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When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic field at its centre is B. When the same wire carrying the same current is bent to form a circular coil of n turns of a smaller radius, the magnetic field at the centre will be

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$n B$

$\frac{B}{n}$

$\frac{B}{n^{2}}$

$n^{2} B$

answer is D.

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Detailed Solution

$B = \frac{μ_{0} I}{2 r}$ For a coil of n·turns, $2 π r = n (2 π r^{'})$

$or r^{'} = \frac{r}{n}$ , where r' is the radius of the coil of n turns.

$∴ B^{'} = \frac{n μ_{0} I}{2 r^{'}} = \frac{n μ_{0} I}{2 r / n} = n^{2} B$

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