Auxiallry Field in a Coaxial Cable

[TFM]

Is you rearrange the formula, that would be:

$$I_{f_encl} = 2\pi sH$$

?

TFM

 

[gabbagabbahey]

True, but that isn't what I meant...what are the free currents that are present in the question...which ones are enclosed by your loop?

 

[TFM]

well you have is going up the inner cable and down the outer, but no mention of free current. i wonder if they is no free current (but I doubt this)

TFM

 

[gabbagabbahey]

You have a current $I$ going up the inner cylinder and down the outer cylinder...are these currents free or bound?

 

[TFM]

well, since the current is causing the magnetization, I wonder if they are bound currents?

TFM

 

[gabbagabbahey]

No, they are free currents:

Free currents are the ordinary currents in wires and conductors.

Bound currents are the tiny loops of current inside the atoms or molecules of the insulator. The moving charges that make up bound currents are all bound to an atom or molecule.

The magnetization of a material is actually determined by both the free and the bound currents in a very complicated way, but can be calculated from the Auxiliary field (and hence from the free current) if you know the magnetic susceptibility of the insulator.

Since the currents given in the question move up and down the length of the cylinders, they are not bound current; they are free currents.

do you follow this?

 

[TFM]

Yes, that makes sense, bound currents are just the moving electrons round atoms (using the bassic), the free current is the current round a circuit, for example.

Thanks for cleasring that bit up for me.


TFM

 

[TFM]

So what would be best to do now?

?

TFM

 

[kitkat27]

Well you need to find the magnetization which is easy.
Then you will need to find the volume bound current density and then the surface bound current density. There are simple formulae for these but you will need to find the magnetization first.

 

[TFM]

Is this the rigght equation to use for magnetization:

$$M = \chi_M H$$

?

If so, the value for H is:

$$H = \frac{I_{f_{encl}}}{2\pi r}$$

Giving:

$$M = \chi_M (\frac{I_{f_{encl}}}{2\pi r})$$

?

TFM

 

[gabbagabbahey]

Is this the rigght equation to use for magnetization:

$$M = \chi_M H$$

?

If so, the value for H is:

$$H = \frac{I_{f_{encl}}}{2\pi r}$$

Giving:

$$M = \chi_M (\frac{I_{f_{encl}}}{2\pi r})$$

?

TFM

Close, remember $\vec{M}$ and $\vec{H}$ are vectors...This is part of why I wanted you to redo part (a) with me; so that you now know not just the magnitude of $\vec{H}$, but also its direction!...Your equations should be:

$$\vec{M} = \chi_M \vec{H}$$

and from part (a)

[tex]\vec{H}= \frac{I_{f_{encl}}}{2\pi s} \hat{\phi}[/itex]

...but what is $I_{f_{encl}}$ ?

 

[TFM]

Sorry, I just kinbd of copied the equations from a previouse post.

Anyway:,

$$I_{f_{encl}}$$ this is the current that is flowing through the Cable (up the middle and down the outer surface)

TFM

 

[gabbagabbahey]

There are two free currents in this question: (1) A current $I$ flowing up the inner cylinder and (2) A current $I$ flowing down the outer cylinder...In Ampere's Law, $I_{f_{encl}}$
is the total free current enclosed by your Amperian loop...So in terms of $I$ how much is that? Is it (a)0, (b)$I \over{2}$, (c)$I$, or (d)$2I$? and more importantly, why?

Remember, you cannot give your final solution in terms of quantities that were not part of the information given in the question; so you cannot give your final answer for $\vec{H}$ in terms of $I_{f_{encl}}$! It must be in terms of the quantities you are given.

 

[TFM]

(1) A current I flowing up the inner cylinder and (2) A current I flowing down the inner cylinder

Do you mean up the inner and down the outer?

My first answer would be 0, since the current going up would cancel the one going down, but this would surely mean that there would be no magnetic field?

So I am wondering if since it is the same current, would the answer be (d) 2I

 

[gabbagabbahey]

Do you mean up the inner and down the outer?

My first answer would be 0, since the current going up would cancel the one going down, but this would surely mean that there would be no magnetic field?

So I am wondering if since it is the same current, would the answer be (d) 2I

How much of the current is actually enclosed by your Amperian Loop?...You do remember what you used for your loop right? Are both of these free currents actually inside the loop?

Hint: it depends on how big the loop is

 

[TFM]

Well the Amperian loop will be of the size yto just have the inner cable, so I should say then that if I is the total current, the current going up the inner would be half that,

b) $$\frac{1}{2}I$$

TFM

 

[gabbagabbahey]

The current going up the wire is defined as $I$ in the problem statement, not $I \over{2}$.

 

[TFM]

That means that $$I_(f_{encl}}$$ must be equal to I

?

TFM

 

[gabbagabbahey]

Yes.What does that make $\vec{H}$ and $\vec{M}$?

 

[TFM]

That would make:

$$\vec{H} = \frac{I}{2\pi r}$$

and

$$\vec{M} = \chi_M \frac{I}{2\pi r}$$

?

TFM

 

[gabbagabbahey]

Why don't your expressions for $\vec{H}$ and $\vec{M}$ have a unit vector to indicate their direction?

 

[TFM]

So:

$$\vec{H} = \frac{I}{2\pi r}\hat{\phi}$$

$$\vec{M} = \chi_M \frac{I}{2\pi r} \hat{\phi}$$

?

TFM

 

[TFM]

The formulas I need for this question I believe are:

$$j_{ind} = M \times \hat{S}$$

for surface bound currents and:

$$J_{ind} = \nabla \times M$$

for volume.

Hence the reason for the directional vectors (Thanks)

So M vector is:

$$\vec{M} = \chi_M \frac{I}{2\pi r} \hat{\phi}$$

so this means that for the Surface bound currents:

$$j_{ind} = (\chi_M \frac{I}{2\pi r} \hat{\phi}) \times \hat{S}$$

I'm not sure the curl matrix for thus one?
and for Volume bound currents:

$$J_{ind} = \nabla \times (\chi_M \frac{I}{2\pi r} \hat{\phi})$$

With the curl Matrix:

S $$\phi$$ Z

d/ds d/d$$\phi$$ d/dZ

0 M 0

Latex doesn't do Matrices (not that I could see)

Does this look right?

?

TFM

 

[gabbagabbahey]

So:

$$\vec{H} = \frac{I}{2\pi r}\hat{\phi}$$

$$\vec{M} = \chi_M \frac{I}{2\pi r} \hat{\phi}$$

?

TFM

Yes.

 

[TFM]

Okay. So, the formulas I need for this question are:

$$j_{ind} = M \times \hat{S}$$

for surface bound currents and:

$$J_{ind} = \nabla \times M$$

for volume.

Hence the reason for the directional vectors (Thanks)

So M is:

$$\vec{M} = \chi_M \frac{I}{2\pi r} \hat{\phi}$$

so this means that for the Surface bound currents:

$$j_{ind} = (\chi_M \frac{I}{2\pi r} \hat{\phi}) \times \hat{S}$$

I'm not sure the curl matrix for thus one?



and for Volume bound currents:

$$J_{ind} = \nabla \times (\chi_M \frac{I}{2\pi r} \hat{\phi})$$

With the curl Matrix:

S $$\phi$$ Z

d/ds d/d$$\phi$$ d/dZ

0 M 0

Latex doesn't do Matrices (not that I could see)

Does this look right?

?

TFM

 

[gabbagabbahey]

The formulas I need for this question I believe are:

$$j_{ind} = M \times \hat{S}$$

for surface bound currents and:

$$J_{ind} = \nabla \times M$$

for volume.

Hence the reason for the directional vectors (Thanks)

So M vector is:

$$\vec{M} = \chi_M \frac{I}{2\pi r} \hat{\phi}$$

so this means that for the Surface bound currents:

$$j_{ind} = (\chi_M \frac{I}{2\pi r} \hat{\phi}) \times \hat{S}$$

I'm not sure the curl matrix for thus one?

This ^^^ is just a an ordinary cross product not a curl...you can find $\hat{\phi} \times \hat{s}$ using the right hand-rule.

and for Volume bound currents:

$$J_{ind} = \nabla \times (\chi_M \frac{I}{2\pi r} \hat{\phi})$$

With the curl Matrix:

S $$\phi$$ Z

d/ds d/d$$\phi$$ d/dZ

0 M 0

Latex doesn't do Matrices (not that I could see)

Does this look right?

?

TFM

The curl of a vector in cylindrical coordinates is actually:

$$\vec{\nabla} \times \vec{v}=\frac{1}{s} \begin{vmatrix} \hat{s} & s \hat{\phi} & \hat{z} \\ \frac{\partial}{\partial s} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ v_s & s v_{\phi} & v_z \end{vmatrix}$$

 

[TFM]

So for:

$$j_{ind} = (\chi_M \frac{I}{2\pi r} \hat{\phi}) \times \hat{S}$$

I need to do:

$$|\mathbf{A}\times\mathbf{B}| = AB \sin \theta$$

but we don't have an angle...?

?

TFM

 

[gabbagabbahey]

The normal to the cylindrical surface of the cable is just $\hat{s}$...The unit vectors $\hat{s}$,$\hat{\phi}$ and $\hat{z}$ are all orthonormal...I repeat, use the right-hand rule...what is $\hat{\phi} \times \hat{s}$?

 

[TFM]

Would it be $$\hat{z}$$

?

TFM

 

[gabbagabbahey]

Close, If I pick a spot on the surface and point my fingers in the direction of $\hat{\phi}$ and pont my palm in the direction of $\hat{s}$, my thumb points downward and so it is $-\hat{z}$...what does that make your surface current?

 

[TFM]

Minus I,

so,

$$j_{ind} = -(\chi_M \frac{I}{2\pi r} \hat{z})$$

?

TFM

 

[gabbagabbahey]

getting there...aren't you also told the value of $r$ for this surface?

 

[TFM]

You are given a as the r for the inner tube and b for the outer tube

?

TFM

 

[gabbagabbahey]

Yes, so doesn't that mean $$j_{ind} = \frac{-\chi_M I}{2\pi a} \hat{z}$$...since this surface current is the induced bound current at r=a ?Now how about your bound volume current, what are you getting for that?

 

[TFM]

I see.

So for the volume bound current:

$$\vec{\nabla} \times \vec{v}=\frac{1}{s} \begin{vmatrix} \hat{s} & s \hat{\phi} & \hat{z} \\ \frac{\partial}{\partial s} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ v_s & s v_{\phi} & v_z \end{vmatrix}$$

So should the middle bottom value be:

$$S \chi_M \frac{I}{2\pi r_{\phi}$$

?

(I tried putting it in the Latex matrix, but it just seemed to screw it up)

TFM