The figure is a cross section through three long wires with linear mass density
ID: 1376754 • Letter: T
Question
The figure is a cross section through three long wires with linear mass density 53.0 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.
What current I will allow the upper wire to "float" so as to form an equilateral triangle with the lower wires?
The figure is a cross section through three long wires with linear mass density 53.0 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. What current I will allow the upper wire to float so as to form an equilateral triangle with the lower wires? Express your answer with the appropriate units.Explanation / Answer
condtiion for a wire to float is that its gravitational force = force between the parallel wires
i,e
mg = upi1i2L/2pid
where force between two parallel wires is given by F = u0i1i2L/2pid
where u0 = constnat = 4pi*10^-7 H/m,
i1, i2 are the currents ,
L s the length o the wire,
d is the distqance between these two wires
if current are in same directiion, there exists , they exert attractive force on each other
and
if currents are in opposite direction, they exert repulsive force
so
0.53 e-3 * 9.8 = 4*3.14 e-7 * i^2 * 0.04/(2*3.14 * 1)
so current i^2 = 25970
i = 161.151 Amps