A long conducting wire having a current flowing through it is bent into a circular coil of turns. Then it is bent into a circular coil of turns. [2023]

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Solution

Q.
A long conducting wire having a current $I$ flowing through it is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is [2023]

1 $N : n$

2 $n^{2} : N^{2}$

3 $N^{2} : n^{2}$

4 $n : N$

Ans.
(3)

$l = (2 π r) n$

$r \propto (\frac{l}{n})$

$B = n (\frac{μ_{0} i}{2 r}) \propto (\frac{μ_{0} i}{2 L}) n^{2}$

$\frac{B_{1}}{B_{2}} = (\frac{N^{2}}{n^{2}})$

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