A coil of $12$ turns made by a constant length current carrying wire. If number of turns makes $3$ then change in magnetic field produced at its centre
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$\mathrm{B} \propto \mathrm{N}^{2} \Rightarrow \frac{\mathrm{B}_{2}}{\mathrm{B}_{1}}=\left(\frac{\mathrm{N}_{2}}{\mathrm{N}_{1}}\right)^{2}=\left(\frac{3}{12}\right)^{2}=\frac{9}{144}$
$\Rightarrow \mathrm{B}_{2}=\frac{9}{144} \mathrm{B}_{1}$
$\%=\frac{\Delta \mathrm{B}}{\mathrm{B}_{1}} \times 100=\frac{\mathrm{B}_{1}-\mathrm{B}_{2}}{\mathrm{B}_{1}} \times 100=\frac{135}{144} \times 100$
$\quad=93.75 \%$
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