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(A) The current $I$ in a conductor is given by the relation $I = n e A v_d$,where $n$ is the number density of electrons,$e$ is the charge of an elect

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$16: 1$

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(A) The current $I$ in a conductor is given by the relation $I = n e A v_d$,where $n$ is the number density of electrons,$e$ is the charge of an electron,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.Since both wires are of the same material,$n$ and $e$ are constant.Thus,$v_d = \frac{I}{n e A} = \frac{I}{n e (\pi r^2)}$.Given the ratio of radii $r_A : r_B = 1 : 2$,the ratio of areas is $A_A : A_B = r_A^2 : r_B^2 = 1^2 : 2^2 = 1 : 4$.Given the ratio of currents $I_A : I_B = 4 : 1$.The ratio of drift velocities is $\frac{v_{dA}}{v_{dB}} = \frac{I_A}{A_A} 	imes \frac{A_B}{I_B} = \left(\frac{4}{1}ight) 	imes \left(\frac{4}{1}ight) = \frac{16}{1}$.Therefore,the ratio is $16:1$.

Two wires $A$ and $B$ of the same material,having radii in the ratio $1: 2$,carry currents in the ratio $4: 1$. The ratio of drift speed of electrons in $A$ and $B$ is .......

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