Two parallel wires separated by distance $d$ carry currents $I_1$ and $I_2$ in the same direction. What is the force per unit length $f$ between them?

Solution

1. The magnetic field $B_1$ produced by the first wire at a distance $d$ is given by $B_1=\frac{{\mu }_0{I}_1}{2\pi d}$.
2. The force per unit length $f$ exerted on the second wire carrying current $I_2$ is $f=I_2{B}_1\sin (\theta )$, where $\theta =90^{\circ }$.
3. Substituting the expression for $B_1$ into the force equation yields $$\begin{array}{rl}f& =I_2\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)\\ & =\frac{{\mu }_0{I}_1{I}_2}{2\pi d}\end{array}$$.
4. Hence the answer is (A).