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(B) The current $I$ flowing through a conductor is given by the relation: $I = n A v_d e$,where $n$ is the number density of electrons,$A$ is the cros

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$\frac{1}{9}$

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(B) The current $I$ flowing through a conductor is given by the relation: $I = n A v_d e$,where $n$ is the number density of electrons,$A$ is the cross-sectional area,$v_d$ is the drift velocity,and $e$ is the charge of an electron.Since the current $I$ is constant throughout the wire,we have $v_d = \frac{I}{n A e}$.Since $I$,$n$,and $e$ are constant,$v_d \propto \frac{1}{A}$.Since the cross-section is circular,$A = \pi r^2$,therefore $v_d \propto \frac{1}{r^2}$.Now,calculating the ratio of drift velocities at points $A$ and $B$:$\frac{(v_d)_A}{(v_d)_B} = \frac{r_B^2}{r_A^2} = \frac{(r)^2}{(3r)^2} = \frac{r^2}{9r^2} = \frac{1}{9}$.Thus,the ratio is $\frac{1}{9}$.

As shown in the figure,a current is passing through a conducting wire. The radius of the cross-section of the wire at point $A$ is $3r$ and at point $B$ is $r$. Find the ratio of the drift velocity at point $A$ to that at point $B$.

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