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EMF Question

  1. Apr 11, 2005 #1

    AKG

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    A square loop is cut out of a thick sheet of aluminum. It is then placed so that the top portion is in a uniform magnetic field B, and allowed to fall under gravity. So the square is in the y-z plane, and the magnetic field is -|B|x say, for z > 0, and the magnetic field is 0 below that. This square loop thus has the top part of it above z=0, and the rest of it is obviously below. If the magnetic field is 1T, find the terminal velocity of the loop. Find the velocity of the loop as a function of time. What would happen if you cut a tiny slit in the loop, breaking the circuit?

    Okay, the force on this ring is

    [tex]-mg\hat{\mathbf{z}} + B(\mathbf{I} + q\mathbf{v}_{falling}) \times (-\hat{\mathbf{x}})[/tex]

    I know that [itex]\mathbf{I}[/itex] will go "counter-clockwise", and we may as well ignore the current in the left and right sides of the square because their effect on the force will cancel out, since the current is in opposite directions for them. So, for the sides, we get:

    [tex]-mg\hat{\mathbf{z}} + Bqv_{falling}(-\hat{\mathbf{z}}) \times (-\hat{\mathbf{x}})[/tex]

    [tex]= -mg\hat{\mathbf{z}} + Bqv_{falling}\hat{\mathbf{y}}[/tex]

    For the top, we get:

    [tex]-mg\hat{\mathbf{z}} + B(-\hat{\mathbf{y}}\mathcal{E} /R -\hat{\mathbf{z}} qv_{falling}) \times (-\hat{\mathbf{x}})[/tex]

    [tex]-mg\hat{\mathbf{z}} + B(-\hat{\mathbf{z}}BLv_{falling} /R + qv_{falling}\hat{\mathbf{y}})[/tex]

    The part outside the field will just experience

    [tex]-mg\hat{\mathbf{z}}[/tex]

    Actually, the values m, q, and L should be dm, dq, and dL (L is the length of the side of the square). If I integrate these force-per-lenghts over the 6 parts in question (the top edge, the top portion of the right side that is in the field, the bottom portion of the right side that isn't in the field, the bottom edge, and both parts of the left side), I should get the total force. Dividing by the mass, I'll get the acceleration, and integrating with respect to time, I'll get the velocity. From this, I can find velocity with respect to time. Also, if after finding the net force, I set it to zero, I should be able to solve for [itex]v_{falling}[/itex] to get the terminal velocity.

    If a snip is made, then no current flows, and the only forces are the gravitational force, and the magnetic force due to the velocity of the charges which now only the velocity from falling (and no longer is there a component to the velocity attributable to current). Is this all right?

    I expect L to cancel out, and if the loop has a cross-sectional area A, that should also cancel out. If aluminum has density D, then using Avogadro's number and the atomic number for aluminum, I can find the charge density. I will thus be able to find things like dq and dm. Is this the right approach to all of this?
     
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