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Physics Question on Magnetism and matter

At some location on earth the horizontal component of earth's magnetic field is 18×10−618 \times 10^{-6} T. At this location, magnetic needle of length 0.12 m0.12\, m and pole strength 1.8 Am1.8\, Am is suspended from its mid-point using a thread, it makes 45∘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 :

A

3.6×10−5  N3.6 \times 10^{-5} \; N

B

6.5×10−5  N6.5 \times 10^{-5} \; N

C

1.3×10−5  N1.3 \times 10^{-5} \; N

D

1.8×10−5  N1.8 \times 10^{-5} \; N

Answer

6.5×10−5  N6.5 \times 10^{-5} \; N

Explanation

Solution

μBsin⁡45∘=Fℓ2sin⁡45∘\mu B\sin45^{\circ}=F \frac{\ell}{2} \sin45^{\circ}
F=2μBF =2\mu B