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A square coil of side 10 cm has 20 turns and carries a current of 12 A. The coil is suspended vertically. The normal to the plane of the coil makes an angle of 30^@ with the direction of a uniform horizontal magnetic field. If the torque experienced by the coil equals 0.96 N - m, find the magnitude of the magnetic field. |
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Answer» Solution :Here side of square `l = 10 cm = 0.1 m`, HENCE area of coil `A = (0.1)^(2) = 0.01 m^(2)`, CURRENT `I = 12 A`, torque `tau = 0.96 N-m, theta = 30^@` and number of turns in coil N = 20. From relation `tau = NIAB sin theta,` we have `B = (tau)/(NIA sin theta) = (0.96)/(20 xx 12 xx 0.01 xx sin 30^(@)) = 0.8 T`. |
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