Current in wire to set magnetic flux

Discussion in 'Homework Help' started by smarch, Jul 4, 2010.

  1. smarch

    Thread Starter Active Member

    Mar 14, 2009
    52
    0
    A solenoid with a circular cross-section has a length of 10 cm, a radius of 3
    cm and is wound with 2,000 turns per metre of wire. How much current must
    flow in the wire to set up a magnetic flux density of 1 T at the centre of the
    solenoid, assuming that the inside of the solenoid is filled with air?


    I know I use the formulas H= nI/2 [cosθ_1- cosθ_2 ] and B = μ0μrH.

    But where do I get θ from?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Presumably at the geometric centre θ_1=θ_2=arctan(3/5)
     
  3. JimmyB

    Member

    Jun 1, 2010
    38
    0
    Hi

    I tried it like this

    L=0.1m CSA=0.0028m2 N=200

    Found S 1st S= L = 0.1 =28420525 A/Wb
    .....................μ A (4∏x10-7)(0.0028)

    Wb=BA = 1x0.0028 = 0.0028Wb

    m.m.f = Wb x S = 0.0028 x 28420525 = 79577 A/t

    therefore I = F = 79577 = 397A
    ...................N 200

    or

    H = B = 1 =795774.7 A/m
    ......U 4pi x 10-7

    F = HL = 795774.7 x 0.1 = 79577 A/t

    I = F = 79577 = 397 A
    .....N 200

    I think this is correct, but I would imagine a 2.5mm2 wire with nearly 400A flowing being quite hot!!!!!!
    you might want to check the answer with someone...
     
    Last edited: Jul 8, 2010
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I believe the formula for B along the coil axis would be of the form

    B=\frac{\mu_o NI}{2L}\[cos(\theta_1)-cos(\theta_2)\]

    Where L is the coil length and the angles are those subtended from the coil axis by a line extended from the point of interest to either extremity of the coil end edges

    Contrary to my earlier post I would have

    θ1=Pi-atan(3/5) and θ2=atan(3/5)

    I get an answer of 46.4A to achieve the required flux density of 1T.
     
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