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Physics Question on Magnetism and matter

A torque of 10−510^{-5} Nm is required to hold a magnet at 90∘90^\circ with the horizontal component of the earth's magnetic field. The torque required to hold it at 30∘30^\circ will be

A

5×10−6Nm5\times10^{-6} Nm

B

12×10−5Nm\frac{1}{2}\times 10^{-5}Nm

C

53×10−6Nm5\sqrt{3}\times10^{-6} Nm

D

Data is insufficient

Answer

5×10−6Nm5\times10^{-6} Nm

Explanation

Solution

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.
τ=MB sin θ\tau=MB\,sin\,\theta
where, M is magnetic dipole moment, B the magnetic field and θ\theta the angle between the two.
Given, τ1=10−5Nm,θ1=90∘,θ2=30∘\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
τ1τ2=10−5τ2=11/2\frac{\tau_1}{\tau_2}=\frac{10^{-5}}{\tau_2}=\frac{1}{1/2}
⇒τ2=10−52\Rightarrow \tau_2=\frac{10^{-5}}{2}
=102×10−6=\frac{10}{2}\times 10^{-6}
=5×10−6Nm=5\times 10^{-6}Nm