Question

A torque of $10^{-5}$ Nm is required to hold a magnet at $90^{\circ }$ with the horizontal component of the earth's 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)

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.
$\tau =MB\sin \theta$
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 Eq. (i) by Eq. (ii) , we ge
$\frac{{\tau }_1}{{\tau }_2}=\frac{10^{-5}}{{\tau }_2}=\frac{1}{1/2}$
$\Rightarrow {\tau }_2=\frac{10^{-5}}{2}$
$=\frac{10}{2}\times 10^{-6}$
$=5\times 10^{-6}Nm$