Cross-sectional area of a copper wire is equal to the area of a square of length $2 \ mm$. If this copper wire draws $8 \ A$ electric current,then find the drift velocity of free electrons. The number density of electrons in the copper wire is $8 \times 10^{28} \ m^{-3}$.

  • A
    $1.56 \times 10^{-4} \ ms^{-1}$
  • B
    $1.56 \times 10^{-2} \ ms^{-1}$
  • C
    $3.12 \times 10^{-3} \ ms^{-1}$
  • D
    $3.12 \times 10^{-2} \ ms^{-1}$

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Similar Questions

Assertion : The electric bulb glows immediately when the switch is on.
Reason : The drift velocity of electrons in a metallic wire is very high.

$A$ copper wire of cross-sectional area $1.0 \ mm^2$ carries a current of $1.34 \ A$. Assuming that each copper atom contributes one free electron,calculate the drift velocity of the free electrons in the wire in $mm/s$. (Given: density of copper = $8990 \ kg/m^3$,atomic mass = $63.50 \ g/mol$)

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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The potential difference between the ends of a straight conductor of length $20 \ cm$ is $16 \ V$. If the drift speed of the electrons is $2.4 \times 10^{-4} \ ms^{-1}$,the electron mobility in $m^2 \ V^{-1} \ s^{-1}$ is

The current density in a cylindrical wire of radius $4 \; mm$ is $4 \times 10^{6} \; A \cdot m^{-2}$. The current through the outer portion of the wire between radial distance $\frac{R}{2}$ and $R$ is $\dots \; \pi \; A$.

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