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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 if $30\degree$ . With the direction of a uniform horizontal magnetic field of magnitude 0.80 T. The magnitude of toroque experienced by the coil?
Option: 1
0.98 Nm

Option: 2
0.96 Nm

Option: 3
0.86 Nm

Option: 4
0.90 Nm

Answers (1)

best_answer

Toroque,

$T= NIA sin \theta$

$=20\times 12\times (10\times10^{-2})\times0.8sin30\degree$

$=0.96 Nm$

Posted by

Riya

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