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A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of $30$ turns and radius $12 cm .$ The coil is in a vertical plane making an angle of $45 °$ with the magnetic meridian when the current in the coil is $0.35 A,$ the needle points magnetic west to east. Determine the horizontal component of earth's magnetic field at the location.

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$3.9 \times 10^{- 7}$ tesla

$3.9 \times 10^{- 5}$ tesla

$8.0 \times 10^{- 5}$ tesla

$7.0 \times 10^{- 7}$ tesla

answer is B.

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Detailed Solution

Here, $n = 30, r = 12 cm = 12 \times 10^{- 2} m . i = 0.35 A, H = ?$

As it is clear from figure shown the needle can point west to east only when $H = B \sin 45 °$

where, $B =$ magnetic field strength due to current in coil $= \frac{μ_{0}}{4 π} \frac{2 π n i}{r} ∴ H = \frac{μ_{0}}{4 π} \frac{2 π n i}{r} \sin 45 °$

$= 10^{- 7} \times \frac{2 π \times 30 \times 0.35}{12 \times 10^{- 2}} \cdot \frac{1}{\sqrt{2}}$

$= 2 \times \frac{22}{7} \times \frac{30 \times 35}{12 \times \sqrt{2}} \times 10^{- 7} = 3.9 \times 10^{- 5} tesla$

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