Question

As a student technician, you are preparing a lecture demonstration on "magnetic suspension." You have a 15-cm long straight, rigid wire that will be suspended by flexible conductive lightweight leads above a long, straight wire. Currents that are equal but are in opposite directions will be established in the two wires so the 15-cm wire "floats" a distance h above the long wire with no tension in its suspension leads. If the mass of the 15-cm wire is 13.4 g and if h (the distance between the central axes of the two wires) is 1.5 mm, what should their common current be?

Answer 1

Answer:

I = 81.0721 A

Explanation:

There are two force that are acting on wire i.e. upward magnetic force and weight of body itself

considering the equilibrium condition

Apply $\sum F$ in vertical direction is 0 thus we have

F_B - mg = 0

repulsive that acting oin wire is

$F_b = 2[\frac{\mu_o I^2 L}{4\pi R}]$

Plugging this value tn above equation

$2[\frac{\mu_o I^2 L}{4\pi R}] - mg = o$

solving for current  I

$I = \sqrt{\frac{4\pi R mg}{2\mu_o L}}$

$I = \sqrt{\frac{13.4\times 10^{-3}9.81(1.5\times 10^{-3}}{2(10^{-7} T.m/A (.15m)}}$

I = 81.0721 A