(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

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

(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, $e = 1.6 \times 10^{-19} \,C$ 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: $v_{d} = \frac{20}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}}$.$v_{d} = \frac{20}{1.6 \times 10^{4}} = \frac{20}{16000} = 1.25 \times 10^{-3} \,ms^{-1}$.

Class 12 Physics Current Electricity Current Density, Drift Velocity and Mobility

Medium

$A$ 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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A
$1.25 \times 10^{-4} \,ms^{-1}$
B
$1.25 \times 10^{-3} \,ms^{-1}$
C
$1.25 \times 10^{-5} \,ms^{-1}$
D
$6.25 \times 10^{-3} \,ms^{-1}$

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Current Electricity

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Current Density, Drift Velocity and Mobility

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$A$ 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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$A$ potential difference is applied across the ends of a metallic wire. If the potential difference is doubled,then the drift velocity

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Assume a hypothetical wire in which free electron density changes with temperature in proportionality $n \propto T$,assuming $\tau$ (relaxation time of collision) and dimensions of the wire remain unchanged with increasing temperature. Which one of the resistance $v/s$ temperature graphs is true?

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