1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Problem involving Ampere's Law and Understanding

  1. Nov 15, 2015 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    Amperes Law
    Integral of Bdl= current enclosed * mu_0

    3. 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!!
  2. jcsd
  3. Nov 15, 2015 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

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

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

    For the small wire, the path for your integral should be a circle centered on the axis of the small wire.
  4. Nov 15, 2015 #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!
  5. Nov 15, 2015 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.

  6. Nov 15, 2015 #5
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted