A) \[6.0\times {{10}^{-4}}N-m\]
B) \[6.0\times {{10}^{-5}}N-m\]
C) \[1.2\times {{10}^{-3}}N-m\]
D) 6.0 N-m
Correct Answer: A
Solution :
When a current carrying coil is placed in a uniform magnetic field B, then torque, \[\tau =MB\text{ }sin\theta \] \[\tau =(NIA)B\text{ }sin\theta \] \[\tau =10\times 1\times 0.2\times 6\times {{10}^{-4}}\times sin\text{ }30{}^\circ \] \[=10\times 1\times 0.2\times 6\times {{10}^{-4}}\times \frac{1}{2}\] \[=6\times {{10}^{-4}}N-m\]
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