The electric current in a circular coil of $2$ turns produces a magnetic induction $B _{1}$ at its centre. The coil is unwound and is rewound into a circular coil of $5$ turns and the same current produces a magnetic induction $B _{2}$ at its centre.The ratio of $\frac{ B _{2}}{ B _{1}}$ is.
JEE MAIN 2022, Medium
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$B=\frac{N \mu_{0} 1}{2 R}$
$B_{1}=\frac{N_{1} \mu_{0} i}{2 R_{1}}$
For $N _{2}=5$
Radius of coil $= R _{2}=\frac{ N _{1} \times R _{1}}{ N _{2}}$
$B _{2}=\frac{ N _{2} \mu_{0} i }{ R _{2}}$
$\frac{ B _{2}}{ B _{1}}=\frac{ N _{2}}{ N _{1}} \frac{ R _{1}}{ R _{2}}=\frac{ N _{2}}{ N _{1}} \times \frac{ N _{2}}{ N _{1}} ; \quad \frac{ B _{2}}{ B _{1}}=\frac{25}{4}$
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