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Question

A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30 turns and radius 12cm. The coil is in a vertical plane making an angle of 450 with the magnetic meridian when the current in the coil is 0.35A, the needle points west to east. Determine the horizontal component of earth's magnetic field at the location.

A
3.9×107 tesla
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B
3.9×105 tesla
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C
8.0×105 tesla
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D
7.0×107 tesla
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Solution

The correct option is B 3.9×105 tesla
Here, n=30,r=12cm=12×102m.i=0.35A,H=?
As it is clear from figure shown the needle can point west to east only when H=Bsin450

where, B = magnetic field strength due to current in coil μ04π2πnir

H=μ04π2πnirsin450=107×2π×30×0.3512×102.12=2×227×30×3512×2×107=3.9×105 tesla

1033449_943195_ans_d96482790aef490e9166a4854f9b4a82.jpg

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