Q114 / 150 · MHT CET 2024 (15 May Shift 1) · Physics

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

  • A
    $\frac{\mu_0}{4 \pi}\left(\frac{\mathrm{I}}{\mathrm{b}^2}\right)$
  • B
    $\frac{\mu_0}{4 \pi}\left(\frac{I^2}{b^2}\right)$.
  • C
    $\frac{\mu_0}{4 \pi}\left(\frac{I^2}{b}\right)$
  • Correct answer
    D
    $\frac{\mu_0}{4 \pi}\left(\frac{2 I^2}{b}\right)$

Solution

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

Hence, the correct answer is option: 'D'.

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