Problem involving Ampere's Law and Understanding

In summary, the conversation is about a problem involving Ampere's Law and a long straight copper wire with a circular hole. The question asks for the magnetic field at a point on the x-axis within the smaller wire, with the same current density but in the opposite direction. The solution involves setting up an Ampere loop within the smaller wire and using the equation μ_0∫J2πrdr to solve for the magnetic field.
  • #1
kbarry2295
3
0

Homework Statement


Hello,
physics.jpg

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!
 
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  • #2
Hello kbarry2295. Welcome to PF!
You get access to various symbols by clicking on the Σ in the green strip above the message box.

kbarry2295 said:

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.
 
  • #3
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!
 
  • #4
kbarry2295 said:
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!
 
  • #5
Thanks!
 

Related to Problem involving Ampere's Law and Understanding

1. What is Ampere's Law and how does it relate to understanding problems?

Ampere's Law is a fundamental law of electromagnetism that relates the magnetic field created by an electric current to the current itself. It is often used to understand and solve problems involving the magnetic field around a current-carrying wire or a loop of wire.

2. How do I apply Ampere's Law to a problem?

To apply Ampere's Law, you must first identify the closed loop over which the integral of the magnetic field will be calculated. Then, you must determine the current enclosed by this loop. Finally, you can use the equation B x dl = μ0I to calculate the magnetic field at any point along the loop.

3. Can Ampere's Law be used to calculate the magnetic field inside a current-carrying wire?

No, Ampere's Law can only be used to calculate the magnetic field outside of a current-carrying wire or loop. Inside the wire, the magnetic field is not constant and varies with distance from the center of the wire.

4. What is the significance of the constant μ0 in Ampere's Law?

μ0, also known as the permeability of free space, is a constant that relates the magnetic field to the electric current. It has a value of approximately 4π x 10^-7 N/A^2 and is essential in determining the strength of the magnetic field in Ampere's Law.

5. How does Ampere's Law differ from Gauss's Law?

Ampere's Law and Gauss's Law are both fundamental laws of electromagnetism, but they are used to understand different aspects of the electromagnetic field. Ampere's Law relates the magnetic field to the current, while Gauss's Law relates the electric field to the charge. Additionally, Ampere's Law is used to calculate the magnetic field around a current-carrying wire, while Gauss's Law is used to calculate the electric field around a charged object.

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