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Self and mutual inductance

  1. May 5, 2015 #1
    1. The problem statement, all variables and given/known data
    problem 5.png
    2. Relevant equations


    ∫BdL = μi

    Φ = BA (flux)

    S= NΦ/i (self inductance)

    M = NΦ/i (mutual inductance)

    3. The attempt at a solution

    a) ∫BdL = μi
    BL = μi
    B = μi/L
    since the solenoid has N1 turns
    B = μiN1/L

    b) flux Φ = BA = AμiN1/L
    self inductance S = N1Φ/i = N1(AμiN1/L)/i = N1^2μA/L

    c) mutual inductance M = N2Φ/i = N1N2μA/L
     
  2. jcsd
  3. May 6, 2015 #2

    ehild

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    If i is the current flowing in the wire of the coil, the formula ∫BdL = μi is not correct. You need to show or explain the integration path .
     
  4. May 6, 2015 #3

    Zondrina

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    For part a), I would assume ##\vec B## is uniform within the solenoid and zero outside of it. If you place a rectangular Amperian loop ##\partial \Sigma## with corners ##a, b, c, d## along a cross section of this ideal solenoid, the left side of Ampere's law will give:

    $$\oint_{\partial \Sigma} \vec B \cdot d \vec S = \int_a^b \vec B \cdot d \vec S + \int_b^c \vec B \cdot d \vec S + \int_c^d \vec B \cdot d \vec S + \int_d^a \vec B \cdot d \vec S$$

    The first integral yields:

    $$\int_a^b \vec B \cdot d \vec S = B \int_a^b dS = BL$$

    Where ##B = |\vec B|## and ##L## is the length of ##\partial \Sigma## from ##a## to ##b##.

    The third integral is taken along a segment of ##\partial \Sigma##, which lies outside of the solenoid. The magnetic field outside the solenoid is zero, hence the integral is zero.

    The second and fourth integrals work out to zero because ##\vec B## is either orthogonal to ##d \vec S##, or is equal to zero outside of the solenoid.

    Hence we can write:

    $$\oint_{\partial \Sigma} \vec B \cdot d \vec S = BL$$

    Now using the right side of Ampere's law:

    $$BL = \mu_0 i_{enc}$$

    The loop ##\partial \Sigma## encloses ##nL## turns where ##n## is the number of turns per unit length of the solenoid. Using this, the enclosed current can be re-written as:

    $$i_{enc} = inL$$

    Where ##i## is the actual current in the solenoid windings. Hence we can write:

    $$BL = \mu_0 inL$$
    $$B = \mu_0 in$$
    $$B = \mu_0 iN_1$$

    This is the magnetic field inside the solenoid.
     
  5. May 6, 2015 #4

    rude man

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  6. May 6, 2015 #5
    Thank you that derivation helped alot. I knew the formula from my book but the derivation of it confused me.

    so its
    BL = μienc
    since ienc = iN1
    so
    BL = μiN1
    B =μiN1/L

    Are parts b) and c) ok? I am pretty sure I did part b) right but I am not as confident about part c)
     
  7. May 6, 2015 #6

    rude man

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    Right on all counts incl. parts b and c!
     
  8. May 6, 2015 #7
    Thank you all
     
  9. May 6, 2015 #8

    Zondrina

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    I always preferred the turns/length version of the derivation, but there is little difference in the final formulas.

    Everything else looked good otherwise.
     
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