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Ampere's Law - finding Magnetic Field

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data
    A long, straight, cylindrical conductor with relative permeability ## \mu ## has a radius a and carries a current I. The current density is uniform. Using Ampere's law in it's integral version, finding the magnetic field B:

    a) Inside, and
    b) outside the cylinder.

    2. Relevant equations

    Amperes law: ## \int \textbf{B}.d\textbf{l} = \mu_0 I ##



    3. The attempt at a solution
    I am using cylindrical geometry so: ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi r l) ##

    Now I think that the magnetic field: ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi a l) ## inside the cylinder.

    And just - ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi r l) ## Outside the cylinder.

    But that seems too simple. What am I missing here? Thanks
     
  2. jcsd
  3. May 13, 2015 #2

    Hesch

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    Well, I have learned:

    circulation H ⋅ ds = N * I ( N=1 as for a long straight conductor ).

    and using this you will find H(r) = N * I / ( 2πr ).

    Outside the conductor you will find: B(r) = μ0 * H(r).

    Inside the conductor you will find: B(r) = μr * μ0 * H(r)
     
    Last edited: May 13, 2015
  4. May 13, 2015 #3

    Hesch

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    Sorry, that was wrong.

    Inside the conductor you will find:

    H(r) = N * I / ( 2πr ) * r2 / R2, ( where R = radius of the conductor and r is the variable radius. )

    You must only integrate around a part of the current, therefore the factor r2 / R2

    B(r) = μr * μ0 * H(r)
     
    Last edited: May 13, 2015
  5. May 13, 2015 #4
    Sorry I am confused - what is H and what is N?
     
  6. May 13, 2015 #5

    Hesch

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    H is the magnetic strength field. Unit = [ A/m ]

    N = number of turns through the integration path ( here N = 1 ).

    B is the magnetic induction. Unit = [ Tesla ].

    B = μ * H where μ is the absolute permeability = μr * μ0 ( In air/vacuum μr = 1 )
     
    Last edited: May 13, 2015
  7. May 13, 2015 #6

    rude man

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    What is ## I \textbf{e_r} ## inside and outside the cylinder? It's not constant. It's a function of r.
     
  8. May 14, 2015 #7
    Sorry guys - I was being silly. I have posted what I think is the correct answer... But I might be completely stupid here. Also apologies for not typing it. I needed the diagram to understand it... Let me know if I'm being completely stupid and it's wrong.

    17441524910_a0eda1756a_b.jpg

    Thanks for all those who have given me hints and help.
     
  9. May 14, 2015 #8

    rude man

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    You have the wrong unit vector for B.
    You have the wrong μ for inside the conductor..
    You should not use I for current density, use j. I is the usual symbol for current.
    See why I'm known by 'rude man'? :rolleyes: Seriously, you're doing OK, main thing is recognizing the right unit vector. You're not 'completely stupid' by a long shot!
     
  10. May 14, 2015 #9
    Actually that's one of the best posts I've got on here. I'd rather people were brutal.

    e(phi) should be unit vector.

    μ should be inside the conductor - although I'm unsure if it should be just μ or μμ_o
     
  11. May 14, 2015 #10

    rude man

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    right.
    You have to be careful with your μ's.
    B = μ0H in vacuum (or other non-magnetic material). B = μH in magnetic material.
    In other words, μ0 is not a dimensionless number. μ0 = μ in a vacuum.
    So you don't want to say μ = μ0μ ever.
    μ > μ0 inside the conductor here.
     
  12. May 14, 2015 #11

    Hesch

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    I don't think you are right here.

    "Normally" you will call the absolute permeability "μ", and the relative permeability "μr".
    μr is dimensionless and μr ≥ 1.
    So μ = μr * μ0 ( at least in Europe ).
     
  13. May 14, 2015 #12

    rude man

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    That in no way contradicts what I said. μr ≠ μ0.
     
  14. May 14, 2015 #13

    Hesch

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    That's right, but there is some confusion in the thread, because in #1 the relative permeability here is called "μ" ( not "μr".

    What I meant was:
    in my terminology:

    So you don't want to say μ = μ0 * μr ever.

    Maybe the problem is that things are called by unconventional names, like if you are reading some text concerning trigonometry, and the symbol "π" is substituted by "θ", the text will indeed be very difficult to read/understand ( though the author has mentioned this substitution ).
     
  15. May 14, 2015 #14

    rude man

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    I am confused by your post. Sorry.
     
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