A circular coil of radius 0.1 m and 100 turns placed in vertical plane, which is perpendicular to magnetic meridian. The angle of dip is zero. When current of 2A is flowing through the coil then the neutral point is obtained at the centre, then horizontal component of earth’s magnetic field at that place will be

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$1.2 \times 10^{- 4} T$

$4 π \times 10^{- 7} T$

$4 π \times 10^{- 4} T$

$4 π \times 10^{- 3} T$

answer is A.

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Detailed Solution

$B_{H} = B_{centre} = \frac{μ_{0} Ni}{2 r} = \frac{4 π \times 10^{- 7} \times 100 \times 2}{2 \times 0.1} = 4 π \times 10^{- 4} T$

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