Physics QOTD
 

Physics Question of the Day

(Yeah, I know, dumb name)
Assume g=9.80 m/s^2
There are two wires of length 5m parallel to each other, 3 meters apart. The plane created by the two wires is normal to the vector from the center of earth to the center of gravity of the system. The currents in both wires is in the same direction, 5 A. The wires are in equilibrium, each held in place; each wire can be treated as a rigid rod. At the instant the wires are let go, what is the magnitude of acceleration of either wire?

Re: Physics QOTD
 

Maybe I'm missing something, but you must have some way of determining the mass of the wires.

Re: Physics QOTD
 

Well... I would say that they both would have an acceleration of 9.8 m/s^2, regardless of mass, just b/c i dont know a thing about electricity. So i stay with my answer of 9.8 m/s^2

Re: Physics QOTD
 

Late night, yes I think you're right. However, I can't exactly picture the orientation of the two wires. I think you're supposed to realize that the wires would repel each other. If placed parallel with the ground, one would be pushed faster toward the ground. A number that would probably be negligable.

Re: Physics QOTD
 

current through a wire creates a circular, nonpolar magnetic field, which could have a force that would affect the acceleration. however, without the masses of the wires the acceleration cannot be determined, so i would have to guess that it would just be normal projectile motion.

Re: Physics QOTD
 

It says the plane they make is normal to this vertical vector so they should fall "side by side." So wouldn't their acceleration have two components? A vertical gravity component and a horizontal attraction/repulsion component? I don’t think you can forget about the horizontal component, but I wouldn't know how to calculate it without mass either...

Re: Physics QOTD
 

no, this is not a normal magnetic field. it does not attract or repulse in a linear way, but rather in a circular way. there are no poles, the magnetic field acts differently.

Re: Physics QOTD
 

They should attract with a force equal to (u0*L*ia*ib)/(2*pi*d). If L is a length vector in the direction of current flow, there must be a perpendicular B vector which means L cross B results in a force that is also perpendicular with the wire. Is that right? PHYS241 was not one of my favorites.

Re: Physics QOTD
 

You need to provide more information in my belief:

If it is not dealing with the magnetic field that is

You need the height of the wires. With that, you can use the following equations:

From sum of forces = m * a

mass * acceleration = mass * acceration due to gravity * HEIGHT

The mass cancels, because all things falling fall at the same acceleration from the same height (assuming no air resistance)

So, provide the hieght, and I can solve it!

Gabe Goldman

Re: Physics QOTD
 

Crazy's right, but you still need to know the mass of the rods!

Re: Physics QOTD
 

Right Caleb :)

As for ggoldman, as long as we are supposed to "Assume g=9.80 m/s^2" the objects should fall independent of height as well. The right-hand side of the equation you provided is just a gravitational potential energy term. All energy terms are relative and therefore useless unless you are trying to create a system and conserve the energies.

Re: Physics QOTD
 

agggh..u are right....It is late and I am still doing homework here so let me try again.


hmm...if you look at the cross section of the two wires, you could see the circular fields around them. Assuming the current is going out of the wires (or out of the page), using the right hand rule, the fields would be pointing in a ccw direction. when u look at where the two fields intersect in the middle, you have two vectors. The left vector is pointed up, and the other in pointed down. So maybe the right wire repels down in addition to gravity, and the other up???

I am guessing now...all my brain power is being used up on my english research paper on robotic parking facilities...


Gabe Goldman

Re: Physics QOTD
 

The rods do have an attractive force and neither rod is above or below the other. They are parallel with each other and directly side-by-side. The axis where the two B fields intersect in the middle just means the B fields cancel out there and the net B field is zero. The force between the rods should be in a plane independent of gravity.

I’ve always wanted a robotic parking facility…

Re: Physics QOTD
 

thats right...forgot the vectors would cancel.

but yeah...robotic parking is interesting...used alot in europe

Re: Physics QOTD
 

Perhaps you could post your document when you are finished? I have to admit, I don't know as much about robotic parking facilities as I should. The idea has gotten me all worked up and now I wanna read it!

Re: Physics QOTD
 

sorry this is not about the physics question, but I will post up my research once I finish.

I am investigating what it would take and cost to adapt European robotic parking structures (mainly the Trevipark underground approach). For example, it would need larger spaces and a more heavy duty lift for the US's larger cars.

Send me a PM if you would like more info, I'd hate to get this thread off topic..


I think the wires dont move...they are too far apart..btw

Re: Physics QOTD
 

The answer should be that in addition to the acceleration of gravity, a force acts between them with magnitude u_0*L*i_a*i_b/(2*pi*d), where d is the distance between them, L is the length in question, and i_a and i_b are the currents in the two wires. This force causes them to ATTRACT, not REPEL... think about it: the two magnetic fields act clockwise around the wires as you go in the direction of the current, so the force is i*L X B, or i*Forwards X Down for the right wire, which pushes it towards the left wire... hard to explain this without a diagram though. But the answer requires the "mass density" which is mass per unit length of the wire... call this k, and our final answer becomes an acceleration of 25Amp^2*k*u_0/(2*pi*3m) towards each other and 9.8m/s^2 down.

Re: Physics QOTD
 

If you don't buy my solution above, test it with the following circuit:

|--l|l|l|---V^V^V^---------------------|
|.............................|________|.......... |
|................................................. .....|
|---------------------------------------|
Where a "." means nothing is there and a "V^V^V^" is a resistor and l|l|l| is an emf source.
A simple resistor on a loop, but important is to observe whether the two branched wires fly towards or away from each other. Careful observation should show that they attract each other, with the current running the same way. Run current in opposite direcitons and they will repel.

Re: Physics QOTD
 

If I remember correctly, the formula you've used for calculating the force between the wires only applies when end effects can be neglected (i.e. when the length of the wires is very large compared to the distance between them). Since this is certainly NOT true in this case, with a length of 5 m and a separation of 3 m, the horizontal component of the acceleration will only be a very rough, order-of-magnitude sort of approximation.

Re: Physics QOTD
 

The equation he used is for wires of known length, that is what L is for.

But Grommit is right, when it comes right down to it, you need mass or mass density or SOMETHING like that to get the second law balance.

Re: Physics QOTD
 

That's quite true... actually I didn't even notice the "5m" length given, I had assumed they were infinite wires... :p. But trying to compute an exact solution might be a bit more pain than this problem merits. ;)

On the other hand, here's an interesting problem. You have a straight wire with radius r and electrons traveling through at velocity v, current i. What is the radial distribution of the electron density in the wire, in terms of r?