A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a…

1. $0.20\mathrm{ }\mathrm{N}\mathrm{m}$
2. $0.24\mathrm{ }\mathrm{N}\mathrm{m}$
3. $0.12\mathrm{ }\mathrm{N}\mathrm{m}$
4. $0.15\mathrm{ }\mathrm{N}\mathrm{m}$

Solution

$\overrightarrow{\mathrm{\tau }}=\mathrm{ }\overrightarrow{\mathrm{M}}\mathrm{ }\times \mathrm{ }\overrightarrow{\mathrm{B}}\mathrm{ }=\mathrm{M}\mathrm{B}\sin {60}^{\mathrm{o}}$
$=\mathrm{N}\mathrm{i}\mathrm{ }\mathrm{A}\mathrm{B}\sin {60}^{\mathrm{o}}$
$=50\mathrm{ }\times 2\times 0.12\mathrm{ }\times 0.1\mathrm{ }\times 0.2\mathrm{ }\times \mathrm{ }\frac{\sqrt{3}}{2}$
$=12\sqrt{3}\mathrm{ }\times {10}^{-2}\mathrm{ }\mathrm{N}\mathrm{m}=0.20748\mathrm{ }\mathrm{N}\mathrm{m}$

Asked in: NEET 2015 (Phase 2)

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