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(B) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 N I}{2R}$.Since the length of the wire $L$ is constant,$L = N(2\p

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$93.75\%$ decreases

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(B) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 N I}{2R}$.Since the length of the wire $L$ is constant,$L = N(2\pi R)$,which implies $R = \frac{L}{2\pi N}$.Substituting $R$ into the magnetic field formula: $B = \frac{\mu_0 N I}{2(L / 2\pi N)} = \frac{\mu_0 \pi N^2 I}{L}$.Thus,$B \propto N^2$.Given $N_1 = 12$ and $N_2 = 3$,the ratio of the magnetic fields is $\frac{B_2}{B_1} = \left(\frac{N_2}{N_1}ight)^2 = \left(\frac{3}{12}ight)^2 = \left(\frac{1}{4}ight)^2 = \frac{1}{16} = 0.0625$.This means $B_2 = 0.0625 B_1$.The percentage decrease is $\frac{B_1 - B_2}{B_1} 	imes 100 = \frac{B_1 - 0.0625 B_1}{B_1} 	imes 100 = (1 - 0.0625) 	imes 100 = 93.75\%$.

$A$ coil of $12$ turns is made from a wire of constant length carrying a current $I$. If the number of turns is changed to $3$,what is the percentage change in the magnetic field produced at its centre?

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