- #1

Steve Spence

- 8

- 3

- 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

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