The figure is a cross section through three long wires with linear mass density

Question

The figure is a cross section through three long wires with linear mass density 54 g/m . They each carry equal currents in the directions shown. The lower two wires are 4.0 cm apart and are attached to a table.

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What current I will allow the upper wire to "float" so as to form an equilateral triangle with the lower wires? (Answer in A)

Explanation / Answer

let legth of wire be L

Magnetic field due to a current carrying wire = u I /2pie*0.04

Force = 2 * I1 * L * u I /2pie*0.04 *cos 30 = 0.054 * L * 9.8 (by equilibrium)

So I1 = 0.2646 *2pie*0.04 / u =52920 / I

I = current in both the wires

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