Question

A wire AB is carrying a steady current 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 $m{s}^2$]

Answer 1

Force per unit length between the current carrying wires is given as :
$F=\frac{{\mu }_0}{4\pi }\frac{2I_1{I}_2}{r}$
Let m be the mass per unit length of wire CD. As the force balances the weight of the wire,
$\therefore \frac{{\mu }_0}{4\pi }\frac{2I_1{I}_2}{r}=mg$
Here, m is mass per unit length.
${10}^{-7}\times \frac{2\times 12\times 5}{1\times {10}^{-3}}=m\times 10$
$m={10}^{-7}\times \frac{2\times 12\times 5}{1\times {10}^{-3}}\times \frac{1}{10}$
$=1.2\times {10}^{-3}kg{m}^{-1}$
The direction of current in CD is opposite to the direction of current AB.