A circular coil of 10 turns and radius 10 ,cm | vedclass

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(D) The torque $	au$ on a current-carrying coil in a magnetic field is given by the formula $	au = |\vec{m} 	imes \vec{B}| = N I A B \sin 	heta$,

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(D) The torque $	au$ on a current-carrying coil in a magnetic field is given by the formula $	au = |\vec{m} 	imes \vec{B}| = N I A B \sin 	heta$, where $	heta$ is the angle between the area vector (normal to the plane of the coil) and the magnetic field vector $\vec{B}$.Given that the magnetic field is normal to the plane of the coil, the area vector (which is also normal to the plane) is parallel to the magnetic field.Therefore, the angle $	heta$ between the area vector and the magnetic field is $0^\circ$.Since $\sin(0^\circ) = 0$, the torque $	au = N I A B \sin(0^\circ) = 0$.

$A$ circular coil of $10$ turns and radius $10 \,cm$ is placed in a uniform magnetic field of $0.1 \,T$ normal to the plane of the coil. If the current in the coil is $5 \,A$, then the magnitude of the torque on the coil is

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