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  • A 100 turns coil shown in fig carries a current of 2 amp in  a magnetic field B =  </

A 100 turns coil shown in fig carries a current of 2 amp in  a magnetic field B =  0.2 Wb/ m^2

 .The torque acting on the coil is 

Option: 1

0.32 Nm  tending to rotate side AD \odot


Option: 2

0.32 Nm  tending to rotate side AD \otimes


Option: 3

0.0032 Nm  tending to rotate side AD \odot


Option: 4

0.0032 Nm  tending to rotate side AD \otimes


Answers (1)

best_answer

As we have learned @5922

\tau = NBIA \sin \theta

\tau = NBIA

\tau =100 \times 0.2\times 2(0.08\times 0.1) = 0.32 Nm tending to rotate side AD 

 

Direction can be found by Fleming's left hand rule

Posted by

SANGALDEEP SINGH

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