Auxiallry Field in a Coaxial Cable

[TFM]

$$\vec{\nabla} \times \vec{v}=\frac{1}{s} \begin{vmatrix} \hat{s} & s \hat{\phi} & \hat{z} \\ \frac{\partial}{\partial s} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ v_s & s \chi_M \frac{I}{2\pi r}_{\phi} & v_z \end{vmatrix}$$

Thats better, helps to makes sure the codes full and correct!

TFM

 

[gabbagabbahey]

Did you mean:

$$\vec{\nabla} \times \vec{M}=\frac{1}{s} \begin{vmatrix} \hat{s} & s \hat{\phi} & \hat{z} \\ \frac{\partial}{\partial s} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ 0 & \frac{\chi_M I}{2\pi } & 0 \end{vmatrix}$$

?

 

[TFM]

Yes, but you seemed to have lost the S for that bottom middle entry. Is that not required?

TFM

 

[gabbagabbahey]

$$s \left( \frac{\chi_M I}{2\pi s } \right) =\frac{\chi_M I}{2\pi }$$

 

[TFM]

Oh, I see

since V_s and V_z = 0

so the curl will be:

$$\frac{1}{S} ((\frac{\partial 0}{\partial \phi} - \frac{\partial \frac{\chi_M I}{2\pi }}{\partial z})\hat{s} + (\frac{\partial 0}{\partial s} - \frac{\partial 0}{\partial z})\hat{\phi} + (\frac{\partial \frac{\chi_M I}{2\pi }0}{\partial s} - \frac{\partial 0}{\partial \phi})\hat{z})$$

Would this not equal 0, since

a) $$\frac{\partial \frac{\chi_M I}{2\pi }}{\partial z})$$= 0

and

b) The curl is for flows in a circle, and this is radial

?

TFM

 

[gabbagabbahey]

Yes, it's zero; and so there is no bound volume current in this scenario.

Now, that you know all of the bound and free currents in the problem, you can apply Ampere's law in terms of $\vec{B}$ (instead of H); and find the magnetic field that way. You should get the same answer as you had before.

 

[TFM]

So would we use:

$$\oint B\cdot dl = \mu_0 I_{encl}$$

where I_encl is the total current (ie free + bound)?

TFM

 

[Vuldoraq]

Hi,

Sorry to muscle in but I'm doing the same problem. If you apply Ampere's law using the sum of the free and bound currents then you end up with a sign difference compared to the earlier result. I got this because the bound surface current is running anti-parallell to the free current. To find the bound current I took the line integral of the bound surface current density. Where have I gone wrong?

$$\hat{B}=\frac{\mu_{0}*(1-\chi_{m})*I}{2\pi * a} \hat{\phi}$$

 

[TFM]

I am not getting quite the same answer :

$$\oint B.dl = \mu_0 I_{encl}$$

$$I_{encl} = I + \frac{-\chi_M I}{2 \pi a} + 0$$

$$\oint B.dl = \mu_0 (I + \frac{-\chi_M I}{2 \pi a})$$

$$B2\pi a = \mu_0 (I + \frac{-\chi_M I}{2 \pi a})$$

$$B = \frac{\mu_0 (I + \frac{-\chi_M I}{2 \pi a})}{2\pi a}$$

?

TFM

 

[gabbagabbahey]

Hi,

Sorry to muscle in but I'm doing the same problem. If you apply Ampere's law using the sum of the free and bound currents then you end up with a sign difference compared to the earlier result. I got this because the bound surface current is running anti-parallell to the free current. To find the bound current I took the line integral of the bound surface current density. Where have I gone wrong?

$$\hat{B}=\frac{\mu_{0}*(1-\chi_{m})*I}{2\pi * a} \hat{\phi}$$

Okay, since the insulator is actually in the region a<r<b, the outward normal at the r=a surface is actually $-\hat{s}$ and so for the inner surface, $\vec{M} \times \hat{n}=\vec{M} \times -\hat{s}=\frac{+\chi_M I}{2\pi a} \hat{z}$

Don't forget to also calculate the surface current for the r=b surface...which direction is the outward normal in that case?

The outer bound surface current won't contribute to $\vec{B}$ though, because it won't be enclosed by your Amperian loop in the a<r<b region...but you should still calculate it anyways.

 

[gabbagabbahey]

I am not getting quite the same answer :

$$\oint B.dl = \mu_0 I_{encl}$$

$$I_{encl} = I + \frac{-\chi_M I}{2 \pi a} + 0$$

$$\oint B.dl = \mu_0 (I + \frac{-\chi_M I}{2 \pi a})$$

$$B2\pi a = \mu_0 (I + \frac{-\chi_M I}{2 \pi a})$$

$$B = \frac{\mu_0 (I + \frac{-\chi_M I}{2 \pi a})}{2\pi a}$$

?

TFM

See my last post about correcting the minus sign...

Also, the bound surface current is spread out uniformly over the r=a surface...what is the circumference of that surface? The total bound current enclosed by your loop is not just $$I_{encl} = \frac{+\chi_M I}{2 \pi a}$$

 

[TFM]

Okay so:

$$B = \frac{\mu_0 (I + \frac{\chi_M I}{2 \pi a})}{2\pi a}$$

$$B = \frac{\mu_0 (I + \frac{\chi_M I}{2 \pi a})}{2\pi a}$$

The Circumference of the loop is $$2 \pi a$$

Making the area of the inner tube

$$2 \pi a z$$

where z is the height of the tube.

Does this help?

TFM

 

[gabbagabbahey]

The total current is not $I + \frac{\chi_M I}{2 \pi a}$ but rather $$I_{f_{total}}+I_{b_{total}}=I + j_{ind} \cdot (2 \pi a)$$

do you follow?

 

[TFM]

I see the induced current is done for just one line going up in the middle cable, but since the cable isn't an infinitesimally thin, it has a circumference, so we multiply the surface current around the circumference of the wire to give the current for the whole cable.

So

$$I_{f_{total}}+I_{b_{total}}=I + j_{ind} \cdot (2 \pi a) = I + \frac{\chi_M I}{2\pi a}*(2 \pi a)$$

This gives:

$$I_{tot} = I + \chi_M I$$

and so:

$$B = \frac{\mu_0 (I + \chi_M I)}{2 \pi a}$$

Edit:

Looking back,we were asking to get:

$$B = \mu_0(1+\chi_m)(\frac{I_{f_{encl}}}{2\pi r})$$

$$B = \frac{\mu_0(1+\chi_m)I_{f_{encl}}}{2 \pi a}$$

From part a)

Which is similar, but not quite the Same - we currently have I, whereas a) had $$I_{f_{encl}}$$

TFM

 

[gabbagabbahey]

Close, Ampere's law should be used for a loop of radius $a<s<b$ not $s=a$ which gives:

$$B = \frac{\mu_0 (I + \chi_M I)}{2 \pi s}=\frac{\mu_0 I(1 + \chi_M )}{2 \pi s}$$

as expected.

 

[TFM]

Is this for the first part?

Because The answer for the first part was originally:

$$B = \mu_0(1+\chi_m)(\frac{I_{f_{encl}}}{2\pi r})$$

with the r, I added the a in just now

loooking back, however, I fiound that we said that $$I_{f_{encl}} = I$$

Making:

$$B = \frac{\mu_0(1+\chi_m)I_{f_{encl}}}{2 \pi a}$$

become:


$$B = \frac{\mu_0(1+\chi_m)I}{2 \pi a}$$

Which equals

$$B = \frac{\mu_0(I+I\chi_m)}{2 \pi a}$$

Which is the answer we were looking for

?

Is this correct?

TFM

 

[gabbagabbahey]

why are you using $a$?... $a$ is the radius on the inner surface...you want to find the magnetis field at any point in the region $a<s<b$ where $s$ is the distance from the axis to the field point.

...And you should also include the direction vector in your final expression for $\vec{B}$

 

[TFM]

So:

$$\vec{B} = \frac{\mu_0(I+I\chi_m)}{2 \pi r} \hat{\phi}$$

but for b) we had:

$$\vec{B} = \frac{\mu_0 (I + \chi_M I)}{2 \pi a}$$

TFM

 

[gabbagabbahey]

No, for (b) you had that...I didn't; see post #85 and look closely...and you really don't like using the letter $s$ do you?

 

[TFM]

I see, so that was for b) then. Sorry, my Mistake

so:

$$B = \frac{\mu_0 (I + \chi_M I)}{2 \pi s}=\frac{\mu_0 I(1 + \chi_M )}{2 \pi s}$$

and you really don't like using the letter $$s$$ do you?

I try, but I've been brought up on $$r$$

TFM

 

[gabbagabbahey]

Usually $r=\sqrt{x^2+y^2+z^2}$ which is the distance from the origin to the field point...In this case you want the distance from the z-axis which is usually given as the cylindrical coordinate $s=\sqrt{x^2+y^2}$...You can call that $r$ if you like, but you need to make sure you state that it is the distance from the axis, not the distance from the origin.