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. $5\times10^{-6} Nm$
• B. $\frac{1}{2}\times 10^{-5}Nm$
• C. $5\sqrt{3}\times10^{-6} Nm$
• 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\,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 \hspace5mm ...(i)$
$ \tau_2 = MB\, sin\, 30^\circ \hspace5mm ...(ii)$
Dividing E (i) by E (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$