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(D) The magnetic field at the center of a circular coil of $N$ turns,radius $r$,and current $I$ is given by $B = \frac{N \mu_0 I}{2r}$.Initially,for o

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$B^{\prime} = n^2 B$

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(D) The magnetic field at the center of a circular coil of $N$ turns,radius $r$,and current $I$ is given by $B = \frac{N \mu_0 I}{2r}$.Initially,for one turn $(N=1)$ and radius $r$,the magnetic field is $B = \frac{\mu_0 I}{2r}$.When the same wire of length $L = 2 \pi r$ is bent into $n$ turns,the new radius $r^{\prime}$ is given by $L = n(2 \pi r^{\prime})$,which implies $r^{\prime} = \frac{r}{n}$.The new magnetic field $B^{\prime}$ is $B^{\prime} = \frac{n \mu_0 I}{2r^{\prime}}$.Substituting $r^{\prime} = \frac{r}{n}$ into the expression for $B^{\prime}$:$B^{\prime} = \frac{n \mu_0 I}{2(r/n)} = n^2 \left( \frac{\mu_0 I}{2r} \right) = n^2 B$.

$A$ long wire carries a steady current. It is bent into a coil of one turn such that the magnetic induction at the centre is $B$. If the same wire is bent to form a coil of smaller radius with $n$ turns,then the new magnetic induction $B^{\prime}$ at the centre is:

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