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(B, C) The length of the wire is $l = 2\pi Rn$,which implies $n = \frac{l}{2\pi R}$.The magnetic field induction at the centre of a circular coil with

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(B, C) The length of the wire is $l = 2\pi Rn$,which implies $n = \frac{l}{2\pi R}$.The magnetic field induction at the centre of a circular coil with $n$ turns is given by:$B = \frac{\mu_0 n i}{2R}$Substituting $n = \frac{l}{2\pi R}$ into the formula:$B = \frac{\mu_0 i}{2R} \left( \frac{l}{2\pi R} ight) = \frac{\mu_0 i l}{4\pi R^2}$This shows that $B \propto \frac{1}{R^2}$. As $R 	o 0$,$B 	o \infty$,and as $R 	o \infty$,$B 	o 0$. This corresponds to the graph shown in option $(b)$.Alternatively,substituting $R = \frac{l}{2\pi n}$ into the formula:$B = \frac{\mu_0 n i}{2 (l / 2\pi n)} = \frac{\mu_0 i}{2l} (2\pi n^2) = \left( \frac{\mu_0 \pi i}{l} ight) n^2$This shows that $B \propto n^2$. The graph of $B$ versus $n$ is a parabola passing through the origin with an increasing slope,which corresponds to the graph shown in option $(c)$.

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