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

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

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$n^2B$

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(B) Let the length of the wire be $L$ and the current be $i$.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 center is $B = \frac{\mu_0 i}{2r} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 i \pi}{L}$.When the 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 center for $n$ turns is $B_n = n 	imes \frac{\mu_0 i}{2r'} = n 	imes \frac{\mu_0 i}{2(r/n)} = n^2 	imes \frac{\mu_0 i}{2r}$.Since $B = \frac{\mu_0 i}{2r}$,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 will be

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