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Finding the current in a solenoid

  1. May 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A 40-turn, 4.0-cm-diameter coil with R = 0.40Ω surrounds a 3.0-cm-diameter solenoid. The solenoid is 20 cm long and has 200 turns. The 60 Hz current through the solenoid is I = Io sin (2Πft). What is Io if the maximum current in the coil is 0.20A?

    2. Relevant equations
    Ncoil = Nc= 40 turns
    Acoil = Ac = 4π*(10^-2) m^2
    Rcoil = 0.40Ω
    n = Nsol/Lsol = 200 turns / 0.2 m = 1000/m
    Bsol = μo*n*Isol
    Isol = Io sin (120Πt)
    Asol = 2.25π*(10^-4) m^2
    φm = Nc*Bsol*Asol (Magnetic flux through coil)
    ∈coil = φm/dt
    Icoil = Nc*∈coil/Rcoil


    3. The attempt at a solution
    Isol is max when Io sin (2Πft) = Io, i.e. when sin 120Πt = 1, i.e. when t ≈ 0.2387s

    From Isol to Bsol
    Bsol = 1000*μo*(Io sin (120Πt)

    From Bsol to Flux through coil (φm)
    φm = 40*(1000*μo*(Io sin (120Πt))*(2.25π*(10^-4) m^2) = 9*(Io sin (120Πt)

    From φm to ∈coil
    ∈coil = (9*(Io sin (120Πt))/dt= 9*Io*120Π* cos (120Πt) = 3392.92*Io*cos (120Πt)

    From ∈coil to Icoil
    Icoil = (40*3392.92*Io*cos (120Πt))/0.40Ω = 339292.0066 *Io*cos (120Πt) = 0.20 A
    5.8946*(10^-7) = Io*cos (120Πt); t ≈ 0.2387 s ⇒ Io = 140.45A

    Is this correct?
     
  2. jcsd
  3. May 7, 2016 #2

    nrqed

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    That step is incorrect. You have to work in radians.
     
  4. May 7, 2016 #3

    So 120*pi*t = 1.570796327
    t = 0.004167 s ....??
    And why in radians?
     
  5. May 7, 2016 #4
    But then in the last step I get :

    (5.8946*10^(-7)) / cos (0.5*pi) = infinity.......
     
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