At some location the horizontal component of earth's magnetic field is $18\times10^{-6}\ \text{T}$. At this location, a magnetic needle of length $0.12\ \text{m}$ and pole strength $1.8\ \text{A m}$ is suspended from its mid-point using a thread; it makes $45^\circ$ angle with the horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:
Answer: (C) $6.5\times10^{-5}\ \text{N}$
The free needle points along the total field, so the dip is $45^\circ$ and $B_V = B_H\tan45^\circ = 18\times10^{-6}\ \text{T}$.
When the needle is horizontal, $B_V$ pushes each pole vertically with force $mB_V$, giving a torque $mB_V\,l$ about the midpoint. A force $F$ at one end gives torque $F\dfrac l2$:
$$F = 2mB_V = 2\times1.8\times18\times10^{-6} \approx 6.5\times10^{-5}\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics