Question

Three coplanar parallel wires, each carrying a current of 10 A along the same direction, are placed with a separation 5.0 cm between the consecutive ones. Find the magnitude of the magnetic force per unit length acting on the wires.

Answer 1

Let wires W₁, W₂ and W₃ be arranged as shown in the figure.

Given:
Magnitude of current in each wire, i₁ = i₂ = i₃ = 10 A
The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
$\frac{F}{l}=\frac{{\mathrm{\mu }}_0{i}_1{i}_2}{2\mathrm{\pi }d}$
So, for wire W₁,
$$\begin{gathered} \frac{F}{l}=\frac{F}{l}\mathrm{by}\mathrm{wire}{W}_2+\frac{F}{l}\mathrm{by}\mathrm{wire}{W}_3 \\ =\frac{{\mathrm{\mu }}_0\times 10\times 10}{2\mathrm{\pi }\times 5\times {10}^{-2}}+\frac{{\mathrm{\mu }}_0\times 10\times 10}{2\mathrm{\pi }\times 10\times {10}^{-2}} \\ =\frac{2\times {10}^{-7}\times {10}^2}{5\times {10}^{-2}}+\frac{2\times {10}^{-7}\times {10}^2}{10\times {10}^{-2}} \\ =4\times {10}^{-4}+2\times {10}^{-4} \\ =6\times {10}^{-4}\mathrm{N} \end{gathered}$$

For wire W₂,
$$\begin{gathered} \frac{F}{l}=\frac{F}{l}\mathrm{by}\mathrm{wire}{\mathrm{W}}_1-\frac{F}{l}\mathrm{by}\mathrm{wire}{\mathrm{W}}_3 \\ =\frac{{\mathrm{\mu }}_0\times 10\times 10}{2\mathrm{\pi }\times 5\times {10}^{-2}}-\frac{{\mathrm{\mu }}_0\times 10\times 10}{2\mathrm{\pi }\times 5\times {10}^{-2}}=0 \end{gathered}$$

For wire W₃,
$$\begin{gathered} \frac{F}{l}=\frac{F}{l}\mathrm{by}\mathrm{wire}{\mathrm{W}}_1+\frac{F}{l}\mathrm{by}\mathrm{wire}{\mathrm{W}}_2 \\ =\frac{{\mathrm{\mu }}_0\times 10\times 10}{2\mathrm{\pi }\times 5\times {10}^{-2}}+\frac{{\mathrm{\mu }}_0\times 10\times 10}{2\mathrm{\pi }\times 10\times {10}^{-2}} \end{gathered}$$
= 6 × 10⁻⁴ N