At some location on earth the horizontal component of earth’s magnetic field is \(18\times10^{-6}\,T\). 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 \(45^o\) 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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Correct Answer

(D) \(6.5\times10^{-5}\,N\)

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Detailed explanation

\(\tau = F \times 0.06 = 1.8 \times 0.012 \times 18 \times {10^{ - 6}}\)

\(F = 6.48\times10^{-5}\)

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