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Induced current in conductor moving circularly in constant B-field

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data

    A light bulb with resistance R is attached on a metal rod which is rotating around the point O on the figure. The metal rod is in contact with an electrical conductor which is a part of a circle with radius d. The metal rod and the circular electrical conductor is a closed circuit. The rod now rotates with angular velocity [itex]\omega[/itex] through the constant magnetic field pointing out from the paper.

    a)

    Find an expression for the induced current through the light bulb, expressed in terms of [itex]\omega[/itex], d, B and R.

    2. Relevant equations

    IR=vBr

    where v is the tangential speed of the rod perpendicular to the B-field (every speed is perpendicular to the B-field, since we are looking at a plane) and r is the length of the rod moving at this speed.

    I=[itex]\frac{\omega Br^{2}}{R}[/itex]

    v substituted for [itex]\omega r[/itex]

    3. The attempt at a solution

    Since the every part of the rod is moving with different linear speeds, we should integrate the RHS from the 0 to d with respect to r and that should be it right?

    i get:

    I=[itex]\int^{d}_{0}\frac{\omega Br^{2}}{R}[/itex]

    I=[itex]\frac{d^{3}B\omega}{3R}[/itex]

    But when i look up the solution it says:

    I=[itex]\frac{Bd^{2}\omega}{2R}[/itex]

    so who's right?

    Edit: Problem solved!
     

    Attached Files:

    Last edited: Nov 8, 2011
  2. jcsd
  3. Nov 8, 2011 #2
    Maybe i should add that RI=vBr is derived from Faradays law of induction, stating that the induced EMF is equal to the closed path integral of E+v X B with respect to l (path of the circuit), and Ohm's law stating that the EMF is equal to RI when looking at the entire circuit. I only integrate over the rod since this is the only thing moving relative to the B-field. The cross product in faradays law reduces to the magnitudes of v and B multiplied, since they are always perpendicular to each other in this problem and since i only need to find the magnitude of the EMF.
     
  4. Nov 8, 2011 #3
    Nevermind i solved it!

    After reading my last post over, i realized that i should use Faradays law of induction as the more general law, rather than IR=vBl which is a solution to Faradays law in a particular situation. I then obtained the same answer as in the solutions sheet.
     
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