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(D) The magnetic field $B$ at the centre of a circular coil with $N$ turns,radius $R$,and current $I$ is given by the formula:$B = \frac{\mu_0 NI}{2R}

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$10^{-3}\,T$

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(D) The magnetic field $B$ at the centre of a circular coil with $N$ turns,radius $R$,and current $I$ is given by the formula:$B = \frac{\mu_0 NI}{2R}$Given values:$N = 1000$$I = 1\,A$$R = 62.8\,cm = 0.628\,m = 62.8 	imes 10^{-2}\,m$$\mu_0 = 4\pi 	imes 10^{-7}\,T\cdot m/A$Substituting the values into the formula:$B = \frac{4 	imes 3.14 	imes 10^{-7} 	imes 1000 	imes 1}{2 	imes 62.8 	imes 10^{-2}}$$B = \frac{12.56 	imes 10^{-4}}{125.6 	imes 10^{-2}}$$B = \frac{12.56 	imes 10^{-4}}{1.256}$$B = 10 	imes 10^{-4} = 10^{-3}\,T$Thus,the magnetic field produced at the centre of the coil is $10^{-3}\,T$.

$A$ closely packed coil having $1000$ turns has an average radius of $62.8\,cm$. If the current carried by the wire of the coil is $1\,A$,the value of the magnetic field produced at the centre of the coil will be (permeability of free space $\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$) nearly:

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