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Homework Help: Finding the inductance of a coil

  1. Dec 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A circular coil has a radius 0.10 m and 30 turns. An external magnetic field 2.60 mT is perpendicular to the coil. a) If there is no current, find the magnetic flux in the coil.
    b) When current = 3.8 A in some direction, there is no more net flux in the coil. Find the inductance.


    3. The attempt at a solution
    I can do a) but I don't understand b. the correct answer of b) is 0.645 mH

    a) flux = NB*PI*r^2 = 2.45 miliweber, this is the correct answer

    b) L = N*flux / current. If I set flux=0 then I get inductance = 0, which makes no sense because inductance only depends of the geometry.

    Please help =(
     
    Last edited: Dec 20, 2009
  2. jcsd
  3. Dec 20, 2009 #2

    ideasrule

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    Homework Helper

    Suppose the external magnetic field didn't exist. What would be the magnetic field created by the solenoid? Net field is 0 when this is equal to 2.60 mT.
     
  4. Dec 20, 2009 #3
    I set B = -2.60 mT

    which means that L = N*flux / current = n*-2.60 mT / current?

    = (30*2.60*10^-3)/3.8 = 0.0205 H

    not correct =(
     
  5. Dec 20, 2009 #4

    ideasrule

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    The magnetic field produced by the solenoid is B=μ0*N*I/L. When B=2.6 mT, net magnetic field is 0, so B=2.6 mT when I=3.8 A.

    Now, what would be the equation for the flux through the solenoid, created by the solenoid itself? You can then find L using L = N*flux / current.
     
  6. Dec 20, 2009 #5
    Ok, B = μ0*n*i

    the length of one turn is 2*Pi*R, the total length of the coil is N times that, so 2*Pi*R*N
    n = number of turns / total length = N /[ (2*Pi*R*N) = 1 / (2*Pi*R)

    so that B = μ0*i / 2*Pi*R

    flux = B*Area = B*Pi*(R^2) = μ0*i*r / 2
    so that L = N*flux / i = μ0*r*N / 2 = 1.88 * 10^-6 W, wrong answer

    I have also tried multiplying the flux by N since there are N turns

    please help =(
     
  7. Dec 20, 2009 #6
    Ideasrule told you that ...


    B=2.6 mT when I=3.8 A ,,,,,, So find L the lenght of the solenoid by this equstion :

    B=μ0*N*I/L

    Where ,

    B = 2.6e-3 T
    N = 30 turns
    I = 3.8 A
    So L = ????

    Then substitute the value of L in this equation to find the Inductance of the Solenoid :

    L Solenoid's Inductance = μ0*N2*A / L Length Found

    Thats it .


    Just for more clarification here are the steps to get the last equation for solenoid's inductance :

    B Solenoid = μ0*n*I

    n = N/L , So B Solenoid = μ0*N*I/L

    ------------------------------------------------------------

    [tex]\phi[/tex] B = N*B*A*Cos(00)

    B Solenoid = μ0*N*I/L

    So , [tex]\phi[/tex] B = N*(μ0*N*I/L )*A*Cos(00)

    Cos(00) = 1

    So , [tex]\phi[/tex] B = μ0*N2*I*A / L


    -------------------------------------------------------------------------

    The emf induced in N turns is N times the emf in one turn :

    [tex]\epsilon[/tex] = - d[tex]\phi[/tex] B / dt

    [tex]\epsilon[/tex] = - ( μ0*N2*A / L ) * (dI/dt)

    --------------------------------------------------------------------------

    Finally the inductance :

    L Solenoid's Inductance = - [tex]\epsilon[/tex] / (dI/dt)

    We will get finally :

    L Solenoid's Inductance = μ0*N2*A / L Length Found

    Good Luck ...
     
    Last edited: Dec 20, 2009
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