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:

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TestHub · Accepted Answer

Magnetic field on the centre of a circular current coil is given by $B = \frac{N μ_{0} i}{2 R}$ , here, $N$ is number of turns, $i$ is current and $R$ is radius of loop.
Now, for, $N = 2$ , magnetic field, $B_{1} = \frac{N_{1} μ_{0} i}{2 R_{1}}$
For $N_{2} = 5$
Radius of coil $= R_{2} = \frac{N_{1} \times R_{1}}{N_{2}}$
Magnetic field, $B_{2} = \frac{N_{2} μ_{0} i}{2 R_{2}}$
Thus, the ratio is $\frac{B_{2}}{B_{1}} = \frac{N_{2}}{N_{1}} \frac{R_{1}}{R_{2}} = \frac{N_{2}}{N_{1}} \times \frac{N_{2}}{N_{1}}$
$\frac{B_{2}}{B_{1}} = \frac{25}{4}$

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