Consider a tightly wound 100 turn coil of rad | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The magnetic field at the center of a circular coil with $N$ turns is given by the formula:$B = \frac{\mu_0 N I}{2 R}$Given:Number of turns $N = 100$Radius $R = 10 \; cm = 0.1 \; m$Current $I = 1 \; A$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 1}{2 	imes 0.1}$$B = \frac{4 \pi 	imes 10^{-5}}{0.2}$$B = 2 \pi 	imes 10^{-4} \; T$$B \approx 6.28 	imes 10^{-4} \; T$

Consider a tightly wound $100$ turn coil of radius $10 \; cm$,carrying a current of $1 \; A$. What is the magnitude of the magnetic field at the centre of the coil?

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