Question
A current carrying circular loop of radius 2 cm with unit normal $\widehat{n} = \frac{\widehat{k} + \widehat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\overrightarrow{B} = B_{0}(3\widehat{i} + 2\widehat{k})$. If $B_{0} = 4 \times 10^{- 3}\text{ }T$ and current $I = 100\sqrt{2}\text{ }A$, the torque experienced by the loop is $\_\_\_\_$ Wb.A. $(\pi = 3.14)$
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Solution
Sol. [Key Concept: Torque $\tau = M \times \ B$ with $M = NIA\ n$ ($n$ the unit normal); evaluating the cross product gives $5024 \times 10^{- 7}$ along $j$.]
$${\overrightarrow{\tau} = \overrightarrow{M} \times \overrightarrow{B} }{\Rightarrow 100\sqrt{2} \times 3.14 \times 4 \times 10^{- 4}\left\lbrack \left( \frac{\widehat{k} + \widehat{i}}{\sqrt{2}} \right) \times (3\widehat{i} + 2\widehat{k}) \right\rbrack B_{0} }{\Rightarrow 16 \times 3.14 \times \lbrack + 3\widehat{j} - 2\widehat{j}\rbrack \times 10^{- 5} }{\Rightarrow 50.24 \times 10^{- 5}\widehat{j} }$$
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