Question

FIGURE EX29.37 is a cross-section through three long wires with a linear mass density $50g/m$. They each carry equal currents in the directions shown. The lower two wires are $4.0cm$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?

Short answer

$240A$current allow the upper wire to “float” so as to form an equilateral triangle with the lower wires.

Step-by-step solution

Given Information

We have given three long wires with a linear mass density $50g/m$and the lower two wires $4.0cm$are apart and are attached to a table.

Simplify

The weight per unit length of the floating wire must be balanced by force per unit length due to the interaction of the current of the floating wire with the other two wires. Set

${\lambda }_m=\frac{m}{l}=50g/m.$

Then weight per unit length is

$f_g=\frac{F_g}{l}=\frac{mg}{l}={\lambda }_{m}g.$

The forces between the wires on the table and the floating wire are repulsive and act at

${30}^{\circ }$away from the vertical. From the figure, we see that their horizontal components out and that their vertical components add up. Since they have the same magnitude of

$F_1=F_2=\frac{{\mu }_0}{2\Omega }\frac{I^{2}l}{d},$

Calculation

The magnitude of the force which points vertically up is simply twice the vertical component of ,say,${\overrightarrow{F}}_1:$