A conductor wire having 10^{29} free electron | vedclass

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(B) The relationship between current $I$ and drift velocity $v_{d}$ is given by the formula: $I = n A e v_{d}$.Here, $n = 10^{29} \,m^{-3}$ is the ele

Vedclass · Accepted Answer

$1.25 	imes 10^{-3} \,ms^{-1}$

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(B) The relationship between current $I$ and drift velocity $v_{d}$ is given by the formula: $I = n A e v_{d}$.Here, $n = 10^{29} \,m^{-3}$ is the electron density, $A = 1 \,mm^{2} = 10^{-6} \,m^{2}$ is the cross-sectional area, $e = 1.6 	imes 10^{-19} \,C$ is the charge of an electron, and $I = 20 \,A$.Rearranging the formula for drift velocity: $v_{d} = \frac{I}{n A e}$.Substituting the values: $v_{d} = \frac{20}{10^{29} 	imes 10^{-6} 	imes 1.6 	imes 10^{-19}}$.$v_{d} = \frac{20}{1.6 	imes 10^{4}} = \frac{20}{16000} = 1.25 	imes 10^{-3} \,ms^{-1}$.

$A$ conductor wire having $10^{29}$ free electrons $/ m^{3}$ carries a current of $20 \,A$. If the cross-section of the wire is $1 \,mm^{2}$, then the drift velocity of electrons will be. $\left(e=1.6 \times 10^{-19} \,C\right)$

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