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(B) The magnetic field $B$ at the centre of a circular coil with $n$ turns is given by the formula:$B = \frac{\mu_0 n I}{2r}$Given:Number of turns,$n

Vedclass · Accepted Answer

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

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(B) The magnetic field $B$ at the centre of a circular coil with $n$ turns is given by the formula:$B = \frac{\mu_0 n I}{2r}$Given:Number of turns,$n = 100$Radius,$r = 8.0 \, cm = 0.08 \, m$Current,$I = 0.40 \, A$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 100 	imes 0.40}{2 	imes 0.08}$$B = \frac{4\pi 	imes 10^{-7} 	imes 40}{0.16}$$B = \frac{160\pi 	imes 10^{-7}}{0.16}$$B = 1000\pi 	imes 10^{-7} \, T = \pi 	imes 10^{-4} \, T$Thus,the magnitude of the magnetic field is $\pi 	imes 10^{-4} \, T$.

$A$ circular coil of wire consisting of $100$ turns,each of radius $8.0 \, cm$,carries a current of $0.40 \, A$. What is the magnitude of the magnetic field $B$ at the centre of the coil?

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