A closely wound circular coil of radius 5,cm | vedclass

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(A) The magnetic field at the center of a circular coil with $N$ turns is given by the formula:$B = \frac{N \mu_{0} I}{2 R}$Given values:$N = 100$$R =

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$3$

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(A) The magnetic field at the center of a circular coil with $N$ turns is given by the formula:$B = \frac{N \mu_{0} I}{2 R}$Given values:$N = 100$$R = 5\,cm = 0.05\,m = 5 	imes 10^{-2}\,m$$B = 37.68 	imes 10^{-4}\,T$$\mu_{0} = 4 \pi 	imes 10^{-7}\,T \cdot m/A$$\pi = 3.14$Substituting the values into the formula:$37.68 	imes 10^{-4} = \frac{100 	imes 4 	imes 3.14 	imes 10^{-7} 	imes I}{2 	imes 5 	imes 10^{-2}}$Simplify the equation:$37.68 	imes 10^{-4} = \frac{400 	imes 3.14 	imes 10^{-7} 	imes I}{10 	imes 10^{-2}}$$37.68 	imes 10^{-4} = \frac{1256 	imes 10^{-7} 	imes I}{10^{-1}}$$37.68 	imes 10^{-4} = 1256 	imes 10^{-6} 	imes I$$37.68 	imes 10^{-4} = 1.256 	imes 10^{-3} 	imes I$$I = \frac{37.68 	imes 10^{-4}}{1.256 	imes 10^{-3}} = \frac{3.768 	imes 10^{-3}}{1.256 	imes 10^{-3}} = 3\,A$Thus,the current through the coil is $3\,A$.

$A$ closely wound circular coil of radius $5\,cm$ produces a magnetic field of $37.68 \times 10^{-4}\,T$ at its center. The current through the coil is $......A$. [Given,number of turns in the coil is $100$ and $\pi=3.14$]

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