Q. At some location on earth the horizontal component of earth's magnetic field is $18 \times 10^{-6}$ T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes $45^{\circ}$ angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :

Solution:

$\mu B\sin45^{\circ}=F \frac{\ell}{2} \sin45^{\circ} $
$ F =2\mu B$

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