Wire Loop falling through uniform magnetic field

Piamedes

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]

So:

[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.

Piamedes

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