Wire Loop falling through uniform magnetic field

In summary, the problem involves a square loop of aluminum being placed in a uniform magnetic field and allowed to fall under gravity. The goal is to find the terminal velocity of the loop, the velocity as a function of time, and the time it takes to reach 90% of the terminal velocity. The solution involves using equations for induced emf and magnetic force, and the dimensions of the loop cancel out. With the help of the natural resistivity and density of aluminum, the force can be calculated.
  • #1
Piamedes
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Homework Statement



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

Homework 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.

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.
 
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  • #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
 

1. How does a wire loop fall through a uniform magnetic field?

As the wire loop falls, it experiences a change in magnetic flux due to the changing area of the loop. This change in magnetic flux induces an electric current in the wire loop, creating a magnetic force that opposes the force of gravity. This results in a slower descent of the wire loop than a simple free-fall.

2. What factors affect the speed at which the wire loop falls through the magnetic field?

The speed at which the wire loop falls through the magnetic field depends on the strength of the magnetic field, the size and shape of the wire loop, and the material of the wire. A stronger magnetic field or a larger, more conductive wire will result in a slower descent.

3. Can the wire loop fall faster than the acceleration due to gravity in this experiment?

No, the wire loop cannot fall faster than the acceleration due to gravity. The induced magnetic force only opposes the force of gravity, it does not exceed it. Therefore, the wire loop will still experience a downward acceleration of 9.8 m/s^2.

4. How does the orientation of the wire loop affect its fall through the magnetic field?

The orientation of the wire loop does not significantly affect its fall through the magnetic field. As long as the loop is falling vertically, it will experience the same magnetic force regardless of its orientation.

5. Is this experiment affected by air resistance?

Yes, air resistance can affect the speed at which the wire loop falls through the magnetic field. However, it would be a negligible factor if the wire loop is falling quickly and the air resistance is not significant compared to the magnetic force. To minimize the effect of air resistance, the experiment can be conducted in a vacuum.

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