InterviewSolution
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InterviewSolution
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A circular current coil with 100 turns and area 20 cm^(2)is placed in a uniform magnetic field in such a manner that magnetic field lines are parallel to the plane of the coil. Torque experienced by the coil is 0.1 Nm when a current of 1 A is passed through the coil. What is the magnitude of magnetic field intensity? |
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Answer» Solution :NUMBER of turns, ` N = 100` Area of coil , `A = 20 cm^(2) = 20 xx 10^(-4) m^(2)` Torque, ` tau = 0.1 ` Nm, Current , `I = 1 A` As the magnetic field lines are parallel to the plane of coil. So, angle between area VECTOR to the plane of coil and magnetic field i.e. ` theta = 90^(@)` As ` tau = NIAB sin theta` ` B = tau/(NIA sintheta)` ` B = (0.1)/(100 xx 1 xx 20 xx 10^(-4) sin 90^(@))` ` B = 0.00005 xx 10^(4)` ` B = 0.5 ` T |
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