(C) The formula for drift velocity is given by $v_d = \frac{I}{nAe}$.Given:$I = 1.344 \, A$$A = 1 \, mm^2 = 10^{-6} \, m^2$$n = 8.4 \times 10^{22} \,

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(C) The formula for drift velocity is given by $v_d = \frac{I}{nAe}$.Given:$I = 1.344 \, A$$A = 1 \, mm^2 = 10^{-6} \, m^2$$n = 8.4 \times 10^{22} \, \text{electrons}/cm^3 = 8.4 \times 10^{28} \, \text{electrons}/m^3$$e = 1.6 \times 10^{-19} \, C$Substituting these values into the formula:$v_d = \frac{1.344}{(8.4 \times 10^{28}) \times (10^{-6}) \times (1.6 \times 10^{-19})}$$v_d = \frac{1.344}{8.4 \times 1.6 \times 10^3}$$v_d = \frac{1.344}{13.44 \times 10^3} = 0.1 \times 10^{-3} \, m/s$$v_d = 0.1 \, mm/s$.

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

Easy

There is a current of $1.344 \, A$ in a copper wire whose area of cross-section normal to the length of the wire is $1 \, mm^2$. If the number of free electrons per $cm^3$ is $8.4 \times 10^{22}$, then the drift velocity would be

A
$1.0 \, mm/s$
B
$1.0 \, m/s$
C
$0.1 \, mm/s$
D
$0.01 \, mm/s$

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Class 12 Questions

Class 12

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Chapter

Current Electricity

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Current Density, Drift Velocity and Mobility

The electric intensity $E$,current density $j$,and specific resistance $\rho$ (often denoted as $k$ in some texts) are related to each other by the relation:

Easy

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$A$ current $I$ is passing through a wire having two sections $P$ and $Q$ of uniform diameters $d$ and $d/2$ respectively. If the mean drift velocity of electrons in sections $P$ and $Q$ is denoted by $v_P$ and $v_Q$ respectively,then

Medium

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Every atom contributes one free electron in copper. If $1.1 \ A$ current is flowing in a copper wire having a $1 \ mm$ diameter,then the drift velocity (approx.) will be (Density of copper $= 9 \times 10^3 \ kg \ m^{-3}$ and atomic weight $= 63$).

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$A$ current of $5 \,A$ is passing through a metallic wire of cross-sectional area $4 \times 10^{-6} \,m^{2}$. If the density of charge carriers of the wire is $5 \times 10^{26} \,m^{-3}$, the drift velocity of the electrons will be:

EasyKCET 2007

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$A$ current $I$ flows through a uniform wire of diameter $d$,when the mean drift velocity is $v_d$. The same current will flow through a wire of diameter $d/2$ made of the same material,if the mean drift velocity of the electrons is

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There is a current of $1.344 \, A$ in a copper wire whose area of cross-section normal to the length of the wire is $1 \, mm^2$. If the number of free electrons per $cm^3$ is $8.4 \times 10^{22}$, then the drift velocity would be

Similar Questions

The electric intensity $E$,current density $j$,and specific resistance $\rho$ (often denoted as $k$ in some texts) are related to each other by the relation:

$A$ current $I$ is passing through a wire having two sections $P$ and $Q$ of uniform diameters $d$ and $d/2$ respectively. If the mean drift velocity of electrons in sections $P$ and $Q$ is denoted by $v_P$ and $v_Q$ respectively,then

Every atom contributes one free electron in copper. If $1.1 \ A$ current is flowing in a copper wire having a $1 \ mm$ diameter,then the drift velocity (approx.) will be (Density of copper $= 9 \times 10^3 \ kg \ m^{-3}$ and atomic weight $= 63$).

$A$ current of $5 \,A$ is passing through a metallic wire of cross-sectional area $4 \times 10^{-6} \,m^{2}$. If the density of charge carriers of the wire is $5 \times 10^{26} \,m^{-3}$, the drift velocity of the electrons will be:

$A$ current $I$ flows through a uniform wire of diameter $d$,when the mean drift velocity is $v_d$. The same current will flow through a wire of diameter $d/2$ made of the same material,if the mean drift velocity of the electrons is

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