Analyzing the Physics of a Falling Square Aluminum Loop in a Magnetic Field

[AKG]

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?

 

[AvroArrow]

Yes, that is the right approach. If you calculate the force on each component due to the magnetic field and the gravity, and then integrate them over the entire loop, you will get the total force on the loop. You can then divide this force by the mass of the loop to get the acceleration, and from that, you can calculate the terminal velocity and the velocity as a function of time. When you cut the tiny slit in the loop, there is no longer any current flowing, so the only force acting on the loop is the gravity. The terminal velocity will be lower than before, and the velocity as a function of time will also be different.

 

[thepatientmental]

Firstly, it is important to note that the force on the loop is dependent on the direction of the magnetic field. In this case, the magnetic field is in the -x direction for z > 0, which means that the loop will experience a force in the +y direction. This will cause the loop to rotate in a counter-clockwise direction as it falls, as you correctly pointed out.

To find the terminal velocity, we can set the net force on the loop to zero, as you mentioned. This will give us an equation for the velocity, which we can solve for to get the terminal velocity. However, since the loop is rotating, we need to consider the rotational inertia of the loop as well. This will affect the velocity and thus the terminal velocity.

To find the velocity as a function of time, we can use the equations of motion for a falling object with a constant force. However, we also need to consider the rotational motion of the loop, so we will need to use the equations of rotational motion as well. This will give us a system of equations that we can solve to find the velocity as a function of time.

If a tiny slit is cut in the loop, breaking the circuit, then no current will flow and the loop will only experience the gravitational force and the magnetic force due to its velocity. This will change the dynamics of the loop and it will no longer rotate as it falls. The velocity of the loop will also be affected, as there will be no current flowing to generate a magnetic force.

Your approach to finding the net force and integrating to find the acceleration and velocity is correct. However, it is important to note that the dimensions of the loop will not cancel out completely. The force on each side of the loop will depend on its length, so the total force will still have a dependence on the dimensions of the loop. Additionally, the charge density of aluminum will not be constant throughout the loop, so the values of dm and dq will also vary. It may be helpful to consider the loop as a collection of small segments, each with its own length and charge density, and then integrate over all of these segments to find the net force.

Overall, your approach seems to be correct, but it may require some fine-tuning and consideration of additional factors such as rotational inertia and varying charge densities. It is a complex problem, but with careful analysis and calculations, the terminal velocity and velocity as a function of time can be determined.