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

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Q. At some location on earth the horizontal component of earth's magnetic field is $18 \times 10^{-6}$ T. At this location, magnetic needle 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 :

JEE MainJEE Main 2019Magnetism and Matter

Solution:

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

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