question_answer

A torque of \[{{10}^{-\,5}}\,Nm\] is required to hold a magnet at \[90{}^\circ \] with the horizontal component H of the earth's magnetic field. The torque to hold it at \[30{}^\circ \] will be

A) \[5\times {{10}^{-6}}\,Nm\]

B) 5 Nm

C) \[\frac{1}{3}\times {{10}^{-5}}\,Nm\]

D) \[5\sqrt{3}\times {{10}^{-6}}\,Nm\]