A horizontal copper conductor with the current density is 3,0 A / mm^2), the wire is perpendicular to the horizontal magnetic field. how big is the magnetic flux density to be the wire float in the magnetic field. The answer is 29mT but how am I supposed to calculate this?

Question

anonymous · Answer

The force on a length of wire L carrying a curren I in a magnetic field of strength B will experience of force F.  If this force is countering the weigh of the wire then F = mg where m is the mass and g is the acceleration due to gravity 8.9 m/sec^2.   therefore
 
 B*I*L = m*g

Dividing both sides by the volume of the wire you get

B*I*L/V = mg/V  =  rho*g   where rho is the density of copper in Kg/m^3

Now look at the left side of the equation and rewrite V as A (area of the wire)*L(wires 

length) you get

B*I*L/A*L =  B*I/A  = B*J  where J is the current density

Therefore                    B*J = rho*g   solve for B