A long wire carries a steady current. It is b | vedclass

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(B) Let the length of the wire be $L$. For a single turn loop of radius $r$,the circumference is $2 \pi r = L$,so $r = \frac{L}{2 \pi}$.The magnetic f

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$n^{2} B$

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(B) Let the length of the wire be $L$. For a single turn loop of radius $r$,the circumference is $2 \pi r = L$,so $r = \frac{L}{2 \pi}$.The magnetic field at the centre of a single turn loop is $B = \frac{\mu_{0} I}{2 r} = \frac{\mu_{0} I}{2 (L / 2 \pi)} = \frac{\mu_{0} I \pi}{L}$.When the same wire is bent into $n$ turns,the new radius $r'$ satisfies $n (2 \pi r') = L$,so $r' = \frac{L}{2 \pi n} = \frac{r}{n}$.The magnetic field at the centre of the $n$-turn loop is $B_{n} = n 	imes \frac{\mu_{0} I}{2 r'} = n 	imes \frac{\mu_{0} I}{2 (r / n)} = n^{2} 	imes \frac{\mu_{0} I}{2 r}$.Since $B = \frac{\mu_{0} I}{2 r}$,we have $B_{n} = n^{2} B$.

$A$ long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B$. It is then bent into a circular loop of $n$ turns. The magnetic field at the centre of the coil for the same current will be

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