Magnet, wire, earth's magnetic field

[thermocleanse]

Homework Statement

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth's magnetic field, B, which is horizontal, perpendicular to the wire, and of magnitude 5.0 x 10^-5 T. What current would the wire carry? Does the answer seem feasible? Explain briefly.

Homework Equations

The Attempt at a Solution

I'm completely lost. Please start in a couple of places so I can get moving on this. Thank you.

 

[Stonebridge]

Take a length of wire = 1m (unit length)
Do you know a formula for the force F, on a wire length L, carrying a current I, in a magnetic field B?

 

[thermocleanse]

Yes, F (max) = IlBsin(theta), because theta = 90 degrees...right?

 

[Stonebridge]

Yes, the wire and field are perpendicular to each other so it's just F=ILB (sin 90 = 1)
Good.

So this is the force on the wire due to the magnetic field. You should be able to show that if the field is horizontal and in a direction north-south, say, and the wire is in the east-west direction, the force can be upwards. There is a "hand" rule for this.

That done, you just need to find out the current when this upwards force is enough to support the weight of the wire.

Spoiler

The wire's weight is mg. To work this out you need to find the mass of the 1m length of wire.

Spoiler

You have its diameter and length, you just need to know the density of the copper.

 

[thermocleanse]

but i don't have the length of the copper. i only have its diameter. is there another approach? thanks.

 

[Stonebridge]

but i don't have the length of the copper. i only have its diameter. is there another approach? thanks.

You take the length to be 1m. (Unit length, as I said in my 1st post) You can take any length you want, but the result will always be the same.
The longer the wire the heavier it is; but the longer the wire, the greater the upward force.

If you take the length as 1m it makes the calculation much easier.
(You could say, let the length be L. It then cancels out in the equation.)

Can you do it now?

 

[thermocleanse]

i'm sorry, i don't get it. are you saying, just use F = IB, because L can be 1m?
Then how do i incorporate the density of Cu?

 

[Stonebridge]

i'm sorry, i don't get it. are you saying, just use F = IB, because L can be 1m?
Then how do i incorporate the density of Cu?

The wire has weight. It's this weight that must be supported by the force F=BIL if the wire is to float.
The weight of the wire is mg
the mass of the wire m= volume x density
the volume of the wire = area of cross section times length = AL
put mg=BIL
As the length of the wire is not given
either
say, let the wire be unit length (1m)
or
say let the length be L
Either way, when you do the maths, the length disappears as it actually doesn't matter what the length of the wire is in this problem.

 

[thermocleanse]

thx 4 spelling it out. i just finished up what I've attached to this thread as you got me this new post. please have a look at these attached pages. i think it's on the right track, maybe even correct, but I'm afraid I've lost sight of one facet near the end, namely knowing how to use the equation in relation to seeing if the force will hold up the copper wire, now with a known weight...thank you...

 

[Stonebridge]

There is an error in the calculation of the mass
Keeping to kg and m
Density of copper = 8940 kg/m³
mass of 1m of wire= 8940 times volume in m³
your value for the volume is correct in m³

The mass of the wire is about 7.0 x 10^-3 kg

If you carry the correct value through you should get the current in the copper wire to be just under 1400 A

 

[thermocleanse]

i was tired and miscalculated density. thanks for correcting that. i get the same math as you do. so, the wire can "float" b/c of its small mass, right? thank you for your help.

 

[Stonebridge]

Yes, it can float if it carries a current of about 1400A.
The question also asks if this is feasible.
Do you think it would be possible to pass a current of 1400A through a wire of 1mm diameter? If not; why not?