Auxiallry Field in a Coaxial Cable

[TFM]

Homework Statement

A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility χm (see Figure). A current I flows down the inner conductor and returns along the outer one, in each case the current distributes itself uniformly over the surface.

a)

Find the magnetic field in the region between the tubes.

b)

As a check, calculate the magnetization and the (surface and volume) bound currents. Confirm that (together with, of course, the free current) they generate the correct magnetic field.

Homework Equations

Just Section A at the moment:

I have a feeling that these equations are useful (hinted in a similar question in textbook)

\oint H .dl =I_{f_encl}

B = \mu H

\mu = \mu_0(1+\chi_m)

The Attempt at a Solution

B = \mu_0(1+\chi_m)H

Does this look okay, what could be a good thing to do next?

TFM

 

[Redbelly98]

You can use the integral expression you wrote to find H. Evaluate the integral around a circular loop.

Once you have H, your final expression can be used to get B.

 

[TFM]

So:

\oint H .dl =I_{f_encl}

I think I am doing the wrong integral, but would it be

H \pi^2 =I_{f_encl}

Or would this be for dA?

TFM

 

[Redbelly98]

No, not sure where you are getting pi^2 from.

The integral is evaluated going around the circumference of a circle.

 

[TFM]

I meant for that to be pi r squared, that's an area integral.

So would it be:

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

?

TFM

 

[Redbelly98]

Yes. So B = ?

 

[TFM]

So:

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

which rearranges to give H:

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

and since:

B = \mu_0(1+\chi_m)H

B equals:

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

B = \mu_0(1+\chi_m)H

B equals:

B = \mu_0(1+\chi_m)(\frac{I_{f_{encl}}}{2\pi r})Does this look correct?

?

TFM

 

[Redbelly98]

Yes.

 

[TFM]

Excellent, thanks.

So for part B,

Question

As a check, calculate the magnetization and the (surface and volume) bound currents. Confirm that (together with, of course, the free current) they generate the correct magnetic field.

Useful Equations:

Magnetization Equation:

M = nm

Surface bound current Equation:

j_{ind} = M x \hat{S}

Volume bound Current Equation:

J_{ind} = \nabla x M

Edit These twi above equations should be curl, not x

I am not sure if these are the right equations to use, though?

Where would be the best place to start?

?

TFM

 

[TFM]

I have a feeling that this equation is very useful for the second part:

M = \chi_M H

where M is the Magnetization

Since we have worked out H to be:

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

Is thuis a useful equation?

TFM

 

[TFM]

So for this question, would the equation for magnetization be:

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

does this look right?

thus would the surface bound current be:

j_{ind} = \chi_M (\frac{I_{f_{encl}}}{2\pi r}) \times \hat{S}

And the Volume bound current:

J_{ind} = \nabla \times (\chi_M (\frac{I_{f_{encl}}}{2\pi r}))

Is this right? If so, what might be the next logical step?

?

TFM

 

[Redbelly98]

It has been a while since I worked on this stuff, so I don't know about part (b). Maybe somebody else can help out here?

 

[gabbagabbahey]

Hmmm... I think you'd better go back to part (a) for a moment, before I can help you through part (b) :

Let's make it a little more obvious which quantities are vectors and which are scalars; and what exactly you are integrating over...Ampere's Law for the Auxiliary Field \vec{H} is:

\int_{\mathcal{P}} \vec{H} \cdot \vec{dl}=I_{f_{enc}}[/itex]<br /> <br /> Where \mathcal{P} is an Amperian loop. While you seem to (thanks to RedBelly&#039;s assistance) have chosen the correct Amperian loop, you have neglected to explicitly state your choice and justify it&#039;s usefulness...<b>Hint:</b> How does the cylindrical symmetry in this problem help you here? Which direction must \vec{H} point in?<br /> <br /> Once you have explicitly stated what your Amperian loop is, you should be able to explicitly determine the amount of free current enclosed in it, I_{f_{enc}} in terms of the I given in the problem.

 

[TFM]

Where {P} is an Amperian loop. While you seem to (thanks to RedBelly's assistance) have chosen the correct Amperian loop, you have neglected to explicitly state your choice and justify it's usefulness...Hint: How does the cylindrical symmetry in this problem help you here? Which direction must \vec{H} point in?

Well the Magnetic field will decrease in strength the further from the wire you get.

the \vec{H} Should point in the direction of be radially outwards using right hand rule on diagram?

TFM

 

[gabbagabbahey]

No, \vec{H} is not radially outwards...what are the free currents in this problem, and which direction do they point? Point your thumb in the direction of the current and your fingers will curl in the direction of the field...are you familiar with cylindrical coordinates s,\phi and z? If you orient your coordinate system so that the axis of the cylinder is along the z-axis (\hat{z} direction), then what direction does that give you for \vec{H}?

 

[TFM]

The free current is the current traveling in the wire. it is going upwards in the inner wire, and downwards in the outer cylinder/cable

I am slightly confused by:

Point your thumb in the direction of the current and your fingers will curl in the direction of the current

If your thumb points in the direction of the current, how can the curl of your fingers also be the current?

TFM

 

[gabbagabbahey]

Sorry, that was a typo; the second "current" was supposed to be "field"...I've edited my post.

...In terms of the cylindrical unit vectors, \hat{s},\hat{\phi} and \hat{z}, which direction do the currents flow (Don't worry about whether they flow in the positive or negative direction)?

 

[TFM]

Judging by this the field is flowing in circles, so in terms of cylindrical coordinates, it must be:

\hat{\phi}

TFM

 

[gabbagabbahey]

Yes, now what is the direction of \vec{dl} If you choose a coaxial circle as your Amperian loop?

 

[TFM]

Would it be:

\hat{s}

Since it is for a circle of radius r through a full 360 degrees?

TFM

 

[gabbagabbahey]

If it were \hat{s}, then that would mean that as you went around the loop, for a small change in the magnitude |\vec{dl}| , s (or r as you have it written) would change by an amount |\vec{dl}| (Since that change would be in the \hat{s} direction)...Does that sound right for a circle of constant radius s?

 

[TFM]

No it doesn't, it must be:

\hat{\phi}

TFM

 

[gabbagabbahey]

Yes, Good! ...so what is the dot product \vec{H} \cdot \vec{dl}?

 

[TFM]

Would it be

H\hat{\phi}

?

TFM

 

[gabbagabbahey]

Well, \vec{H}=H\hat{\phi} and \vec{dl}=dl\hat{\phi}, so \vec{H} \cdot \vec{dl}=Hdl (\hat{\phi} \cdot \hat{\phi})

What is the dot product of the unit vectors \hat{\phi} \cdot \hat{\phi}?

 

[TFM]

Would that be 1?

?

TFM

 

[gabbagabbahey]

Yes, since after all; they are unit vectors...What does that make \vec{H} \cdot \vec{dl}?

 

[TFM]

Since:

\vec{H}=H\hat{\phi}

and

\vec{dl}=dl\hat{\phi}

Would that make

\vec{H} \cdot \vec{dl}

1?

TFM

 

[gabbagabbahey]

Nope,

<br /> \vec{H} \cdot \vec{dl}=Hdl (\hat{\phi} \cdot \hat{\phi})=Hdl *(1)=Hdl<br />

Now what about the magnitude of |\vec{dl}|=dl...what is that for the circle? (remember, the only quantity that changes along the path is \phi)

 

[TFM]

Would this be:

\hat{s}

Since the magnitude will decrease the further out you get from the current?

TFM

 

[gabbagabbahey]

no, dl is just a magnitude; specifically it is the infinitesimal displacement as you move along the Amperian loop. Given that your loop is a circle of constant radius s, dl=sd\phi

...do you follow?

 

[TFM]

Ah yes, that makes sense.

TFM

 

[gabbagabbahey]

Good, so now you have:

\int_{\mathcal{P}} Hsd\phi=I_{f_{encl}}

What should the limits of integration be? Does H vary over this integral? Why or why not? How about s? How much free current is enclosed by the loop?

 

[TFM]

The limits should be from 0 to 2pi

H should be constant over the integral, since the Magnetic field will only decrease radially, not round the circle - the top of the circle will have the same H field as the bottom.

TFM

 

[gabbagabbahey]

Yes, great! ...This is of course the whole point of choosing a coaxial circle as your Amperian loop: this means that you can pull the Hs out of the integral since they are constant over \phi...And so you get:

Hs \int_0^{2\pi} d \phi =2 \pi s H=I_{f_{enc}} \Rightarrow H=\frac{I_{f_{enc}}}{2 \pi s}

...which is what you had before (Although hopefully you now understand a little better Why you get this result)

How about my last question: How much free current is enclosed by the loop?

 

[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)

\vec{H}= \frac{I_{f_{encl}}}{2\pi s} \hat{\phi}[/itex]<br /> <br /> ...but what <i>is</i> 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