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Magnetostatics: Finding B field using Amperes Circuital Law

  1. Jan 16, 2016 #1
    I am preparing for an exam and I am going through a past paper which has solutions given for the questions but I need help understanding how the answer comes about. I suspect it may be just the algebra I don't get, but it may be the physics too.

    Wasn't sure if this was the correct forum either so move if needed.

    The question asks:
    A long non-magnetic cylindrical conductor with inner radius R1 and outer radius R2 carries a current I. Using Ampere’s law and some symmetry arguments, find the magnitude of the magnetic field within the conductor and outside it. Assume that the current I is uniformly distributed across the cross-section of the conductor.

    For outside, i.e. r > R2 then I personally got that its simply just
    [itex]\oint_c \vec{B} \cdot d\vec{r} = \vec{B} 2 \pi r = \mu_0 I \\
    \vec{B} = \frac{\mu_0 I}{2 \pi r}
    [/itex]

    Which was the correct answer in the solutions. But for R1 < r < R2 I struggle to understand where the final answers comes from as the solution given misses a few small steps.
    The solution given is:
    [itex]
    \oint_c \vec{B} \cdot d \vec{r} = \vec{B} 2 \pi r = \mu_0 \vec{j_0} \pi (r^2 - R_1^2) \\
    \vec{B} = \frac{\mu_0 I}{2 \pi r} \cdot \frac{r^2 - R_1^2}{R_2^2 - R_1^2 }
    [/itex]
    I think I understand the first line, where the ##\pi (r^2 - R_1^2) ## on the very RHS is the area of the cross-section of the contour, but I cant see how the second line follows from that? Any help is much appreciated!

    Thanks :)
     
  2. jcsd
  3. Jan 16, 2016 #2
    Inside and at R1 the current does not cause a field. At the outer radius R2, the field is equal to the field outside.
     
  4. Jan 16, 2016 #3
    Yeah I understand that bit, I just dont understand how algebraically to get to the second line from the first line r.e. R1 < r < R2. There was an image included but im having trouble getting it on here, but essentially it is a cylindrical wire such that from the very centre up to R1 is thin air, the the wire bit with current is between R1, and R2, The second bit of the question that I am stuck on, that has the solution I included, had their contour right in the middle between R1 and R2. I can see how the first line comes about, but algebraically cant see how to get to the second line.
     
  5. Jan 16, 2016 #4
    It is just going from the current density j to the current I.
     
  6. Jan 16, 2016 #5
    Ok yeah I understand that is what has been done. The bit I dont get is how the second line comes about algebraically, as it looks (to me) as if there has been a factor of ##\frac{1}{\pi (R_2^2 - R_1^2)} ## has been introduced. I cant see where that came from?
     
  7. Jan 16, 2016 #6
    That is the problem of studying solutions.
    Unless you have reasons to expect the same problem, it is better to work the problem yourself.
     
  8. Jan 16, 2016 #7
    First I do try the problems myself and then go back to check the solutions to see if I was correct. With this particular one one I had the first answer correct (for r > R2) but the second bit I was way off from what the solution says and could not see how the final answer comes about, hence asking on here.
     
  9. Jan 16, 2016 #8

    Philip Wood

    User Avatar
    Gold Member

    The second factor in the product in line 2 is the area of the conductor inside a circle of radius r expressed as a fraction of the whole cross-sectional area of the conductor. So that gives the fraction of the total current, I that is enclosed by the circle of radius r.
     
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