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A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30^@ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil? |
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Answer» SOLUTION :It is given that each SIDE of square coil `l = 10 cm = 0.1 m,` hence area `A = l^2 = (0.1)^(2) = 0.01 m^2` `N = 20 , I = 12 A , theta = 30^@ and B = 0.80 T` `:.` Torque `TAU = N B A I sin theta = 20 xx 0.8 xx 0.01 xx 12 xx sin 30^@` `= 20 xx 0.8 xx 0.01 xx 12 xx 1/2 = 0.96 N m^(-1)`. |
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