pi(5\ cm)\cancel{wL}}=159\ Pa$$

So the energy density for the averaged field value of 10mT is 79.6 Pa, and the pressure is 159 Pa ie they still are not equal .

 

[Kavya Chopra]

Here's a perspective on it: consider a hollow cylinder with a current flowing along its length, now consider a small element on this cylinder. It's easy to see that it would experience a force due to the other elements on the same wire. I like to think of this as a small element (which is capable of generating its own magnetic field) immersed in an external magnetic field, experiencing the force due to the external field only. (It can't experience a force due to itself, evidently, but it can experience a force due to the other elements that constitute the system it is a part of). In your solution, the value you calculated is the force experienced by an element of the outer wire due to the inner wire AND due to the other elements on the same wire. And you only need the first term in your answer

 

[Steve Spence]

Hi Kavya,
Which of the 2 solutions are you referring to as erroneous, the one using the energy density equation or the one using P=F/A=FLB/A?

 

[haruspex]

Recently we faced a similar problem: the force between platas of charged capacitor. Let me repeat (badly) Feynman's reasoning. You ha assumed the outer wire is completely immersed in the inner wire's field. But, just outside the outer wire, the field drops to zero. Feyman said "lets use the average value of those fields" and so he explained the 1/2 factor.

I feel you may have misunderstood Feynman's argument.
There are two ways to think about the field in a capacitor.
Each plate generates a field $\frac{\sigma}{2\epsilon_0}$ each side, the fields having opposite sign.
Between the plates the fields add, making $\frac{\sigma}{\epsilon_0}$; beyond them they cancel.
1. For each plate, the total field varies from $\frac{\sigma}{\epsilon_0}$ on one side to 0 on the other, giving an average $\frac{\sigma}{2\epsilon_0}$.
2. For each plate, its action on itself is immaterial. We only care about the force on it due to the field from the other, i.e. $\frac{\sigma}{2\epsilon_0}$.

 

[Kavya Chopra]

Hi Steven, I meant to say that the answer of 318 Pa using ilB seems right to me

 

[haruspex]

The magnetic field strength at the outer cylinder is
...
Plugging this value into the energy density formula I get :

You only seem to have considered the energy density due to one wire. The force is a consequence of both currents, no? OTOH, I can't see that it is just a matter of adding them. What if there were no current in the outer wire?
(You may discern that I know very little about magnetic fields.)

 

[Tom.G]

How about considering two single conductors, each carrying 5000A in opposite directions (supply and return), spaced 5cm apart.

Conductor A acts on conductor B with a force. Does Conductor B also act on conductor A with a force?

Hope it helps!

Cheers,
Tom

 

[Steve Spence]

Thanks for your help guys, but I'm still thinking about this one...may take a while !

 

[TSny]

Thanks for your help guys, but I'm still thinking about this one...may take a while !

The key point brought out by @Kavya Chopra is that the question is asking for the pressure on the outer cylinder due to the magnetic interaction between the current in the inner conductor and the current in the outer cylinder. This pressure is only a part of the total pressure experienced by the outer cylinder. The current in the outer cylinder alone creates a pressure on the outer cylinder.

Suppose we have just the outer cylinder carrying a current $I$. Inside the cylinder, $B = 0$. On the outer surface of the cylinder, $B = B_0 = \mu_0 I/(2 \pi a)$, where $a$ is the radius of the outer cylinder.

The force per unit area on the outer cylinder is $\vec f_1 = -\frac { B_0^2} {2 \mu_0} \hat r$, where $\hat r$ is a unit vector pointing radially outward. The pressure acts inward.

Next consider the case where we have current $I$ in both the outer cylinder and the inner conductor with the currents in opposite directions.

Now, $B = 0$ everywhere outside the outer cylinder and $B = B_0 = \mu_0 I/(2 \pi a)$ at the inner surface of the outer cylinder. We then get a net force per unit area on the outer cylinder that points outward: $\vec f_2 = \frac { B_0^2} {2 \mu_0} \hat r$. This net pressure is the sum $\vec f_1 + \vec f_3$ where $\vec f_1$ is the pressure shown in the first figure and $\vec f_3$ is the pressure due to just the interaction between the current in the inner conductor and the current in the outer cylinder. The question is asking for $\vec f_3$.

Thus, $\vec f_3 = \vec f_2 - \vec f_1 = 2\left( \frac { B_0^2} {2 \mu_0} \right) \hat r$.

 

[Steve Spence]

Hallelujah ! Many thanks TSny, that is a brilliant explanation.

My mistake had been assuming the net pressure, f2, was zero and/or not appreciating the double minus in f2-f1.

 

[rude man]

The key point brought out by @Kavya Chopra is that the question is asking for the pressure on the outer cylinder due to the magnetic interaction between the current in the inner conductor and the current in the outer cylinder. This pressure is only a part of the total pressure experienced by the outer cylinder. The current in the outer cylinder alone creates a pressure on the outer cylinder.

Suppose we have just the outer cylinder carrying a current $I$. Inside the cylinder, $B = 0$. On the outer surface of the cylinder, $B = B_0 = \mu_0 I/(2 \pi a)$, where $a$ is the radius of the outer cylinder.

View attachment 288796

The force per unit area on the outer cylinder is $\vec f_1 = -\frac { B_0^2} {2 \mu_0} \hat r$, where $\hat r$ is a unit vector pointing radially outward. The pressure acts inward.

Next consider the case where we have current $I$ in both the outer cylinder and the inner conductor with the currents in opposite directions.

View attachment 288797

Now, $B = 0$ everywhere outside the outer cylinder and $B = B_0 = \mu_0 I/(2 \pi a)$ at the inner surface of the outer cylinder. We then get a net force per unit area on the outer cylinder that points outward: $\vec f_2 = \frac { B_0^2} {2 \mu_0} \hat r$. This net pressure is the sum $\vec f_1 + \vec f_3$ where $\vec f_1$ is the pressure shown in the first figure and $\vec f_3$ is the pressure due to just the interaction between the current in the inner conductor and the current in the outer cylinder. The question is asking for $\vec f_3$.

Thus, $\vec f_3 = \vec f_2 - \vec f_1 = 2\left( \frac { B_0^2} {2 \mu_0} \right) \hat r$.

Hm.
Magnetic pressure applies only when a high-$\mu$ material is in the presence of a B field in a low- $\mu$ region, with a finite $\bf B$ gradient. So if the cylinder were made of high-mu material of thickness $d$ then the pressure on it would be $P = \bf B \nabla \bf B d/\mu$. If there is B on one side and B=0 on the other side, $\nabla \bf B = B/d$ in the radial direction. Then we can say that $P = B^2/2\mu_0$ but this formula assumes an infinite-mu cylinder.

But the problem did not state that the outer cylinder is a high-mu material. Default assumption is that it's just $\mu_0$.

Must be missing something.

 

[TSny]

Then we can say that $P = B^2/2\mu_0$ but this formula assumes an infinite-mu cylinder.

This simple formula for the "magnetic pressure" can be obtained from the Maxwell stress tensor for the specific case where the magnetic field is tangent to the surface of interest, as in this problem with the cylinder. The formula does not require the cylinder to have any specific magnetic properties. In particular, it does not require the material of the cylinder to have a large permeability $\mu$.

 

[rude man]

This simple formula for the "magnetic pressure" can be obtained from the Maxwell stress tensor for the specific case where the magnetic field is tangent to the surface of interest, as in this problem with the cylinder. The formula does not require the cylinder to have any specific magnetic properties. In particular, it does not require the material of the cylinder to have a large permeability $\mu$.

The Maxwell stress tensor theory seems to be dealing with electromagnetism, implying time-changing electric and magnetic fields, which are not given in this problem. Just dc.

While I admit to not being conversant with this stress tensor I am certain that what I wrote about the "suction" caused by a B field with gradient on a high-mu body is correct, and the fact that the pressure formula (B^2/2mu) is identical to this stress tensor derivation is more than a bit remarkable.

If you're familiar with the "suction" concept perhaps you can explain that remarkable coincidence. It would be very interestging and illuminating. And - no insult intended at all! I really would like to know.

 

[TSny]

The Maxwell stress tensor theory seems to be dealing with electromagnetism, implying time-changing electric and magnetic fields, which are not given in this problem. Just dc.

The stress tensor is useful whether or not the fields are changing with time. For time dependent fields, the tensor is helpful in analyzing the flow of momentum in the system. For static fields, the tensor can be used to determine electric and magnetic forces, as in this problem.

While I admit to not being conversant with this stress tensor I am certain that what I wrote about the "suction" caused by a B field with gradient on a high-mu body is correct, and the fact that the pressure formula (B^2/2mu) is identical to this stress tensor derivation is more than a bit remarkable.

If you're familiar with the "suction" concept perhaps you can explain that remarkable coincidence. It would be very interestging and illuminating. And - no insult intended at all! I really would like to know.

I am not very familiar with the physics of magnetic materials. So, I'm not confident that I can be of any help here. It doesn't seem too surprising to me that you could get a result $B^2/(2 \mu)$ for the pressure in a magnetic material and also get $B^2/(2 \mu_0)$ for the pressure in a non-magnetic material. If the pressure can be expressed in terms of $B$, then by dimensional analysis I would expect the pressure to be proportional to $B^2/\mu$.

 

[rude man]

Good points, especially the dimensional argument. This is a new one on me, for sure. Homework ahead. Would it even be covered in Jackson or Purcell I wonder.

In any case my analysis was worthless since the direction of the B field was not normal to the surface.

The "suction" I described is pretty elementary, It's why a hi-mu object is attracted to a magnet while a low-mu isn't.

In case you're interested: at an interface between high-mu and low-mu regions, a pressure exists normal to the interface plane from the hi-mu to the lo-mu region. The B field must have a finite gradient $\nabla \bf B$..

So a cube of sides d of say iron facing a permanent magnet pole will see a net force of $\bf F = dB\nabla B/2\mu_0$, both sides seeing a pull force with the net force as above. This assumes infinite $\mu$ for the cube. If the cube's $\mu = \mu_0$ e.g copper then there is no force.

All this is derivable from virtual work and magnetic energy density.