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Ampere's Law

  1. Apr 16, 2009 #1
    We have an infinite coil of N turns per unit length with radius a, carrying a current I and we want to find the magnetic field using Ampere's Law.

    If we put the loop outside the coil it can be established that B=0

    If we use the loop to enclose the coil and we apply Ampere's Law we get that [itex]\vec{B}=\mu_0 NI \mathbf{\hat{z}}[/itex]

    apparently this means that the field is [itex]\mu_0 NI \mathbf{\hat{z}} \forall r < a[/itex] and [itex]0 \forall r>a[/itex]

    two questions:

    (i)why, if we take an amperian loop INSIDE the coil (i.e. not enclosing any current, doesn't the field =0 by ampere's law?

    (ii) what is the field at r=a?

    thanks
     
  2. jcsd
  3. Apr 16, 2009 #2

    dx

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    For (i), Ampere's law would say that the integral of the field around the loop is zero, not the field itself.
     
  4. Apr 16, 2009 #3
    ok so for a rectangular loop inside the coil, the contributions from each of the 4 sides would actually cancel out wouldn't they because the field is uniform. This would mean we'd just end up with 0=0 and wouldn't be able to establish anything!

    any ideas for (ii)?

    Also, see when we talk about magnetostatics, I'm struggling to see how anything is "static"? we need a current which is a movement of charges and for the case of bulk currents we have [itex]\vec{J}=\rho \vec{v}(\vec{r})[/itex] and so the velocity term there implies something is moving does it not?
     
  5. Apr 16, 2009 #4

    dx

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    For (ii), we're considering an idealized case, and there's a discontinuity r = a.

    In magnetostatics, "static" means that the fields are static, not the charges. The charges can move, but the charge distribution and the current density everywhere must be static to make the fields static.
     
  6. Apr 16, 2009 #5
    so we just can't tell what the field at r=a is because of said discontinuity?
     
  7. Apr 16, 2009 #6

    dx

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    It doesn't even make sense because were assuming that the wires are infinitely thin. In a realistic example, you would have to consider the thickness of the wires etc, and the field at the axis of the wire will be something like half the field inside the solenoid.
     
  8. Apr 16, 2009 #7
    ok. should really start a new thread i guess but just quickly, why does

    [itex](m \times \nabla) \times B=-m(\nabla \cdot B) + \nabla (m \cdot B)[/itex]???
     
  9. Apr 16, 2009 #8

    dx

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    Well you could just calculate the components on each side to check that they're equal, or you could write the LHS in tensor notation and use properties of the Levi-Civita symbol etc. I don't know of any simpler way to prove it.
     
    Last edited: Apr 16, 2009
  10. Apr 16, 2009 #9
    yeah my problem is it doesn't work!

    [itex]-\epsilon_{ijk} B_j \epsilon_klm} m_l \partial_m = (\delta_{il} \delta_{jm} -\delta_{im} \delta_{jl}) B_j m_l \partial_m = B_j m_i \partial_j - B_j m_j \partial_i = (B \cdot \nabla)m - \nabla(B \cdot m)[/itex] my problem lies in the fact that the dot product doesn't commute when a differential operator is involved...
     
  11. Apr 16, 2009 #10

    dx

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    You just have to keep track of what the differential operator is acting on. In this case, its always acting on B.

    [tex] [m \times ( \nabla \times B )]_i = \epsilon_{ijk} \epsilon_{klm} m_j \partial_l B_m = m_j(\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) \partial_l B_m = [\nabla (m \cdot B)-m(\nabla \cdot B)]_{i} [/tex]

    m is a constant vector here. In your work, you moved the derivative to the end, which is not valid. It must act on the B components.
     
    Last edited: Apr 16, 2009
  12. Apr 16, 2009 #11
    why isnt it valid to be at the end? just because then it isn't acting on anything at all?
     
  13. Apr 16, 2009 #12

    dx

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    No, it's just that it was originally acting on B, and you can't change the order. It's just like writing [tex] \nabla \cdot B = \partial_i B_i [/tex] as [tex] B_i \partial_i [/tex], which is not valid.
     
  14. Apr 17, 2009 #13
    yeah but you changed the original expression, you moved the brackets didn't you?
     
  15. Apr 18, 2009 #14

    dx

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    Where? You mean [tex] (m \times \nabla) \times B = m \times (\nabla \times B) [/tex]? They mean the same thing.
     
  16. Apr 18, 2009 #15
    how do they mean the same thing i though when we had a vector triple product you couldn't move the brackets without changing the meaning?
     
  17. Apr 18, 2009 #16

    dx

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    They're the same in this case:

    [tex] [(m \times \nabla) \times B]_i = \epsilon_{ijk} (m \times \nabla)_j B_k = \epsilon_{ijk} \epsilon_{jlm} m_l \partial_m B_k = - \epsilon_{jik} \epsilon_{jlm} m_l \partial_m B_k [/tex]

    [tex] = -m_l(\delta_{il}\delta_{km} - \delta_{im} \delta_{kl}) \partial_m B_k = -m_i \partial_k B_k + m_k \partial_i B_k = -m_i \partial_k B_k + \partial_i m_k B_k = -m_i (\nabla \cdot B) + \partial_i (m \cdot B)= [-m (\nabla \cdot B) + \nabla (B.m)]_i [/tex]
     
    Last edited: Apr 18, 2009
  18. Apr 18, 2009 #17
    ok. thanks. see if we have a wire of radius a with uniform current density and total current I with potential difference across a length L of the wire is V and we want to find E and B inside and outside the wire.

    i used ampere to get [itex]\vec{B}(\vec{r})=\frac{\mu_0 I}{2 \pi r} \mathbf{\hat{\phi}}[/itex] if the current is in the z direction. anyway will this apply to both the inside and outside of the wire - i couldn't get any other result but i reckoned it should be different inside???

    for teh electric field, is it just [itex]\vec{E}(\vec{r})=\frac{V}{L} \mathbf{\hat{z}}[/itex] where i just used the fact the E is in the same direction as the current flow??

    outside however, i have been told to show its 0 using gauss but i don't know what sort of gaussian sruface to take?

    thanks for your help again.
     
  19. Apr 18, 2009 #18

    dx

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    B will be different inside. If you draw a circular loop whose radius is smaller than a, the current through the loop will no longer be I, it will be smaller. Find out how much this is, and use Ampere's law as before.
     
  20. Apr 19, 2009 #19
    ah of course so the enclosed current will now be [itex]\frac{a^2 I}{r^2}[/itex] yes?

    how can i find the electric field outside though?
     
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