A solenoid of length 80 , cm and radius 3 , c | vedclass

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(C) The magnetic field inside a solenoid is given by $B = \frac{\mu_0 N i}{l}$,where $N$ is the total number of turns,$i$ is the current,and $l$ is th

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$2.4 	imes 10^3 \, m$

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(C) The magnetic field inside a solenoid is given by $B = \frac{\mu_0 N i}{l}$,where $N$ is the total number of turns,$i$ is the current,and $l$ is the length of the solenoid.Given: $B = 0.2 \, T$,$l = 0.8 \, m$,$i = 10 \, A$,$\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$.Substituting the values: $0.2 = \frac{4\pi 	imes 10^{-7} 	imes N 	imes 10}{0.8}$.Solving for $N$: $N = \frac{0.2 	imes 0.8}{4\pi 	imes 10^{-6}} = \frac{0.16}{4\pi 	imes 10^{-6}} = \frac{4 	imes 10^4}{\pi}$.The total length of the wire $L$ is given by $L = (2\pi r) 	imes N$,where $r = 3 \, cm = 3 	imes 10^{-2} \, m$.$L = 2\pi 	imes (3 	imes 10^{-2}) 	imes \frac{4 	imes 10^4}{\pi}$.$L = 6 	imes 10^{-2} 	imes 4 	imes 10^4 = 24 	imes 10^2 = 2.4 	imes 10^3 \, m$.

$A$ solenoid of length $80 \, cm$ and radius $3 \, cm$ produces a magnetic field of $B = 0.2 \, T$ when a current of $10 \, A$ is passed through it. What is the total length of the wire used in the solenoid?

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