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

Answer: (A)

$5\times {{10}^{-6}}Nm$

Answer 2

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
$\frac{{{\tau }_{1}}}{{{\tau }_{2}}}=\frac{{{10}^{-5}}}{{{\tau }_{2}}}=\frac{1}{1/2}$
$\Rightarrow$ ${{\tau }_{2}}=\frac{{{10}^{-5}}}{2}$
$=\frac{10}{2}\times {{10}^{-6}}$
$=5\times {{10}^{-6}}Nm$