A compass needle of magnetic moment 60 ampere-metre² is pointing geographical north at a certain place. It experiences a torque of 1.2x 10^-3Nm. The horizontal component of earth’s magnetic field at that place is 40 micro weber/m². What is the angle of declination at that place?

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15⁰

60⁰

30⁰

45⁰

answer is A.

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Detailed Solution

$∵ τ = {MB}_{H} sinθ \Rightarrow sinθ = \frac{τ}{B_{H} M} = \frac{1.2 \times 10^{- 3}}{40 \times 10^{- 6} \times 60} = \frac{1}{2}; \Rightarrow θ = 30^{0}$

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