Question

Figure shows a part of an electric circuit. The wires AB, CD and EF are long and have identical resistance. The separation between the neighbouring wires is 1.0 cm. The wires AE and BF have negligible resistance and the ammeter reads 30 A. Calculate the magnetic force per unit length of AB and CD.

Answer 1

Since wires AB, CD and EF have identical resistance, the current (30 A) gets equally distributed in them, that is, 10 A in each wire.

The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
$\frac{F}{l}=\frac{{\mathrm{\mu }}_0{i}_1{i}_2}{2\mathrm{\pi }d}$
∴ Magnetic force per unit length of AB = Force due to current in CD + Force due to current in EF
$$\begin{gathered} \frac{F}{l}=\frac{{\mathrm{\mu }}_0\times 10\times 10}{2\mathrm{\pi }\times 1\times {10}^{-2}}+\frac{{\mathrm{\mu }}_0\times 10\times 10}{2\mathrm{\pi }\times 2\times {10}^{-2}} \\ =\frac{2\times {10}^{-7}\times {10}^2}{{10}^{-2}}+\frac{2\times {10}^{-7}\times {10}^2}{2\times {10}^{-2}} \\ =2\times {10}^{-3}+{10}^{-3} \\ =3\times {10}^{-3}\mathrm{N}/\mathrm{m} \end{gathered}$$
Similarly,
Magnetic force per unit length of CD = Force due to current in AB − Force due to current in EF
∵ Force on CD due to current in AB = Force due to current in EF
∴ Magnetic force per unit length of CD = 0