A)
B)
C)
D)
Correct Answer: C
Solution :
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 \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi ni}{R}=\frac{{{\mu }_{0}}l\,i}{4\pi {{R}^{2}}}\] Þ \[B\propto \frac{1}{{{R}^{2}}}\] It means, when \[R\to 0,\ B\to \infty \]and \[R\to \infty ,\ B\to 0,\] 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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