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 $1.8 A m$ 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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$3.6 \times 10^{- 5} N$

$1.3 \times 10^{- 5} N$

$1.8 \times 10^{- 5} N$

$6.5 \times 10^{- 5} N$

answer is C.

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Detailed Solution

$B_{H} = 18 \times 10^{- 6} T, m = 1.8 A m, l = 0.12 m$

$|τ_{F}| = |τ_{F B}| \Rightarrow F \times \frac{l}{2} \sin 45 ° = μ B \sin 45 °$

$F = 2 μ B / l = 2 m B = 2 \times 1.8 \times 18 \times 10^{- 6} = 6.48 \times 10^{- 5} N$

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