question_answer

A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field \[B\hat{i}\]. The torque on the coil due to the magnetic field is:             [JEE Main 8-4-2019 Morning]

A) \[B\pi {{r}^{2}}IN\]

B) \[\frac{B{{r}^{2}}I}{\pi N}\]

C) Zero                

D) \[\frac{B\pi {{r}^{2}}I}{N}\]

Correct Answer: A

Solution :

            Magnetic moment of coil \[=NIA\hat{j}\] \[=NI(\pi {{r}^{2}})\hat{j}\] Torque on loop (coil)\[=\vec{M}\times \vec{B}\] \[=NI\left( \pi {{r}^{2}} \right)B\sin {{90}^{o}}\left( -\hat{k} \right)\] \[=NI\pi {{r}^{2}}B\left( -\hat{k} \right)\]