Two wires $A$ and $B$ of same material having length $L_A, L_B$ and radii $R_A, R_B$ and drift velocity $v_A…
Two wires $A$ and $B$ of same material having length $L_A, L_B$ and radii $R_A, R_B$ and drift velocity $v_A, v_B$ respectively carries same current. If $L_A=L_B$ and $R_A=2 R_B$ then the value of $\left(\frac{v_A}{v_B}\right)$ is
$0.25$
$0.5$
$2.0$
$1.0$
Solution
Current through a wire is given by
$I=n e v_d A=n e v_d \pi R^2$
Here, $n$ = number density of free charge carriers
$v_d=$ drift speed of electrons
$R=$ radius of wire
$e=$ charge on an electron
Here, for two wires $A$ and $B$
$\begin{aligned} & I_A=I_B \text { (given) } \\ & n_A=n_B \text { (wires have same material) }\end{aligned}$
$R_A=2 R_B$ (given)
so we have,
$I_A=I_B$
$\Rightarrow \quad n e v_A \pi R_A^2=n e v_B \pi R_B^2$
$\begin{array}{ll}\Rightarrow & \frac{v_A}{v_B}=\left(\frac{R_B}{R_A}\right)^2=\left(\frac{R_B}{2 R_B}\right)^2 \\ \Rightarrow & \frac{v_A}{v_B}=\frac{1}{4}=0.25\end{array}$