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Determine the maximum magnetic flux through an inductor

  1. Nov 30, 2003 #1
    I'm sure I'm gonna kick myself when someone shows me the way, but there's a limit to how long I can stare at this question...

    Determine the maximum magnetic flux through an inductor connected to a standard outlet (ΔVrms = 120 V., f = 60 Hz.)

    That's all that's given.

    I know that ΔVmax = 120*√2 and ω = 2Πf = 120Π and I think the magnetic flux will be at a maximum when the current is at a maximum, and
    Imax = (ΔVmax) / (ω L) so

    Imax = (120*√2) /(120ΠL)

    Also, ΦB = I*L/N so I get

    ΦB = (120*√2) /(120Π N)

    but I have no idea what the value of N is for the inductor, & I can't see any way to get rid of that N. :frown:

    (edited to correct 60 x 2 = 120, not 100 )
    Last edited: Dec 1, 2003
  2. jcsd
  3. Dec 1, 2003 #2
  4. Dec 1, 2003 #3
    Yeah, thanks Arcnets, that's basically the answer. My objection is that ΦB = IL if there's only one turn.

    As you found in that Wolfram link, it comes from Faraday's
    ε = -dΦB/dt
    dΦB/dt ~ dI/dt (using ~ to mean proportional to)

    From that, the inductance L is defined as the proportionality constant, so

    dΦB/dt = L*dI/dt

    and then, by integrating we get

    ΦB = L* I

    My objection is that in a coil with N turns, the inductance is
    L = (NΦB)/I
    and any inductor in a circuit is going to have more than 1 turn, so it seems to me that the question was, to say the least, ambiguous.

    But, I asked the professor about it today and, yes, his answer was that the question "meant" the total flux through all the turns, in other words, they were looking for the value of N * ΦB, which of course is just
    (120*√2) /(120Π) = .450 T*m2
  5. Dec 2, 2003 #4
    Did you kick yourself?
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