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Ampere's Law-cylindrical shell

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A conducting cylindrical shell has it's axle on Ox axis, inner radius R1 and outer radius R2. If the volume current density is given by the formula [itex]\vec{j}[/itex](r)=J0[itex]\frac{r}{R1}[/itex][itex]\hat{i}[/itex] where J0 is a constant and r the distance from the cylinder's axle, calculate the magnetic field [itex]\vec{B}[/itex] : 1) Everywhere in space and 2) On the Oy. Oz axis in a position y>R2 and z>R2



    3. The attempt at a solution

    Using ampere's law we can easy say that for r<R1 the magnetic field is zero beacause there is no current passing through the amperian loop. For R1<r<R2 the region that the not uniform density takes place.
    I draw an amperian loop of radius r. The whole point as i see it is to calculate the Ienc . So step by step what i did :


    1) dI=J*da ,da=2πrdr so dI= [itex]\frac{2π}{R1}[/itex]j0 r2dr

    2)so Ienc= [itex]\frac{2π}{R1}[/itex]j0 ∫r2dr from R1 to r.

    3)so Ienc= [itex]\frac{2π}{3R1}[/itex]j0 (r3-R13)

    4) Thus the amperes law gives ∫Bdl=μ0[itex]\frac{2π}{3R1}[/itex]j0 (r3-R13) and B2πr=μ0[itex]\frac{2π}{3R1}[/itex]j0 (r3-R13) so B=μ0[itex]\frac{2π}{3R1}[/itex]j0 [r2-(R13/r)]

    5) For r>R2, outside the cylinder, the amperian loop will include the total current which the same as Ienc but for r=R2 so Itotal= [itex]\frac{2π}{3R1}[/itex]j0 (R23-R13)

    6) And applying the ampere's law again into a loop radius r>R2 it gives B= (μ0j0/3r)[(R23/R1)-R12]

    7) It seems that in the Oy,Oz axis the magnetic field is the same and outside the cylinder it decreases with a ratio 1/r. The magnetic field is doing loops around the cylinder and inside the area with the current in the [itex]\hat{φ}[/itex] direction.

    I am not sure if what i did is correct and the conclusions i made. Can you comment please?
     
  2. jcsd
  3. May 22, 2013 #2

    rude man

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    Right.
    Wrong. Why did you not integrate dI = 2 pi j r' dr' from R1 to r?
     
  4. May 23, 2013 #3
    I thought that if the current is in the green area, then the integration is from R1 o r.

    Cylinder.jpg

    EDIT: i saw it now. I did integrate from R1 to r..i forgot to put the 'd' in the r2..so it is dr2 in the first step
     
    Last edited: May 23, 2013
  5. May 23, 2013 #4

    rude man

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    Not sure I understand that.

    Actually however, you did everything right in your original post except when it came to evaluating B for case 2 where you forgot to divide by 2 pi.

    I'm sorry, your expressions fooled me for a while but they were mostly OK all along.
     
    Last edited: May 23, 2013
  6. May 24, 2013 #5
    never mind that, it was a mistake i thought i did :)

    thanx for your reply you really helped me
     
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