JEE Mains · Physics · STD 12 - 5. Magnetism and matter
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
- A \(3.6\times10^{-5}\,N\)
- B \(1.8\times10^{-5}\,N\)
- C \(1.3\times10^{-5}\,N\)
- D \(6.5\times10^{-5}\,N\)
Answer & Solution
Correct Answer
(D) \(6.5\times10^{-5}\,N\)
Step-by-step Solution
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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