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Mutual inductance in concentric cylinders

  1. Jul 8, 2011 #1
    Friends:

    I have been thinking about this question for a while, so I hope you can help! Here it is:

    Lets say you have two concentric, conducting cylinders.

    The outer cylinder ("Cylinder 2") has the following properties: radius R2, carries a current of I2, relative magnetic permeability mu2, impedance of Z2 = R2 + jwL2.

    The inner cylinder ("Cylinder 1") has the following properties: radius R1, carries a current of I1 (in the same direction as I2), relative magnetic permeability mu1, impedance of Z1 = R1 + jwL1.

    My question is: how does the mutual inductance between these cylinders increase their impedance?

    My guess is the following:

    "Cylinder 1" creates a magnetic field B = (mu0)(mu1)(I1)/(2*pi*R1)
    This creates a mutual inductance on "Cylinder 2". The mutual inductance created on "Cylinder 2" is M21 = [(mu0)(mu1)(I1)/(2*pi*R1)]*LN(R2/R1).
    Thus the impedance of "Cylinder 2" increases to Z2 = R2 + jwL2 + jwM21.

    "Cylinder 2" creates a magnetic field B = (mu0)(mu2)(I2)/(2*pi*R2). However, this field exists only outside the cylinder. Inside the cylinder, the magnetic field is 0. Therefore, the mutual inductance created on "Cylinder 1" by "Cylinder 2" is M12 = 0.
    Thus the impedance of "Cylinder 1" remains unchanged at Z1 = R1 + jwL1.

    Please let me know if this is correct or if I'm terribly mistaken! Thank you kindly!
     
  2. jcsd
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