JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 12)
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 :
$ N^{2}: n^{2}$
$\mathrm{N}: \mathrm{n}$
$\mathrm{n}: \mathrm{N}$
$n^{2}: N^{2}$
Explanation
$I=(2 \pi r) n$
$$ \begin{aligned} & r \propto\left(\frac{I}{n}\right) \\\\ & B=n\left(\frac{\mu_{0} i}{2 r}\right) \propto\left(\frac{\mu_{0} i}{2 L}\right) n^{2} \\\\ & \frac{B_{1}}{B_{2}}=\left(\frac{N^{2}}{n^{2}}\right) \end{aligned} $$
$$ \begin{aligned} & r \propto\left(\frac{I}{n}\right) \\\\ & B=n\left(\frac{\mu_{0} i}{2 r}\right) \propto\left(\frac{\mu_{0} i}{2 L}\right) n^{2} \\\\ & \frac{B_{1}}{B_{2}}=\left(\frac{N^{2}}{n^{2}}\right) \end{aligned} $$
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