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(A) The torque $	au$ on a current-carrying loop in a magnetic field is given by $	au = N i A B \sin 	heta$,where $N$ is the number of turns,$i$ is

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$\frac{B i L^2}{4 \pi}$

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(A) The torque $	au$ on a current-carrying loop in a magnetic field is given by $	au = N i A B \sin 	heta$,where $N$ is the number of turns,$i$ is the current,$A$ is the area,$B$ is the magnetic field,and $	heta$ is the angle between the normal to the loop and the magnetic field.For a wire of length $L$ formed into a circular coil of $N$ turns with radius $r$,the circumference is $L = N(2 \pi r)$,so $r = \frac{L}{2 \pi N}$.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 = N i \left( \frac{L^2}{4 \pi N^2} ight) B \sin 	heta = \frac{i L^2 B \sin 	heta}{4 \pi N}$.To maximize the torque,we set $\sin 	heta = 1$ and choose the minimum number of turns,$N = 1$.Therefore,the maximum torque is $	au_{\max} = \frac{i L^2 B}{4 \pi}$.

$A$ wire of length $L$ carries a current $i$. If the wire is turned into a circular coil and kept in a uniform magnetic field $B$,the maximum magnitude of torque in the given magnetic field will be

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