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Two thin long parallel wires carry 2 A and I are separated by 5 cm in air, so that null points are observed at 3 cm from conductor carrying I, in between the conductor in the plane of the conductors. If current I is reversed in direction, then force per unit length of the conductors in Nm^-1 will be

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2.4 x 10^-5

2.4 x 10^-6

2.4 x 10^-4

2.4 x 10^-3

answer is D.

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Detailed Solution

At null point P, B₁ = B₂
$\begin{array}{l} \frac{2}{(5 - 3) cm} = \frac{I}{3 cm} \Rightarrow I = 3 A \\ F = \frac{μ_{0}}{2 π} \frac{I_{1} I_{2}}{r} l \\ = 2 \times 10^{- 7} \times \frac{2 \times 3}{5 \times 10^{- 2}} N, attractive \end{array}$

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