A) \[2\pi \times {{10}^{-7}}N\times m\]
B) \[2\pi \times {{10}^{-5}}N\times m\]
C) \[0.5N\times m\]
D) \[0.5\times {{10}^{2}}N\times m\] (\[{{\mu }_{0}}=4\pi \times {{10}^{-7}}weber/amp\times m\])
Correct Answer: A
Solution :
Torque \[\tau =M{{B}_{H}}\sin \theta \] \[=0.1\times {{10}^{-3}}\times 4\pi \times {{10}^{-3}}\times \sin {{30}^{o}}={{10}^{-7}}\times 4\pi \times \frac{1}{2}\] \[=2\pi \times {{10}^{-7}}N\times m\]
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