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

EAMCETEAMCET 2006

Solution:

$ \frac{F}{l}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{i}_{a}}{{i}_{b}}}{r} $ $ \therefore $ $ 0.144=\frac{{{10}^{-7}}\times 2\times 96\times 24}{r} $ $ \therefore $ $ r=\frac{{{10}^{-7}}\times 2\times 96\times 24}{0.144} $ $ =3.2\times {{10}^{-3}}m $ $ =3.2\,mm $