The radius of a circular ring of wire is R an | vedclass

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(A) The magnetic field $B$ produced by the larger ring at its center is given by $B = \frac{\mu_0 I}{2R}$.The magnetic moment of the smaller coil with

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$Ni\pi r^2 	imes \left( \frac{\mu_0 I}{2R} ight)$

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(A) The magnetic field $B$ produced by the larger ring at its center is given by $B = \frac{\mu_0 I}{2R}$.The magnetic moment of the smaller coil with $N$ turns,current $i$,and radius $r$ is $M = N i A = N i (\pi r^2)$.The torque $	au$ acting on a magnetic dipole in an external magnetic field is given by $	au = M B \sin 	heta$,where $	heta$ is the angle between the magnetic moment vector and the magnetic field vector.Since the planes of the two rings are perpendicular,the magnetic moment vector of the smaller ring (which is perpendicular to its plane) is perpendicular to the magnetic field vector produced by the larger ring (which is perpendicular to the larger ring's plane). Thus,$	heta = 90^{\circ}$.Substituting the values,we get $	au = (N i \pi r^2) 	imes \left( \frac{\mu_0 I}{2R} ight) 	imes \sin 90^{\circ}$.Since $\sin 90^{\circ} = 1$,the torque is $	au = Ni\pi r^2 \left( \frac{\mu_0 I}{2R} ight)$.

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