(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

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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

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

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Current Electricity

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Current Density, Drift Velocity and Mobility

We are able to obtain fairly large currents in a conductor because

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$A$ current of $10 \,A$ is maintained in a conductor of cross-section $1 \,cm^2$. If the number density of free electrons is $9 \times 10^{28} \,m^{-3}$,the drift velocity of free electrons is .......... $m/s$.

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Assertion: Free electrons always keep on moving in a conductor,even then no magnetic force acts on them in a magnetic field unless a current is passed through it.
Reason: The average velocity of free electrons is zero.

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Current density in a cylindrical wire of radius $R$ varies with radial distance as $J(r) = \beta(r + r_0)^2$. The current through the shaded section of the wire shown in the figure is:

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$A$ conducting wire of cross-sectional area $1 \,cm^2$ has $3 \times 10^{23}$ charge carriers per $m^3$. If the wire carries a current of $24 \,mA$, then the drift velocity of the carriers 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

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We are able to obtain fairly large currents in a conductor because

$A$ current of $10 \,A$ is maintained in a conductor of cross-section $1 \,cm^2$. If the number density of free electrons is $9 \times 10^{28} \,m^{-3}$,the drift velocity of free electrons is .......... $m/s$.

Assertion: Free electrons always keep on moving in a conductor,even then no magnetic force acts on them in a magnetic field unless a current is passed through it.Reason: The average velocity of free electrons is zero.

Current density in a cylindrical wire of radius $R$ varies with radial distance as $J(r) = \beta(r + r_0)^2$. The current through the shaded section of the wire shown in the figure is:

$A$ conducting wire of cross-sectional area $1 \,cm^2$ has $3 \times 10^{23}$ charge carriers per $m^3$. If the wire carries a current of $24 \,mA$, then the drift velocity of the carriers is:

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Assertion: Free electrons always keep on moving in a conductor,even then no magnetic force acts on them in a magnetic field unless a current is passed through it.
Reason: The average velocity of free electrons is zero.