Two wires A and B of the same material,having | vedclass

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(A) The drift velocity $v_d$ is related to the current $I$ by the formula $I = n e A v_d$,where $n$ is the electron density,$e$ is the charge of an el

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$16: 1$

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(A) The drift velocity $v_d$ is related to the current $I$ by the formula $I = n e A v_d$,where $n$ is the electron density,$e$ is the charge of an electron,and $A$ is the cross-sectional area.Since the wires are of the same material,$n$ is the same for both.The area $A = \pi r^2$. Given the ratio of radii $r_A : r_B = 1 : 2$,the ratio of areas is $A_A : A_B = (1)^2 : (2)^2 = 1 : 4$.Given the ratio of currents $I_A : I_B = 4 : 1$.Using the formula $v_d = \frac{I}{n e A}$,we get the ratio of drift speeds:$\frac{v_{dA}}{v_{dB}} = \frac{I_A}{I_B} 	imes \frac{A_B}{A_A} = \frac{4}{1} 	imes \frac{4}{1} = 16 : 1$.

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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