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:

Engineering

Physics

Magnetic Field

Question

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:

6.5 × 10^–5 N

3.6 × 10^–5 N

1.8 × 10^–5 N

1.3 × 10^–5 N

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τ = F × 0.06 = 1.8 × 0.012 × 18 × 10^–6

F = 6.48 × 10^–5