When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction 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 induction at the centre will be

$\frac{B}{n}$

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

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$n^{2} B$

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Solution

The correct option is D
$n^{2} B$

$B = \frac{μ_{0} I}{2 r}$ . 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.
$∴ B^{'} = \frac{n μ_{0} I}{2 r^{'}} = \frac{n μ_{0} I}{\frac{2 r}{n}} = n^{2} B$
Hence the correct choice is (d)

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When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction 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 induction at the centre will be

The correct option is D n2B B=μ0I2r. For a coil of n turns, 2πr=n(2πr′) or r′=rn, where r' is the radius of the coil of n turns. ∴ B′=nμ0I2r′=nμ0I2rn=n2B Hence the correct choice is (d)