Two wires A and B of the same material have l | vedclass

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(A) The current $I$ flowing through a conductor is given by the formula: $I = n e v_d A = n e v_d \pi R^2$.Here,$n$ is the number density of free char

Vedclass · Accepted Answer

$0.25$

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(A) The current $I$ flowing through a conductor is given by the formula: $I = n e v_d A = n e v_d \pi R^2$.Here,$n$ is the number density of free charge carriers,$e$ is the elementary charge,$v_d$ is the drift velocity,and $R$ is the radius of the wire.Since both wires are made of the same material,the number density $n$ is the same for both.Given that the current $I$ is the same for both wires $(I_A = I_B)$,we can write:$n e v_A \pi R_A^2 = n e v_B \pi R_B^2$Dividing both sides by $n e \pi$,we get:$v_A R_A^2 = v_B R_B^2$Rearranging to find the ratio $\frac{v_A}{v_B}$:$\frac{v_A}{v_B} = \left(\frac{R_B}{R_A}\right)^2$Given $R_A = 2R_B$,substitute this into the equation:$\frac{v_A}{v_B} = \left(\frac{R_B}{2R_B}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25$.

Two wires $A$ and $B$ of the same material have lengths $L_A, L_B$ and radii $R_A, R_B$ and drift velocities $v_A, v_B$ respectively. Both wires carry the same current. If $L_A = L_B$ and $R_A = 2R_B$,then the value of $\left(\frac{v_A}{v_B}\right)$ is:

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