A copper wire of radius 0.1 text{ mm} and res | vedclass

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(B) Given: Radius of copper wire $r = 0.1 	ext{ mm} = 1 	imes 10^{-4} 	ext{ m}$.Resistance $R = 2 	ext{ k}\Omega = 2 	imes 10^3 \Omega$.Voltage $

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$1.25 	imes 10^{17}$

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(B) Given: Radius of copper wire $r = 0.1 	ext{ mm} = 1 	imes 10^{-4} 	ext{ m}$.Resistance $R = 2 	ext{ k}\Omega = 2 	imes 10^3 \Omega$.Voltage $V = 40 	ext{ V}$.Using Ohm's law,the current $I$ flowing through the wire is:$I = \frac{V}{R} = \frac{40}{2 	imes 10^3} = 2 	imes 10^{-2} 	ext{ A}$.The charge $q$ flowing per second is equal to the current $I$ (since $q = I 	imes t$ and $t = 1 	ext{ s}$):$q = 2 	imes 10^{-2} 	ext{ C}$.The number of electrons $n$ transferred per second is given by $n = \frac{q}{e}$,where $e = 1.6 	imes 10^{-19} 	ext{ C}$ is the elementary charge:$n = \frac{2 	imes 10^{-2}}{1.6 	imes 10^{-19}} = 1.25 	imes 10^{17} 	ext{ electrons}$.

$A$ copper wire of radius $0.1 \text{ mm}$ and resistance $2 \text{ k}\Omega$ is connected across a power supply of $40 \text{ V}$. The number of electrons transferred per second between the supply and the wire at one end is

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