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Electromagnetic Induction and Alternating Current
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A 20-turn square coil of side 8.0 mm is pivoted at the centre and placed in a magnetic field of flux density 0.010 T, such that two sides of the coil are parallel to the field and two sides are perpendicular to the field. A current of 5.0 mA is passed through the coil. What is the magnitude of the torque acting on the square coil?
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A
6.4 x 10-8 N m
💡 Explanation:
Magnetic force on asingle coil = BIL
0.01 X 5.0 X 10-3 X 8 X 10-3
4.0 X 10-7
Torque acting on the coil = 20 x 4.0 x 10-7 x 8.0 x 10-3= 6.4 x 10-8 N m