Question

A long straight conductor PQ, carrying a current of 75 A , is fixed horizontally. Another long conductor XY is parallel to PQ at a distance of 5mm,in air. Conductor XY is free to move and carries a current I. Calculate the magnitude and direction of’ ‘I’ for which the magnetic repulsion just balances the weight of conductor XY ( mass per unit length for conductor XY is 10-2 kg/m).
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Answer 1

Dear Student,
$$\begin{gathered} Hereforceonsecondconductorduetofirstwillbeequaltoweightperunitlengthofsecondwire \\ F=\frac{{\mu }_0}{4\pi }\frac{2\times 75\times I}{5\times {10}^{-3}}=mg \\ orI=\frac{{10}^{-2}\times 10\times 5\times {10}^{-3}}{2\times 75\times {10}^{-7}}=\frac{5\times {10}^3}{150}=33.33A \end{gathered}$$
Regard