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Magnetic force from BIL

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    There is a rectangular loop submerged in uniform magnetic field, where all but one edge is in the field, and I am given W (length), Bo, R (resistance), L (self-inductance), m.

    I am to find v(t), y(t), and i(t). when R = 0 and L =/ 0 (L is NOT =0)

    2. Relevant equations

    Fnet = m(dv/dt) = Fexternal - Fmagnetic

    Fexternal = mg

    Emfinduced = Vinduced= -(dflux/dt) = v*B*W

    Fmagnetic = Iinduced*W*Bo


    Emfinduced = -L*(dI/dt)

    3. The attempt at a solution

    So I think the result is that two parallel sides with induced currents in opposite directions will cancel out, and only one side, length W will contribute to force, because if total loop was submerged into B field region then force would be zero.

    And, so, I solved this for the conditions R=/0 and L = 0 (R is NOT =0), so I have not included any equations with R because I am not using it, so what I am trying to do is solve it in similar fashion as before, using the equation of motion.

    So I think I am to use again the Fmagnetic = Iinduced*W*Bo

    What I am having trouble with is finding Fmagnetic in terms of Iinduced in terms of the given variables.
    I see that there is an equation relating flux to Iinduced, as well as Emfinduced to Iinduced

    I'm not using the flux equation, because I don't see a differential area, because I am dealing with a wire here, or do I use the differential area covered by the whole rectangle even though there is no contribution from the two paralell wires?

    My guess would be to again use:

    Fmagnetic = Iinduced*W*Bo

    and (dI/dt) = (vBoW/L) = (dy/dt)*(BoW/L)

    and then divide out dt, because current over a change in time is not a current?
    so I get:

    (dI) = (dy)*(BoW/L)

    and so then I = y*(BoW/L)

    and then use:

    m(dv/dt) = Fexternal - Fmagnetic = mg - (y*(BoW/L)*WBo)

    and integrate?

    but I don't know if I can do this...

    any help would be graetly appreciated
  2. jcsd
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