Levitating wire in magnetic field.

[SherlockOhms]

Homework Statement

A copper wire of diameter d carries a current density at the Earth’s equator where the Earth’s magnetic field is horizontal, points north, and has magnitude 5.yz × 10^-5 T . The wire lies in a plane that is parallel to the surface of the Earth and is oriented in the east west direction. The density and resistivity of copper are ρm = 8.9 × 10^3 kg/m3 and ρ = 1.7 × 10^−8 Ω-m, respectively. (x,y and z are given constants).

(a) How large must J be, and which direction must it flow in order to levitate the wire? Use g= 9.81 m/s

.(b) When the wire is floating, how much power will be dissipated per cubic meter due to resistive heating in the wire?

Homework Equations

F = ILB (F = JALB = JVB.)
F = mg.
m = density(Volume)
So, F = density(volume)(g)

The Attempt at a Solution

JVB = density(V)(g)
cancel the V's and solve for J (Positive so it flows in the east direction?)

I think that's the correct way to do (a), please point out if I've made an error. I can't seem to find the relative equation for part (b). Any help would be great!

 

[SherlockOhms]

Anyone have any clue? I can't see a way to do it without some extra numerical value for the dimensions of the wire.

 

[haruspex]

The specific dimensions should cancel out. Create unknowns for cross-sectional area etc. as necessary and post you working.

 

[SherlockOhms]

Well, I said that resistance = (resistivity x L) / A. Then JA = I. Power = (JA)^2 x (restivity x L) / A. This simplifies to be P = J^2 x resistivity x V. I don't know the value of V though. I'm not sure what way you had in mind. Any ideas?

 

[haruspex]

This simplifies to be P = J^2 x resistivity x V. I don't know the value of V though.

You are not asked to find P. What does the question ask for?