The drift velocity of electrons in a current- | vedclass

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Question

(C) The formula for electric current in terms of drift velocity is $i = neAv_d$,where $A = \pi r^2$ is the cross-sectional area.Thus,$i = ne \pi r^2 v

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$i/2$

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(C) The formula for electric current in terms of drift velocity is $i = neAv_d$,where $A = \pi r^2$ is the cross-sectional area.Thus,$i = ne \pi r^2 v$.For the new wire,the radius is $r' = r/2$ and the required drift velocity is $v' = 2v$.The new current $i'$ is given by $i' = ne \pi (r')^2 v'$.Substituting the values: $i' = ne \pi (r/2)^2 (2v)$.$i' = ne \pi (r^2/4) (2v) = \frac{1}{2} (ne \pi r^2 v)$.Since $i = ne \pi r^2 v$,we get $i' = i/2$.

The drift velocity of electrons in a current-carrying wire of radius $r$ is $v$. If we want a drift velocity of $2v$ in a wire of the same material but with half the radius,what current must be passed through it?

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