Three long straight wires carrying current are arranged mutually parallel as shown in the figure. Th

Question

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire $Q$ is $\_\_\_\_$ .
$
\left(\mu_{\mathrm{o}}=4 \pi 	imes 10^{-7} \mathrm{~T} . \mathrm{m} / \mathrm{A}ight)
$

Professor PC Thomas Classes · Answer

The force per unit length between two parallel current-carrying wires is $F/L = \frac{\mu_0 I_1 I_2}{2\pi d}$. The wire Q experiences a force from P and a force from R. Calculating the individual attractive/repulsive forces using the specific currents and separation distances from the figure, we find the net force by vector addition. The resultant force on a 15 cm segment of wire Q computes exactly to $6 	imes 10^{-6}$ N, directed towards wire R.