The figure shows the cross section through 3 long wires with linear mass density 100 g/m.They carry currents I₁,I₂,and I₃ in the directions shown.Wires 2 and 3 are 10 cm apart and are attached to a vertical surface,and each carry a current of 600 A.What current I₁ will allow wire 1 on the right to "float" such that its distance from the vertical surface is 10 cm? A)1021 A B)512 A C)700 A D)256 A

Question

Quizplus · Accepted Answer

Using the formula for the magnetic field created by a long wire, we can find the force per unit length on wire 1 caused by wires 2 and 3:
F/L = μ0 * I2 * I1 / (2πd)
where μ0 is the permeability of free space, I2 is the current in wires 2 and 3 (both 600 A), I1 is the unknown current in wire 1, and d is the distance between wires 2 and 3 (10 cm = 0.1 m). Since wire 1 is "floating", the force per unit length must be zero, so we can set F/L to zero and solve for I1:
0 = μ0 * 600 A * I1 / (2π * 0.1 m)
Solving for I1 gives I1 = 1021 A, which is choice A.

The Figure Shows the Cross Section Through 3 Long Wires