A thin wire of length l is carrying a constan | vedclass

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$(b, c)$ Since length of the wire is equal to $l$, therefore, $2\pi Rn = l$ or $n = \frac{l}{{2\pi R}}$.
Magnetic induction at centre of a circular c

Vedclass · Accepted Answer

Vedclass · Answer

$(b, c)$ Since length of the wire is equal to $l$, therefore, $2\pi Rn = l$ or $n = \frac{l}{{2\pi R}}$.
Magnetic induction at centre of a circular coil is given by $B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni}}{R} = \frac{{{\mu _0}l\,i}}{{4\pi {R^2}}}$ $==>$ $B \propto \frac{1}{{{R^2}}}$
It means, when $R 	o 0,\;B 	o \infty $ and $R 	o \infty ,\;B 	o 0,$
Hence $(b)$ is correct and $(d)$ is wrong.
Substituting $R = \frac{l}{{2\pi n}}$ in $B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni}}{R}$
$B \propto {n^2}$. It means graph between $B$ and $n$ will be a parabola having increasing slope and passing through origin. Hence $(c)$ is correct and $(a)$ is wrong.

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