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Q.

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 the centre. The ratio of $\frac{B_{2}}{B_{1}}$ is

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answer is 6.25.

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Detailed Solution

Magnetic field at centre of a circular coil is $B = \frac{N μ_{0} i}{2 R}$
So , $B_{1} = \frac{μ o i}{2 R_{1}}$ and $B_{2} = \frac{μ o N_{2} i}{2 R_{2}}$

$\frac{B_{2}}{B_{1}} = \frac{N_{2}}{N_{1}} . \frac{R_{1}}{R_{2}} \Rightarrow \frac{B_{2}}{B_{1}} = \frac{N_{2}}{N_{1}} . \frac{R_{1}}{\frac{N_{1} R_{1}}{N_{2}}} = \frac{{N_{2}}^{2}}{{N_{1}}^{2}} = \frac{25}{4}$

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