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Magnetic field of a wire within a hollow cylinder.

  1. Oct 30, 2013 #1
    1. Problem statement.
    29yfsb5.jpg

    2. Known equations
    Amperes law, biot-savart law

    3. Attempt.

    Just taking steps 1 at a time, I first drew a diagram then I found Jo in terms of I and R, and since it is a non uniform current I know that to find the current I use

    J = I / A

    substitute A for 2(pi)rdr, and then integrate and solve for Jo,

    w98nc7.jpg

    Am I correct as of this far?
     
  2. jcsd
  3. Oct 30, 2013 #2

    TSny

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    Looks good to me. :smile:
     
  4. Oct 30, 2013 #3
    Awesome!, now this is what I did for the magnetic field inside the wire. Since there is s non uniform current I have to integrate the i.

    28k5w20.jpg
     
  5. Oct 31, 2013 #4

    TSny

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    You need a correction in your upper limit of integration. But you are close to setting it up correctly.

    (Also, you might want to be more careful in writing your equations. It is not true, for example, that ##B (2 \pi r) = \mu_0 J\, 2 \pi r dr##. The right side needs to include the integration in order for the left side to equal the right side.)
     
  6. Oct 31, 2013 #5
    Oh, my upper bound goes to little r. So that changes my value of

    B(inside wire) = (mu-knot) * I / 4(pi) *R(wire)

    For next question, magnetic field in space between cylinder and wire I believe would be the same as part a.

    Part c is a little tricky, since I am working with a non uniform and uniform current, so to tackle this would I just have to first find the current of the whole wire, then the current of the hollow cylinder then just sum them up?

    I also had trouble deriving the radius for Ra < r < Rb

    2nkhus3.jpg
     
  7. Oct 31, 2013 #6

    TSny

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    Yes, that's right.

    No, this can't be right. The magnetic field inside the wire should depend on the distance r from the center of the wire. So, you should recheck your calculation.

    No, the field between the wire and cylinder will have a different dependence on r than the field inside the wire.

    Yes, you will need to combine the current of the wire with the part of the current in the cylinder that is enclosed by your path of integration. I don't understand your calculation of ##I_{wire}##. The current in the wire is given to be ##I## so you don't need to calculate ##I_{wire}##. But your expression for the current in the cylinder that is enclosed by your path of integration of radius r looks correct.

    I'm not sure what you mean here by "deriving the radius".
     
  8. Oct 31, 2013 #7
    Oh I see that I made some arithmetic error for part a)

    What I got was

    B = (mu-knot * I * (r^4)) / (2pi * (R^5))

    Part b)

    I am unclear on how to solve this one, if I put a circle just outside of the wire wouldn't the current from the wire still be the same, which is what I found in part a? I think the only thing different will be I am dividing part a by a different r, lets say R(outside) so I have

    B = (mu-knot * I * (r^4)) / (2pi * (R^5) * R(outside))
     
  9. Oct 31, 2013 #8

    TSny

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    Yes, I think that's right.

    For part (b), pick any radius r between ##R## and ##R_a##. Apply Ampere's law to a circular path of that radius.
     
  10. Nov 1, 2013 #9
    Hmm, I am still unsure, with your suggestion I feel like my answer would still be the same since I am just looking at what is enclosed in my r, which is just the wire alone. But this is the steps I took,

    First I need to find the current of the wire, but this time I separated the r that was given and r'. Integrated that to R which is the radius of the whole wire. Then I did some algebraic manipulation to find the current in terms of I and r, and then plugged it in amperes law.

    avkdg5.jpg
     
  11. Nov 1, 2013 #10

    TSny

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    In part (a) you are looking for B(r) for some radius r that is less than R. So, you chose an "Amperian" path that is a circle of radius r. You had to find the current inside the path, so you integrated J(r') over the area enclosed by the path of radius r. If r' is the integration variable, then you let r' go from 0 to r.

    In part (b) you want B(r) for R < r < Ra. So, now the path is outside the wire. So, think about the current enclosed by this path. You should see that no integration is needed here. (Of course you could integrate over the whole wire if you wanted to, but you already essentially did this in your first post when you were finding an expression for Jo.)
     
  12. Nov 1, 2013 #11

    TSny

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    In your last post where you set up the integral ∫dI for the whole wire, you should of course get an answer that represents the total current in the wire. This is given to be I, so your integral should reduce to I. The integral should be ##\int_0^R J(r')2\pi r'dr'##, but you set it up as ##\int_0^R J(r)2\pi r'dr'## where you used ##r## rather than ##r'## for the argument of ##J##.
     
  13. Nov 1, 2013 #12
    Ohh... So for part b since the current is just I and only the wire has that current my magnetic field should just be..

    B = (mu-knot * I ) / ( 2*pi*r)
     
  14. Nov 1, 2013 #13

    TSny

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    Yes. That's it.
     
  15. Nov 1, 2013 #14
    Awesome! Now for part c, using the same logic as before the total current in the wire would just be I, and so I just need to find the distribution of I within the cylinder, which was found above. But this is what I have so far,

    2h71jd2.jpg
     
  16. Nov 1, 2013 #15

    TSny

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    Your first equation is correct if ##I_{cylinder}## is the part of the current in the cylinder that is enclosed in your dotted path and if ##I_{total}## is the total current in the cylinder. But remember, the total current in the cylinder is given to be ##I## (the same as in the wire).

    In your second equation, did you include the current in the wire?
     
  17. Nov 2, 2013 #16
    No I did not, I figured since the radius r is bigger than the wire, the current would've already been included in that I. To account for the current in the wire, do I just add in the current for the whole wire? Which is

    Mu-knot * I / 2 * pi * Rwire ??
     
  18. Nov 3, 2013 #17

    TSny

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    Yes. Whenever you use Ampere's law [itex]\oint \vec{B} \cdot d\vec{s} = \mu_0I_{enclosed}[/itex], the current on the right hand side is the total current passing through the path of integration that you are using on the left side.
     
  19. Nov 3, 2013 #18
    So my final answer for part C, since I am adding in the current of the wire my equation would be

    jj1ffk.jpg
     

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  20. Nov 3, 2013 #19

    TSny

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    Not quite. The very last term inside the brackets on the right is incorrect. Note how the first term in the brackets is dimensionless but the second term is not.

    Backing up to ##B 2\pi r = \mu_0 I_{enclosed}##, how would you express ## I_{enclosed}##?
     
  21. Nov 3, 2013 #20

    TSny

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    This expression is not the current due to the whole wire. Note that this expression does not have the correct dimensions for a current. It has the dimensions of a magnetic field.

    The current in the whole wire is given in the problem to be ##I##.
     
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