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Coaxial solenoids

  1. Feb 20, 2006 #1
    This one is from Griffiths.

    Two coaxial long solenoids each carry current I, but in opposite directions.
    The inner solenoid has radius 'a' and has 'n1' turns per unit length.
    The outer solenoid has radius 'b' and has 'n2' turns per unit length.
    Find the magnetic field [itex]\vec B[/itex] in three regions:
    1] inside the inner solenoid
    2] between them
    3] outside both

    My work:
    I worked out the solution for these. Someone verify if my answers are correct.
    General formula for magnetic field for a solenoid of 'n' turns is:
    [tex]\vec B = \mu_0 nI \hat k[/tex]

    1] For inner solenoid:
    [tex]\vec B = \mu_0 I n_1 \hat k[/tex]

    2]Between the solenoids:
    [tex]\vec B = \mu_0 I n_1\hat k - \mu_0 I n_2\hat k[/tex]

    [tex]\vec B = \mu_0 I \left(n_1 - n_2\right)\hat k[/tex]

    3]Outside both:
    [tex]\vec B = 0[/tex]
     
    Last edited: Feb 20, 2006
  2. jcsd
  3. Feb 20, 2006 #2
    not quite right.
    why would the fields add between them but cancel outside both?
    [especially what is the field of the "inner" solenoid?
    use superposition.]
     
  4. Feb 20, 2006 #3
    You mean the fields superimpose at the inner solenoid and not between them?
     
  5. Feb 20, 2006 #4
    the fields superimpose everywhere.
    but the fields are only nonzero inside
    the respective solenoids.
     
  6. Feb 21, 2006 #5
    So the field inside the inner solenoid would be:
    [tex]\vec B = \mu_0 I \left(n_1 - n_2\right)\hat k[/tex]
    & the field between them would be:
    [tex]\vec B = -\mu_0 I n_2\hat k[/tex]

    Hope, I got it right & thanks for the help!!
     
  7. Feb 24, 2006 #6
    No clarifications so far....then I suppose my answer is correct :biggrin:.
     
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