A) \[BNiA\theta \]
B) \[BNAi\cos \theta \]
C) \[BNAi\sin \theta \]
D) \[BNAi\tan \theta \]
Correct Answer: C
Solution :
The magnitude of force experienced by a current carrying coil of area A, kept in magnetic field of intensity B is \[F=iBl\] where\[l\]is length of coil. Since, Torque\[(\tau )=\]force\[(F)\times \]perpendicular distance \[\tau =iBl\times b\sin \theta \] where \[lb=Area=A\] \[\therefore \] \[\tau =iBA\text{ }\sin \theta \] For N turns, \[\tau =i\text{ }NAB\text{ }sin\theta \]You need to login to perform this action.
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