# Magnetic Levitation

1. May 2, 2015

### Angie K.

1. The problem statement, all variables and given/known data

A straight 2.34-mm-diameter copper wire can just 'float' horizontally in air because of the force of the Earth's magnetic field B, which is horizontal, perpendicular to the wire, and of magnitude 5 × 10-5 T. What current I does the wire carry? (The density of copper is 8.96 g/cm3).

•The wire feels a downward gravitational force of magnitude mg, where m is mass and g = 9.80 m/s2 is the gravitational field strength near the Earth's surface.

2. Relevant equations

F = I*L*B sin θ

3. The attempt at a solution

F = I*L*B
F = mg = mass of copper * earth's gravitational field strength
(63.546*9.8) = I*(.234)(5*10^-5)

then solve for I

but I think I'm doing something wrong with the F (plugging in wrong values?)

2. May 2, 2015

### Staff: Mentor

What is 63.546 and where does it come from?
Why did you use 0.234 (units?) as length?

If you use numbers that are not given in the problem statement (for exactly this quantity - like the magnetic field strength) you should explain what you are doing.

3. May 2, 2015

### Angie K.

63.546 is the mass of copper (used in equation F=ma)
and I convered 2.34mm to cm

Last edited by a moderator: May 3, 2015
4. May 2, 2015

### Staff: Mentor

There is no 63.546 given in the problem statement, where does it come from? If you calculated it, how?
The diameter of the copper wire is 0.234cm, fine, but that is not its length (also, where is the point in using cm?).

5. May 2, 2015

### Angie K.

It's just the atomic mass of copper and I thought I could use it to figure out some unknowns.

Last edited by a moderator: May 3, 2015
6. May 2, 2015

### Angie K.

Length is 2pi*r which will be 1.17

7. May 2, 2015

### Staff: Mentor

That is the circumference, not the length.
Plugging random numbers into random formulas does not help.

8. May 2, 2015

### Angie K.

So I got the volume to be: 6.7088

Then the mass is 60.11

And the length I'm having a hard time with. I guess it's more of a math issue for me than physics.

9. May 3, 2015

### Staff: Mentor

You keep posting numbers without an explanation how you got them, and without units.
Sorry, it is impossible to help like that.

10. May 3, 2015

### Staff: Mentor

It's just a straight piece of copper wire, not a copper ring. We don't know its length, so denote it simply as $\ l\$metres.

11. May 3, 2015

### Angie K.

Isn't there a way to find the length by findung the volume or something? Because there's the given density of copper.

12. May 3, 2015

### Staff: Mentor

You don't know the length and there is no way to find it. The length does not matter, it will cancel out if you simplify the equations.

13. May 16, 2015

### Angie K.

(8.96E3kg)/m3(Π*.001883m)*L=m
.094267kg/m*L=m

Fg=mg
Fg = (.094267kg/m*L)(9.80m/s)= .9238kg/s*L

FB = I*L*B which = Fg

magnetic field,

B = 2Πx10-7(2.08/.0571) sin 60
= 4.1610*10-5kg/A*s

So now FB=Fg
.9238kg*L = I*L(4.1610*10-5kg/A*s)

14. May 16, 2015

### Staff: Mentor

Large, yes. I'm expecting it will be because the earth's field is weak.

Where does the sin 60 come from?

Does the textbook give the correct answer?

15. May 16, 2015

### Angie K.

It is an online homework assignment so I don't know the answer until I get the right solution.

16. May 16, 2015

### Staff: Mentor

17. May 16, 2015

### Angie K.

It is an equilateral triangle so each angle is 60.

18. May 16, 2015

### Staff: Mentor

One wire perpendicular to the field, where is the triangle?

19. May 16, 2015

### Angie K.

For current going into the page the field goes to the right and the for current out of the page the field is to the left.

Is sin 60 the wrong angle ? Would it be 30?

20. May 16, 2015

### Staff: Mentor

I see only right angles. The wire is horizontal and perpendicular to the field.