A long horizontal rigidly supported wire carries a current i_a=96A. Directly above it and parallel to it at a distance, another wire of 0.144N weight per metre carrying a current i_b=24A, in a direction opposite to that of i_a. If the upper wire is to float in air due to magnetic repulsion, then its distance (in mm) from the lower wire is

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4.8

1.6

3.2

9.6

answer is C.

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Detailed Solution

$\frac{Mg}{l} = \frac{μ_{0} i_{1} i_{2}}{2 πr}; 0 .144 = \frac{4 π \times 10^{- 7} \times 24 \times 96}{2 π \times r}$
r = 3.2mm

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