Two parallel wires separated by distance 'b' are carrying equal current 'I' in the same direction. The force…

Two parallel wires separated by distance 'b' are carrying equal current 'I' in the same direction. The force per unit length of the wire is
  1. $\frac{\mu_0}{4 \pi}\left(\frac{\mathrm{I}}{\mathrm{b}^2}\right)$
  2. $\frac{\mu_0}{4 \pi}\left(\frac{I^2}{b^2}\right)$.
  3. $\frac{\mu_0}{4 \pi}\left(\frac{I^2}{b}\right)$
  4. $\frac{\mu_0}{4 \pi}\left(\frac{2 I^2}{b}\right)$

Solution

$\begin{aligned} & \frac{\mathrm{F}}{l}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}} \\ \therefore \quad & \frac{\mathrm{F}}{l}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{I}^2}{b} \\ \therefore \quad & \phi=60^{\circ}=\pi / 3\end{aligned}$ .

Asked in: MHT CET 2024 (15 May Shift 1)

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