A long conducting wire having a current I flowing through it, is bent into a circular coil of $\mathrm{N}$ turns. Then it is bent into a circular coil of $\mathrm{n}$ 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 :
Solution
$I=(2 \pi r) n$
<br/><br/>$$
\begin{aligned}
& r \propto\left(\frac{I}{n}\right) \\\\
& B=n\left(\frac{\mu_{0} i}{2 r}\right) \propto\left(\frac{\mu_{0} i}{2 L}\right) n^{2} \\\\
& \frac{B_{1}}{B_{2}}=\left(\frac{N^{2}}{n^{2}}\right)
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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