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(B) The magnetic field $B$ at the centre of a circular coil is given by the formula: $B = \frac{\mu_0 Ni}{2r}$.Given values are:Number of turns $N = 5

Vedclass · Accepted Answer

$1.25 	imes 10^{-4}\,T$

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(B) The magnetic field $B$ at the centre of a circular coil is given by the formula: $B = \frac{\mu_0 Ni}{2r}$.Given values are:Number of turns $N = 50$Radius $r = 0.5\, m$Current $i = 2\, A$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 50 	imes 2}{2 	imes 0.5}$$B = \frac{4 	imes 3.14159 	imes 10^{-7} 	imes 100}{1}$$B = 12.566 	imes 10^{-5} = 1.2566 	imes 10^{-4}\, T$.Rounding to the nearest option,we get $1.25 	imes 10^{-4}\, T$.

Magnetic field intensity at the centre of a coil of $50$ turns,radius $0.5\, m$ and carrying a current of $2\, A$ is:

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