Two long, straight conductors are separated by a distance of $0.04,\text{m}$. One carries a current of $9,\text{A}$ and the other $7,\text{A}$, both flowing in opposite directions. What is the magnitude of the force per unit length exerted on each conductor?

Step-by-step solution

1. Recall the magnetic force per unit length between two parallel currents: $\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$, where ${\mu }_0=4\pi \times 10^{-7},\text{H/m}$.
2. Compute the product of currents: $I_1{I}_2=9,\text{A}\times 7,\text{A}=63,{\text{A}}^2$.
3. Insert the values into the formula: $\frac{F}{L}=\frac{4\pi \times 10^{-7},\text{H/m}\times 63}{2\pi \times 0.04,\text{m}}$.
4. Simplify: the $4\pi$ and $2\pi$ cancel, leaving $\frac{F}{L}=\frac{2\times 10^{-6}\times 63}{0.04}=\frac{1.26\times 10^{-4}}{0.04}$.
5. Calculate: $1.26\times 10^{-4}/0.04=3.15\times 10^{-3},\text{N/m}$? Wait recompute: $\frac{1.26\times 10^{-4}}{0.04}=3.15\times 10^{-3},\text{N/m}$.
6. Correct the arithmetic: $2\times 10^{-6}\times 63=1.26\times 10^{-4}$, divide by $0.04$ gives $3.15\times 10^{-3},\text{N/m}$. Actually earlier we had $3.15\times 10^{-4},\text{N/m}$. Re‑check: $1.26\times 10^{-4}÷0.04=3.15\times 10^{-3}$. So the correct value is $3.15\times 10^{-3},\text{N/m}$.
7. State the final result $F/L=3.15\times 10^{-3},\text{N/m}$.

Concept tested: Magnetic force between parallel currents

Published 22 September 2026 · How this solution was prepared and checked