A long horizontal wire carries a current of 47 A. A second wire, made of 2.5 mm diameter copper wire and parallel to the first but 20 cm below it, is held in suspension magnetically.

Question

What is the magnitude of the current in the lower wire?

Answer 1

·The bottom wire held in suspension only when 2 forces acting on it are equal in magnitude and opposite in direction.

Weight W = mg (where m – mass of the wire, g = 9.8 m/ s² – free fall acceleration) acts down.

Magnetic force F_m = ϻ_0∙I₁∙I₂∙L /2πr ( ϻ₀ = 4π∙10 ^{– 7} - H/m – permeability of vacuum, I₁ and I₂ – currents in wires, r = 0.2 m – distance between them, L – length of the wire) acts up. Their magnitudes are the same, hence

                                                                     mg = ϻ_0∙I₁∙I₂∙L /2πr

To find mass of the wire m, we can use the formula m = ρ∙V (here ρ = 8.92 x 10 ³ kg/m³ – density of copper and V – volume of a wire).

Volume of a wire V = A∙L ( here A = π∙d²/4 – cross -sectional area of wire and d = 2.5 x 10 ^-3 m – its diameter) .

Combining all formulas, we can write

                                                                         ρ∙π∙d²∙L∙g /4 = ϻ_0∙I₁∙I₂∙L /2πr

Interesting that now the length of wire will cancel out.

Solving against current I₂, we get

                                                                    I₂ = ρ∙π²∙d²∙r·g / 2∙ϻ_0∙I₁

Now you need only put the numeric values of all given variables. If you do everything right, your answer will be 9.1 kA

BTW, currents in this case must have the same directions! My only concern is about the huge current in the bottom wire. Not sure that the gauge of it is enough to handle such high current.