Problem 5 4 points Suspending a wire A long straight wire is

Problem 5. (4 points) Suspending a wire. A long straight wire is held fixed in a horizontal position. A second parallel wire is 1.2 mm below the first but is free to fall under its own weight. The second wire is copper (density 8920 kg/m3), with diameter 1.0 mm. What equal current in both wires will suspend the lower wire against gravity? (A) 10 A (B) 20 A (C) 30 A (D) 40 A

Solution

The bottom wire has a certain fixed mass per meter, or \"linear density\". In this case,

linear density = density * cross-sectional area
= 8920 kg/m^3 * (1mm/2)^2 * pi = 0.007 kg/m

This gives a weight-per-meter of

weight per meter = mass per meter * g
= 0.007 kg/m * 9.81 N/kg = 0.0686 N/m

We want to balance this weight-per-meter against the force-per-meter exerted by the current in both wires. Ampere\'s force law is applies directly here:

inductive force per meter = 2k * I_1 * 1_2 / r,

where 2k = 2e-7 N/A^2 and r = 1.2mm. Thus

0.0686N/m = 2e-7N/A^2 * I^2 / 1.2mm, so

I = sqrt(0.0686N/m * 0.0012m / (2e-7N/A^2))
= 20 A