A closely wound flat circular coil of 25 turn | vedclass

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Question

(C) The magnetic field $B$ at the centre of a circular coil is given by the formula: $B = \frac{\mu_0 n i}{2r}$.Given:Number of turns $n = 25$.Diamete

Vedclass · Accepted Answer

$1.257 	imes 10^{-3}\, T$

Vedclass · Answer

(C) The magnetic field $B$ at the centre of a circular coil is given by the formula: $B = \frac{\mu_0 n i}{2r}$.Given:Number of turns $n = 25$.Diameter $D = 10\, cm$,so radius $r = 5\, cm = 5 	imes 10^{-2}\, m$.Current $i = 4\, A$.Permeability of free space $\mu_0 = 4\pi 	imes 10^{-7}\, T\cdot m/A$.Substituting the values:$B = \frac{(4\pi 	imes 10^{-7}) 	imes 25 	imes 4}{2 	imes 5 	imes 10^{-2}}$$B = \frac{4\pi 	imes 10^{-7} 	imes 100}{10 	imes 10^{-2}}$$B = 4\pi 	imes 10^{-7} 	imes 10^3 = 4\pi 	imes 10^{-4}\, T$$B \approx 4 	imes 3.14159 	imes 10^{-4} = 12.566 	imes 10^{-4} = 1.257 	imes 10^{-3}\, T$.

$A$ closely wound flat circular coil of $25$ turns of wire has a diameter of $10\, cm$ and carries a current of $4\, A$. Determine the magnetic flux density at the centre of the coil.

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