If the radius of a coil is halved and the number of turns doubled, then the magnetic field at the centre of the coil, for the same current will
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$\mathrm{B}_{\text {centre }}=\frac{\mu_{0}}{4 \pi} \frac{2 \pi \mathrm{ni}}{\mathrm{r}}$
or $\mathrm{B}_{\text {centre }} \propto \frac{\mathrm{n}}{\mathrm{r}}$
or $\frac{\mathrm{B}^{\prime}}{\mathrm{B}}=\frac{\mathrm{n}^{\prime}}{\mathrm{r}^{\prime}} \times \frac{\mathrm{r}}{\mathrm{n}}=\frac{2 \mathrm{n}}{\mathrm{r} / 2} \times \frac{\mathrm{r}}{\mathrm{n}}$
or $\mathrm{B}^{\prime}=4 \mathrm{\,B}.$
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