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The perpendicular distance between two conductors of 12m each is 0.15cm. They carry 300A in same direction. The force acting between them is

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36N repulsion

14N attraction

144N attraction

288N repulsion

answer is B.

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Detailed Solution

i₁=i₂=300A, l=12m
r=0.15cm=0.15 x 10^-12m
$\begin{array}{l} ∴ F = \frac{μ_{0}}{2 π} \frac{i_{1} i_{2}}{r} \times l \\ = 2 \times 10^{- 7} \times \frac{300 \times 300}{0.15 \times 10^{- 2}} \times 12 \end{array}$
F=144N attraction

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