The electric current in a circular coil of 2 turns produces a magnetic induction B₁ 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 B₂ at its centre. The ratio of $\frac{B_{2}}{B_{1}}$ is :

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$\frac{5}{2}$

$\frac{5}{4}$

$\frac{25}{2}$

$\frac{25}{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} B = \frac{{Nμ}_{0} i}{2 R} \\ B_{1} = \frac{N_{1} μ_{0} i}{2 R_{1}} \end{array}$

For $N_{2} = 5$

Radius of coil $= R_{2} = \frac{N_{1} \times R_{1}}{N_{2}}$

$\begin{array}{l} B_{2} = \frac{N_{2} μ_{0} i}{R_{2}} \\ \frac{B_{2}}{B_{1}} = \frac{N_{2}}{N_{1}} \cdot \frac{R_{1}}{R_{2}} = \frac{N_{2}}{N_{1}} \times \frac{N_{2}}{N_{1}}; \frac{B_{2}}{B_{1}} = \frac{25}{4} \end{array}$

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