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(C) The magnetic field $B$ at the centre of a circular coil with $n$ turns is given by the formula: $B = \frac{\mu_0 n I}{2R}$.Given values:$B = 3.14

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$100$

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(C) The magnetic field $B$ at the centre of a circular coil with $n$ turns is given by the formula: $B = \frac{\mu_0 n I}{2R}$.Given values:$B = 3.14 	imes 10^{-4} \ T$$I = 0.4 \ A$$R = 8 \ cm = 0.08 \ m$$\mu_0 = 12.56 	imes 10^{-7} \ T \cdot m/A$Substituting these values into the formula:$3.14 	imes 10^{-4} = \frac{(12.56 	imes 10^{-7}) 	imes n 	imes 0.4}{2 	imes 0.08}$$3.14 	imes 10^{-4} = \frac{12.56 	imes 10^{-7} 	imes n 	imes 0.4}{0.16}$$3.14 	imes 10^{-4} = 7.85 	imes 10^{-7} 	imes 4 	imes n 	imes 10^{-1} / 0.16$ (simplifying the fraction $\frac{12.56}{0.16} = 78.5$)$3.14 	imes 10^{-4} = (78.5 	imes 10^{-7}) 	imes 0.4 	imes n$$3.14 	imes 10^{-4} = 31.4 	imes 10^{-7} 	imes n$$n = \frac{3.14 	imes 10^{-4}}{31.4 	imes 10^{-7}} = \frac{3.14 	imes 10^{-4}}{3.14 	imes 10^{-6}} = 10^2 = 100$.

$A$ circular coil of wire consisting of $n$ turns,each of radius $8 \ cm$,carries a current of $0.4 \ A$. The magnitude of the magnetic field at the centre of the coil is $3.14 \times 10^{-4} \ T$. The value of $n$ is (Take $\mu_0 = 12.56 \times 10^{-7} \ T \cdot m/A$)

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