# Homework Help: Magnetic fields at earth equator

1. Mar 24, 2004

### pattiecake

Near the earth's equator, the earth's magnetic field is approximately horizontal. What current would have to flow through a long wire with a mass density of 10g/m to keep the wire floating in mid-air? What would be the direction of the wire, and the current in the wire?

Well, I know I'd need a very LARGE current. Not sure what equation to go with. Can anyone point me in the right direction?

Also I'm having trouble picturing how the earth's magnetic field is horizontal. The field lines go in a circle, right?

Last edited by a moderator: Apr 20, 2017
2. Mar 24, 2004

### pattiecake

yo whoever moved my thread here to k-12 please move it back to where it was- classical physics. This was from my college honors physics II lab, and I've noticed that not many people respond to this k-12 forum.

3. Mar 24, 2004

### chroot

Staff Emeritus
Well, this is a college level problem -- not a K-12 one. No matter. We do not answer homework questions of any sort in the general physics forums -- only here in the homework help section.

We're working on getting more people to contribute to this part of the forum.

The field lines don't go in circles, though they do make complete loops. The field lines emerge from the earth's surface at its geomagnetic south pole (somewhat close to the geographical south pole), make a large loop outside the earth, then re-enter the earth at the geomagnetic north pole.

Look at the picture on the upper-left of the website you linked. Draw a sphere where the bar magnet is. That's the earth.

Why don't you start by describing the magnetic field produced by running a current of A amps through a long wire? Your textbook should provide this formula and its derivation.

- Warren

4. Mar 25, 2004

### pattiecake

Thanks for your support! Anyways, at this point, the magnetic force should be equal and opposite the force of gravity (right??). So the force of a conductor of length L when placed in a uniform magnetic field B is Fb=IL x B. But that's another thing- I'm not sure exactly what to do with a cross product? Do you just take the sine of an angle? Which angle? Also I think I'm suposed to integrate this equation...help!

5. Mar 25, 2004

### Staff: Mentor

Yes. The net force on the wire is zero for equilibrium.
Right.
Yes, in general, the magnitude of the cross product iL x B is iLB sin&theta;, where &theta; is the angle between the direction the current is flowing in the wire and the direction of the magnetic field. In this case, you want maximum force (why not make it easy?) so you would orient the wire perpendicular to the magnetic field.

For more on cross products, check your text. Here are two sites to get you started: http://cyclo.mit.edu/~schol/802x/howtos/rhr_howto.pdf [Broken]
http://www.math.arizona.edu/~vector/Block1/vectors/node22.html [Broken]
No need to integrate. Write your equation for equilibrium for a length of wire L. Then divide both sides by L to get things in terms of mass/length.

Last edited by a moderator: May 1, 2017
6. Mar 25, 2004

### Chen

$$F_m = \ell \vec v \times \vec B = IB\ell \sin \alpha$$
Where $$\alpha$$ is the angle between the current direction and the magnetic field. To find the direction of the force, use this method:
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/MagneticField/RightHandRule.html [Broken]

You need that force to cancel the gravitational force:
$$F_m = F_g$$
$$IB\ell \sin \alpha = mg$$
$$I = \frac{mg}{B\ell \sin \alpha}$$
Basically you decide what $$\alpha$$ is, but the bigge it is the less current you will need, so you should try to make it 90 degrees. And $$\ell$$, the length of the wire, is your call. Again, the longer the wire is, the less current you will need.

Last edited by a moderator: May 1, 2017
7. Mar 25, 2004

### Staff: Mentor

Careful. The longer the wire, the greater its mass. It turns out that the current needed to support the wire is independent of the length of the wire.

8. Mar 25, 2004

### Chen

Also, the longer the wire, the more resistance it has. So you might want it to be short, so that you need less voltage to create the current you want.

9. Mar 25, 2004

### Severian596

Interesting! Nice stuff guys...

10. Mar 25, 2004

### pattiecake

Wow! Thanks for all the great links guys! I actually wrote the letters v, B, and Fb on my hand so I can get the right hand rule down!