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

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$