Magnetic energy inside a coaxial cable

[happyparticle]

Homework Statement

Magnetic energy inside a coaxial cable

Relevant Equations

$E_b = \frac{1}{2\mu_0} \int\int\int B^2 dv$

Hi,

I have to find the magnetic energy inside a coaxial cable of inner radius $a$ and outer radium $b$, $I = I$

By using Ampere's law
if $r<a$
$B = \frac{\mu_0Ir}{2\pi a^2}$

if $a<r<b$
$B = \frac{\mu_0I}{2\pi r}$

if $r>b$
$B = 0$

Then, the energy in a magnetic field $E_b = \frac{1}{2\mu_0} \int\int\int B^2 dv$

Since I have 2 different $B$ inside the cable, I'm not sure how to use this formula.

 

[hutchphd]

Your fields inside and out have different units. You should worry.

 

[happyparticle]

Your fields inside and out have different units. You should worry.

I'm not sure to understand. Of course the fields inside and out have different units since $B = 0$ if $r>b$
Is $E_{tot} = E_{b1} + E_{b2}$

For example,
if $a<r<b$ and for a length = l
$E_{b1} = \frac{\mu_0 I^2 l}{4\pi} \cdot ln\frac{b}{a}$

 

[hutchphd]

I see you have corrected it. Good.

 

[happyparticle]

I see you have corrected it. Good.

All right, but is it correct to say that $E_{tot} = E_{b1} + E_{b2}$

 

[hutchphd]

What is $b_1$ ? $b_2$? You need the integral as defined (notice $B^2$). You will need to divide up the integral and use the value for B in each region.

 

[happyparticle]

I have to find the energy stored by the magnetic field.

$b_1$ is the magnetic field where $r<a$ and $b_2$ is the magnetic field where $a<r<b$

So, I would like to know If I have to find $E_b$ in each region?

 

[hutchphd]

Oh, I see, yes you do. So there will be two (easy) integrals one for each region. You've already done the in between ...did you do r<a yet?

 

[hutchphd]

You have also called the regions several different ways! And look at he constants in the energy.

 

[happyparticle]

You have also called the regions several different ways! And look at he constants in the energy.

I didn't notice, sorry.

Thus,
$E_{tot} = \frac{\mu_0 I^2 L}{4\pi}(\frac{1}{4} + ln(\frac{b}{a}))$

Does it makes sense?

$r<a$
$E_{b2} = \frac{\mu_0 I^2 L}{16\pi}$

 

[hutchphd]

I think the units are correct ($\mu_0$ is Henrys /m ) . Check the r<a it seems off by a factor of 2 to me.

 

[hutchphd]

Yeah I guess its right...up to you now!

 

[happyparticle]

All right, so my $r<a$ is not off by a factor of 4?

 

[hutchphd]

My original objection was incorrect. I haven't looked at this problem before and the 1/4 seems weird to me but I don't see anything wrong so call it good.

 

[happyparticle]

All right, I trust you more than I trust myself.

Thanks!

 

[hutchphd]

When it comes to Algebra, probably not a good decision on your part.

 

[rude man]

EDIT: sorry about the mix-up.

What you are describing is not a coax cable. Looks like it's just the outer conductor of a coax cable.
Assuming that and a current flowing through the conductor, use Ampere's law as you describe for a<r<b, then your volume integral.
(Question to you: what is the field 0<r<a?). An easy integration (think cylindrical coordinales).
Don't know what all the E fields are doing in the foregoing posts.