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(D) The torque acting on a current-carrying coil in a magnetic field is given by $	au = NIAB \sin 	heta$. For maximum torque,$\sin 	heta = 1$,so $\

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(D) The torque acting on a current-carrying coil in a magnetic field is given by $	au = NIAB \sin 	heta$. For maximum torque,$\sin 	heta = 1$,so $	au_{\max} = NIAB$.Let the length of the wire be $L$. If the coil has $N$ turns and radius $r$,then $L = N(2\pi r)$,which implies $r = \frac{L}{2\pi N}$.The area of the coil is $A = \pi r^2 = \pi \left( \frac{L}{2\pi N} ight)^2 = \frac{L^2}{4\pi N^2}$.Substituting $A$ into the torque equation: $	au_{\max} = Ni \left( \frac{L^2}{4\pi N^2} ight) B = \frac{i L^2 B}{4\pi N}$.From this expression,it is clear that $	au_{\max} \propto \frac{1}{N}$.Therefore,to maximize the torque,the number of turns $N$ should be as small as possible. Since the minimum number of turns for a coil is $1$,the torque is maximum when $N = 1$.

$A$ wire of length $L$ is bent in the form of a circular coil and current $i$ is passed through it. If this coil is placed in a magnetic field,then the torque acting on the coil will be maximum when the number of turns is

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