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Wire Loop falling through uniform magnetic field

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    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. (B is perpendicular to the loop) If the magnetic field is 1 T, find the terminal velocity of the loop. Find the velocity as a function of time. How long does it take to reach 90% of the terminal velocity? What would happen if you cut a tiny slit in the loop, breaking the circuit? [Note: The dimensions of the loop cancel out; determine the actual number.]

    My Note: It's from Griffiths

    2. Relevant equations

    [tex]\epsilon = - \frac{d\phi_{B}}{dt} = - \frac{d}{dt} \int_S B \cdot da [/tex]

    [tex] F_{mag} = \int I \times B dl = I \int dl \times B [/tex]

    [tex]\epsilon = IR [/tex]

    I think I'm missing another equation or two, but I don't know which ones.

    3. The attempt at a solution

    I first set one side of the square as length l.
    The resistance of the wire is R.
    B is parallel to the unit vector normal to the area of the loop, so:

    [tex]\epsilon = - \frac{d}{dt} \int_S B \cdot da = -B \frac{da}{dt} = -Bl \frac{dy}{dt} = -Blv [/tex]

    [tex]\epsilon = IR [/tex]


    [tex] I = \frac{\epsilon}{R} = \frac{-Blv}{R} [/tex]

    Now for the force. The cross product will cancel out on the two legs of the square, so only the component from the top of the square will contribute. Since I is clockwise as the square falls, the direction of the resulting force will be upwards.

    [tex] F_{mag} =I \int dl \times B = I \int Bdl = BIl [/tex]

    Plugging in for I:

    [tex]F_{mag} = (Bl)(\frac{-Blv}{R}) = \frac{-B^2 l^2 v}{R} [/tex]

    And this is where I get stuck. I can't seem to eliminate the length of the wire, nor its resistance from the equation. I assume there's some equation relating the natural resistivity of aluminum to its length, but I have no idea what it is or how to go about solving for this force without the dimensions of the wire. Any advice would be great.
  2. jcsd
  3. Feb 1, 2010 #2
    Never mind, I worked it out with some help. the l squared over resistance quantity reduces to a ratio of the resistivity and density of aluminum
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