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calculat the torque on a 20 turns square coil of side 5 cm carrying a current of 10 A when placed making an angle of 300 with the magnetic field of 0.8T
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Explanation:
torque = MBsin$
torque = NAi *Bsin$ = 20*(0.05)²*10*0.8*sin30°
= 20*1/2*8*0.0025
= 0.2 N/m
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