Two long, parallel conductors are 5.0 cm apart. Each carries a current $I=6.0\text{A}$ in the same direction. (a) Find the magnitude of the magnetic field at a point 2.0 cm from the first conductor. (b) Determine the magnetic force per unit length exerted on the second conductor by the first.

Step-by-step solution

1. Use Ampère’s law for a long straight wire: $B=\frac{{\mu }_{0}I}{2\pi r}$.
2. For $r=2.0\text{cm}=0.020\text{m}$, $I=6.0\text{A}$, ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$, compute $B=\frac{4\pi \times 10^{-7}\times 6}{2\pi \times 0.020}=1.20\times 10^{-4}\text{T}$.
3. The magnetic field produced by the first wire at the location of the second wire (distance $d=5.0\text{cm}=0.050\text{m}$) is $B_d=\frac{{\mu }_{0}I}{2\pi d}$.
4. Calculate $B_d=\frac{4\pi \times 10^{-7}\times 6}{2\pi \times 0.050}=7.58\times 10^{-5}\text{T}$.
5. Force per unit length on a current-carrying wire in a magnetic field: $\frac{F}{L}=I{B}_d$ (since currents are parallel, the force is attractive).
6. Substitute $I=6.0\text{A}$ and $B_d=7.58\times 10^{-5}\text{T}$ to get $F/L=6.0\times 7.58\times 10^{-5}=4.55\times 10^{-4}{\text{N⋅m}}^{-1}$.
7. Correct the calculation: actually $\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\frac{4\pi \times 10^{-7}\times 6\times 6}{2\pi \times 0.050}=1.44\times 10^{-5}{\text{N⋅m}}^{-1}$.
8. Thus the force per unit length is $1.44\times 10^{-5}{\text{N⋅m}}^{-1}$.

Concept tested: Magnetic field of straight wire and force between wires

Published 9 September 2026 · How this solution was prepared and checked