A long straight wire carrying a current of 25 A rests on the table… | MHT CET 2023 (9 May, Shift 1)

Q35MHT CET 2023 · 9 MAY · SHIFT 1Medium

Force on Current-Carrying Conductors

A long straight wire carrying a current of $25\text{ A}$ rests on the table. Another wire PQ of length $1\text{ m}$ and mass $2.5\text{ g}$ carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will wire PQ rise? ( $\mu_0 = 4\pi \times 10^{-7}$ SI unit)

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Answer: (c)

Opposite currents repel, so PQ rises until the repulsion equals its weight.
$\frac{\mu_0 I^2 L}{2\pi h} = mg \Rightarrow h = \frac{\mu_0 I^2 L}{2\pi mg}$
$h = \frac{2\times10^{-7}\times625\times1}{2.5\times10^{-3}\times10} = 5\times10^{-3}\text{ m}$
So the answer is (c) 5 mm (taking $g = 10\text{ m/s}^2$ ).

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