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

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inversely proportional to $N$

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(B) The torque $	au$ on a current-carrying coil in a magnetic field is given by $	au = N I A B \sin 	heta$. For maximum torque,$\sin 	heta = 1$,so $	au_{max} = N I A B$.The length of the wire $L$ is used to form $N$ turns of a circular coil of radius $r$. Thus,$L = N(2 \pi r)$,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 the expression for $A$ into the torque formula: $	au_{max} = N I B \left( \frac{L^2}{4 \pi N^2} ight) = \frac{I B L^2}{4 \pi N}$.Since $I$,$B$,and $L$ are constants,$	au_{max} \propto \frac{1}{N}$.Therefore,the maximum torque is inversely proportional to $N$.

$A$ circular coil having $N$ turns is made from a wire of length $L$ meters. If a current of $I$ amperes is passed through it and placed in a magnetic field of $B$ tesla,the maximum torque on it is

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