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Question

A wire AB is carrying a steady current 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the directions of the current flowing in CD with respect to that in AB [Take the value of g=10ms2]

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Solution

Force per unit length between the current carrying wires is given as:
F=μ04π2I1I2r=mg
Here, m is mass per unit length
107×2×12×51×103=m×10
m=107×2×12×51×103×110
=1.2×103kgm1

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