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Finding Charge From Changing Magnetic Flux

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    In the figure, a flexible square loop 0.700 m on a side is made of wire of resistance 0.600 Ω. A magnetic field with magnitude 1.70 T is directed onto the plane of the loop. A student crushes the wires together forming a loop of zero area in 0.350 s. Find the total amount of charge flowing through the wire.

    2. Relevant equations

    Maxwell's equations such as
    ∫E[itex]\cdot[/itex]dA = Qenclosed/epsilon nought

    3. The attempt at a solution
    Now I am having trouble starting this question. I am thinking that because the loop is being crushed, the area is going to be changing, which means magnetic flux is changing because [itex]\Phi[/itex]B=BA. And I know that magnetic flux somehow influences electric flux (forget the name of the law for this), so I'll have to use the equation listed above. Any hints please?

    Attached Files:

  2. jcsd
  3. Dec 4, 2011 #2
    Here's what I tried. I calculated PhiB = 0.833. This next part is where I'm not sure I can actually do this. I just divided .833 by the time .35 and got dPhi/dt = 2.38, so the EMF induced is -2.38 V. Then I went V=Ed and got E= .85 Vm-1. Then used Gauss' Law Qenclosed = epsilon nought * E * A = 3.69 pC. This seems like an extremely small amount so I haven't entered my answer yet. Also I'm not sure if I can do half the things I did here lol.
  4. Dec 4, 2011 #3
    I calculated the flux through the loop.
    Induced emf = change in flux per second
    Current = emf/R
    Current = coulombs per second..... should give you the charge in coulombs (I got 1.38)
  5. Dec 4, 2011 #4
    Okay so that's right. So I was right about being able to divide B by time. But I thought current was the rate of change of charge? So why can we just use it as charge per unit time?
  6. Dec 4, 2011 #5
    You are correct.... amps = coulombs per second.
    I calculated the current from V/R .... V = induced emf, R =0.600ohms
    I got the voltage to be 2.38.... the same as you.
    I am not certain what your V = Ed means !!
  7. Dec 4, 2011 #6
    I was mistaken and thought I would have to use Gauss' Law because the assignment name is 'Maxwell's Equations"... so I tried solving for the electric field induced but it definitely didn't work.
  8. Dec 4, 2011 #7
    Are you convinced by what I did !!!!?
  9. Dec 4, 2011 #8
    Yes I am. I did and and the answer was correct. I was just making sense of something we did.
  10. Dec 4, 2011 #9
    great.... well done
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