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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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The correct answer is:
$\mathrm{n}^2 \mathrm{~B}$
$\mathrm{n}^2 \mathrm{~B}$
$B^{\prime}=\frac{\pi \mu_0 I}{2 r^{\prime}}=n^2 \frac{\mu_0 \pi \pi}{\ell}=n^2 B$.
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