A 100-turn closely wound circular coil of rad | vedclass

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(C) The magnetic field $B$ at the centre of a circular coil is given by $B = \frac{\mu_0 N I}{2r}$.Given: $B = 3.14 	imes 10^{-3} 	ext{ T}$,$N = 100

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$2.5 	ext{ A}, 2 	ext{ A m}^2$

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(C) The magnetic field $B$ at the centre of a circular coil is given by $B = \frac{\mu_0 N I}{2r}$.Given: $B = 3.14 	imes 10^{-3} 	ext{ T}$,$N = 100$,$r = 5 	ext{ cm} = 0.05 	ext{ m}$,$\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.Substituting the values: $3.14 	imes 10^{-3} = \frac{4 	imes 3.14 	imes 10^{-7} 	imes 100 	imes I}{2 	imes 0.05}$.$3.14 	imes 10^{-3} = \frac{4 	imes 3.14 	imes 10^{-5} 	imes I}{0.1}$.$10^{-3} = \frac{4 	imes 10^{-5} 	imes I}{0.1} = 4 	imes 10^{-4} 	imes I$.$I = \frac{10^{-3}}{4 	imes 10^{-4}} = \frac{10}{4} = 2.5 	ext{ A}$.The magnetic moment $M$ is given by $M = N I A = N I (\pi r^2)$.$M = 100 	imes 2.5 	imes 3.14 	imes (0.05)^2$.$M = 250 	imes 3.14 	imes 0.0025 = 1.9625 \approx 2 	ext{ A m}^2$.Thus,the current is $2.5 	ext{ A}$ and the magnetic moment is $2 	ext{ A m}^2$.

$A$ $100$-turn closely wound circular coil of radius $5 \text{ cm}$ has a magnetic field of $3.14 \times 10^{-3} \text{ T}$ at its centre. The current flowing through the coil,and the magnitude of the magnetic moment of this coil are,respectively: (Take $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$)

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