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(A) The relationship between current $i$ and drift velocity $v_d$ is given by $i = neA v_d$,where $n$ is the electron density,$e$ is the charge of an

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$16 : 1$

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(A) The relationship between current $i$ and drift velocity $v_d$ is given by $i = neA v_d$,where $n$ is the electron density,$e$ is the charge of an electron,and $A$ is the cross-sectional area.Since the material is the same,$n$ is constant. Thus,$i \propto A v_d$.Given $A = \pi r^2$,we have $i \propto r^2 v_d$,which implies $v_d \propto \frac{i}{r^2}$.Therefore,the ratio of drift velocities is $\frac{v_{d1}}{v_{d2}} = \left( \frac{i_1}{i_2} ight) 	imes \left( \frac{r_2}{r_1} ight)^2$.Substituting the given values: $\frac{i_1}{i_2} = \frac{4}{1}$ and $\frac{r_1}{r_2} = \frac{1}{2} \Rightarrow \frac{r_2}{r_1} = \frac{2}{1}$.$\frac{v_{d1}}{v_{d2}} = \left( \frac{4}{1} ight) 	imes \left( \frac{2}{1} ight)^2 = 4 	imes 4 = 16$.Thus,the ratio is $16 : 1$.

If the ratio of the radii of two wires of the same material is $1 : 2$ and the ratio of the current flowing through them is $4 : 1$,what is the ratio of their drift velocities?

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