(B) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.Given:$I = 24 \, mA = 24 \times 10^{-3} \, A$$n = 3 \times 10^{23} \, m^{-3}$$e =

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(B) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.Given:$I = 24 \, mA = 24 \times 10^{-3} \, A$$n = 3 \times 10^{23} \, m^{-3}$$e = 1.6 \times 10^{-19} \, C$$A = 1 \, cm^2 = 10^{-4} \, m^2$Substituting these values into the formula:$v_d = \frac{24 \times 10^{-3}}{3 \times 10^{23} \times 1.6 \times 10^{-19} \times 10^{-4}}$$v_d = \frac{24 \times 10^{-3}}{4.8 \times 10^{0}}$$v_d = 5 \times 10^{-3} \, m/s$.

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

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In a wire with a cross-sectional area of $1 \, cm^2$ carrying a current of $24 \, mA$,the electron number density is $3 \times 10^{23} \, m^{-3}$. What is the drift velocity?

A
$5 \times 10^{-6} \, m/s$
B
$5 \times 10^{-3} \, m/s$
C
$0.5 \, m/s$
D
$5 \times 10^{-2} \, m/s$

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Current Electricity

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Current Density, Drift Velocity and Mobility

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$A$ copper wire of length $10 \, m$ and radius $(10^{-2} / \sqrt{\pi}) \, m$ has an electrical resistance of $10 \, \Omega$. The current density in the wire for an electric field strength of $10 \, V/m$ is:

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Is the motion of a charge across a junction momentum conserving? Why or why not?

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In a wire with a cross-sectional area of $1 \, cm^2$ carrying a current of $24 \, mA$,the electron number density is $3 \times 10^{23} \, m^{-3}$. What is the drift velocity?

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