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?

[MikeDubreuil]

Originally Posted by SilverStar

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?

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

[JoshJ]

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

[MikeDubreuil]

Originally Posted by JoshJ

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

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.

[Solace]

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.

[CrazyCarl461]

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...

[Solace]

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.

[CrazyCarl461]

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.

[ggoldman]

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

[Caleb Fulton]

Originally Posted by CrazyCarl461

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.

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

[CrazyCarl461]

Originally Posted by Caleb Fulton

but you still need to know the mass of the rods!

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.

[ggoldman]

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

[CrazyCarl461]

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…

[ggoldman]

thats right...forgot the vectors would cancel.

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

[CrazyCarl461]

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!