question_answer

A current carrying coil is placed in a unifonr. magnetic field of induction B. If the current in the coil is i, it has N turns and A is the force area of the coil and normal to the surface makes an angle \[\theta \] with B, then the torque experienced by the coil is :

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, \[\tau =i\text{ }NAB\text{ }sin\theta \]