A) \[20{}^\circ \]
B) \[45{}^\circ \]
C) \[~60{}^\circ \]
D) \[30{}^\circ \]
Correct Answer: D
Solution :
(d) \[30{}^\circ \] In stable equilibrium, a compass needle points along magnetic north and experiences no torque. When it is turned through declination a, it points along geographic north and experiences torque. \[\tau =MB\,\sin \alpha \] \[\therefore \sin \alpha =\frac{\tau }{MB}=\frac{12\times {{10}^{-3}}}{60\times 40\times {{10}^{-6}}}=\frac{1}{2}\] Or \[\alpha =30{}^\circ \]You need to login to perform this action.
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