A conductor wire having ${10^{29}}$ free ${\rm{electrons}}\,\,/\,\,{{\rm{m}}^3}$ carries a current of ${\rm{20}}\,{\rm{A}}$. If the cross-section of the wire is $1\,\,{\rm{m}}{{\rm{m}}^2}$, then the drift velocity of electrons will be

$1.25\,\, \times \,\,{10^{ - 3}}\,\,{\rm{m}}{{\rm{s}}^{ - 1}}$

$6.25\,\, \times \,\,{10^{ - 3}}\,\,{\rm{m}}{{\rm{s}}^{ - 1}}$

$1.25\,\, \times \,\,{10^{ - 5}}\,\,{\rm{m}}{{\rm{s}}^{ - 1}}$

$1.25\,\, \times \,\,{10^{ - 4}}\,\,{\rm{m}}{{\rm{s}}^{ - 1}}$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

A conductor wire having

{10^{29}}

free

\mathrm{electrons}\,\,/\,\,{\mathrm{m}^3}

carries a current of

\mathrm{20}\,\mathrm{A}

. If the cross-section of the wire is

1\,\,\mathrm{m}{\mathrm{m}^2}

, then the drift velocity of electrons will be

View answer and explanation

Correct option: C

1.25\,\, \times \,\,{10^{ - 3}}\,\,\mathrm{m}{\mathrm{s}^{ - 1}}

Option C (

1.25 \times {10^{ - 3}}\,\mathrm{m}{\mathrm{s}^{ - 1}}

) is correct because it is the drift velocity calculated using the formula

{v_d} = \frac{I}{{nAe}}