A winding wire which is used to frame a solen | 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: $B = 0.2\, T$,$i = 10\, A$,$l = 0.8\, m$,and $\mu_0 = 4\pi 	imes 10^{-7}\, T\cdot m/A$.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 length of the winding wire $L$ is equal to the circumference of one turn multiplied by the total number of turns: $L = (2\pi r) 	imes N$.Given $r = 3\, cm = 3 	imes 10^{-2}\, m$.$L = 2\pi 	imes (3 	imes 10^{-2}) 	imes \frac{4 	imes 10^4}{\pi} = 6 	imes 10^{-2} 	imes 4 	imes 10^4 = 24 	imes 10^2 = 2.4 	imes 10^3\, m$.

$A$ winding wire which is used to frame a solenoid can bear a maximum $10\, A$ current. If the length of the solenoid is $80\, cm$ and its cross-sectional radius is $3\, cm$,then the required length of the winding wire is $(B = 0.2\, T)$.

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