The magnetic field at the center of a circula | vedclass

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(C) The magnetic field $B$ at the center of a circular coil is given by the formula: $B = \frac{\mu_0 Ni}{2r}$.Given values are:Number of turns $N = 1

Vedclass · Accepted Answer

$6.28 	imes 10^{-4}\,T$

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(C) The magnetic field $B$ at the center of a circular coil is given by the formula: $B = \frac{\mu_0 Ni}{2r}$.Given values are:Number of turns $N = 1000$Current $i = 0.1\, A$Radius $r = 0.1\, m$Permeability of free space $\mu_0 = 4\pi 	imes 10^{-7}\, T\cdot m/A$.Substituting these values into the formula:$B = \frac{4\pi 	imes 10^{-7} 	imes 1000 	imes 0.1}{2 	imes 0.1}$$B = \frac{4\pi 	imes 10^{-7} 	imes 100}{0.2}$$B = 2\pi 	imes 10^{-4}\, T$Using $\pi \approx 3.14$,we get:$B = 2 	imes 3.14 	imes 10^{-4} = 6.28 	imes 10^{-4}\, T$.

The magnetic field at the center of a circular coil of radius $0.1\, m$ having $1000$ turns and carrying a current of $0.1\, A$ is:

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