Question

A torque of $ {{10}^{-5}}Nm $ is required to hold a magnet at $ 90{}^\circ $ with the horizontal component of the earths magnetic field. The torque required to hold it at $ 30{}^\circ $ will be

• A. $ 5\times {{10}^{-6}}Nm $
• B. $ \frac{1}{2}\times {{10}^{-5}}Nm $
• C. $ 5\sqrt{3}\times {{10}^{-6}}Nm $
• D. Data is insufficient

Answer 1

The Correct Option is A

The magnet in a magnetic field experiences a torque which rotates the magnet to a position in which the axis of the magnet is parallel to the field.
$ \tau =MB\text{ }sin\,\theta $
where, $M$ is magnetic dipole moment, $B$ the magnetic field and $\theta$ the angle between the two.
Given, $ {{\tau }_{1}}={{10}^{-5}}Nm,{{\theta }_{1}}=90{}^\circ ,{{\theta }_{2}}=30{}^\circ , $
$ {{\tau }_{1}}=MB\sin 90{}^\circ $ ...(i)
$ {{\tau }_{2}}=MB\sin 30{}^\circ $ ...(ii)
Dividing E (i) by E i(ii), we get
$ \frac{{{\tau }_{1}}}{{{\tau }_{2}}}=\frac{{{10}^{-5}}}{{{\tau }_{2}}}=\frac{1}{1/2} $
$ \Rightarrow $ $ {{\tau }_{2}}=\frac{{{10}^{-5}}}{2} $
$ =\frac{10}{2}\times {{10}^{-6}} $
$ =5\times {{10}^{-6}}Nm $