InterviewSolution
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InterviewSolution
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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 vartically and the normal to the plane of the coil, makes an angle `theta` with the direction of a uniform horizontal magnetic field of 0.80 T. if the torque, experienced by the coil, equals 0.N-m , the value of ` theta ` isA. `0^(@)`B. `pi/2 ` radC. `pi/3` radD. `pi/6` rad |
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Answer» Correct Answer - d Area of coil A =`side^(2) =(0.1)^(2) =0.01 m^(2)`, Number of turns, N=20 current I=12 A Normal to the coil make an angle `theta` with the direction of B, magnetic field , B=0.80 T. torque, experienced by the coil , `tau=0.96 N-m` since, total torque on the coil , `tau=(NIA) B sin theta` Substituting the values in above formula, we get 0.96 N-m =20x12 A x `0.01 m^(2) xx0.80 T xx sin theta` `sin theta=0.96/1.92 =1/2` `theta=(pi)/6` rad |
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