Question

The wire carrying 425 A to the motor of a commuter train feels an attractive force of 3.00 ✕ 10−3 N/m due to a parallel wire carrying 5.00 A to a headlight.(a) How far apart (in m) are the wires? m

Answer 1

$F=\frac{\mu_0\cdot I_1\cdot I_2}{2\pi\cdot d}\cdot l$

Based on the formula we have two currents with the same direction, we know this due to the attractive force.

Also, we don't have information about the length l of the wires, but thanks to the force which is expressed in terms of Newtons per length we can avoid it.

$\begin{gathered} \frac{F}{L}=3\cdot10^{-3}N/m=\frac{425A\cdot5A}{2\pi\cdot d}\cdot4\pi\cdot10^{-7}N/A^2 \\ d=0.1417m \end{gathered}$