Problem involving Ampere's Law and Understanding

[kbarry2295]

Homework Statement

Hello,

This is a multi-part problem but I am stuck on just the third part, but I am a little confused in general with Ampere's Law. I know that the equation is mu_0*Ienc=Integral of B dl, but this problem really confused me.

A long straight copper wire lies along the z axis and has a circular cross section of radius R. In addition, there is a circular hole of radius a running the length of the copper wire, parallel to the z axis. The center of thhe hole is located at x=b with a+b<R.

The first part was What is the current density J and the answer is that J=1/pi(R^2-a^2)

The second part said if the hole is not there but has the same current density J, which I was able to solve through Ampere's law and the answer is B=1/2 mu_0 J*r

Now the third part is where I have had troubles. It saws suppose the wire of radius R is not there, and in place of the hole there is a smaller circular wire of radius a, with center at x=b, carrying the same current density J but into the page. What would the magnetic field be on the x-axis at a distance b<r<a+b from the origin, inside the smaller wire?

Homework Equations

Amperes Law
Integral of Bdl= current enclosed * mu_0[/B]

The Attempt at a Solution

Alright, so my thoughts on this problem were to ignore the large copper wire loop like it said, and I drew a Ampere loop inside the smaller circle. On the left side of Ampere's law is B*2pir.
On the right hand side, Ienc= to the integral of Jda, but J is claimed to be the same so I can take that out of the integral, and then I just have the integral of da.

Now, I set up the integration to go from the middle of the small loop, which is B, to my Ampere Loop, which is R, to get a current enclosed of Jpi(r-b)^2 and I set that equal to the other side. But I am not getting the right answer which my textbook says it is B=0.5(mu_0)J(b-r). I am not really concerned with the direction at this point, more on where am I messing up in my thinking of Ampere's Law and why am I not getting to this same answer? Any help or suggestions is greatly appreciated!

 

[SammyS]

Hello kbarry2295. Welcome to PF!
You get access to various symbols by clicking on the Σ in the green strip above the message box.

Homework Statement

Hello,

[ ATTACH=full]91897[/ATTACH]
This is a multi-part problem but I am stuck on just the third part, but I am a little confused in general with Ampere's Law. I know that the equation is mu_0*Ienc=Integral of B dl, but this problem really confused me.

A long straight copper wire lies along the z axis and has a circular cross section of radius R. In addition, there is a circular hole of radius a running the length of the copper wire, parallel to the z axis. The center of thhe hole is located at x=b with a+b<R.

The first part was What is the current density J and the answer is that J=1/pi(R^2-a^2)

The second part said if the hole is not there but has the same current density J, which I was able to solve through Ampere's law and the answer is B=1/2 mu_0 J*r

I suppose this second part actually asked for the magnetic field at a distance, r, from the z-axis, where 0 ≤ r ≤ R .

Now the third part is where I have had troubles. It says suppose the wire of radius R is not there, and in place of the hole there is a smaller circular wire of radius a, with center at x=b, carrying the same current density J but into the page. What would the magnetic field be on the x-axis at a distance b<r<a+b from the origin, inside the smaller wire?

Homework Equations

Amperes Law
Integral of Bdl= current enclosed * mu_0[/B]

The Attempt at a Solution

Alright, so my thoughts on this problem were to ignore the large copper wire loop like it said, and I drew a Ampere loop inside the smaller circle. On the left side of Ampere's law is B*2pir.
On the right hand side, Ienc= to the integral of Jda, but J is claimed to be the same so I can take that out of the integral, and then I just have the integral of da.

Now, I set up the integration to go from the middle of the small loop, which is B, to my Ampere Loop, which is R, to get a current enclosed of Jpi(r-b)^2 and I set that equal to the other side. But I am not getting the right answer which my textbook says it is B=0.5(mu_0)J(b-r). I am not really concerned with the direction at this point, more on where am I messing up in my thinking of Ampere's Law and why am I not getting to this same answer? Any help or suggestions is greatly appreciated!

For the small wire, the path for your integral should be a circle centered on the axis of the small wire.

 

[kbarry2295]

Hello, thank you. I am still a little confused because wouldn't my circle be entirely inside the small wire? The initial circle that is in the picture is what I did for the first part.I set up the right side of the equation by μ_0∫J2πrdr by taking integration limits from b to r, and then divided by 2πr, but I am not getting the answer of 1/2μ_0*J(b-r). Am I missing something?
Thank you!

 

[SammyS]

Hello, thank you. I am still a little confused because wouldn't my circle be entirely inside the small wire?

Yes. They might not have asked this very clearly, but ... They ask for the magnetic field at a point on the x-axis with r between b and b+a. r is the distance from the z-axis. That puts you at a point on the x-axis anywhere from the center of the hollow region (small wire) to its right hand edge.

Much like using Gauss's Law for E-field, Ampere's Law is only useful for computing the B-field when taking advantage of the symmetry of the situation. You need to integrate over a path on which the magnitude of the B-field is constant as well as being suitably oriented to the path.

The initial circle that is in the picture is what I did for the first part.I set up the right side of the equation by μ_0∫J2πrdr by taking integration limits from b to r, and then divided by 2πr, but I am not getting the answer of 1/2μ_0*J(b-r). Am I missing something?
Thank you!

 

[kbarry2295]

Thanks!