Question

Two cylindrical rods of uniform cross-section area $A$ and $2A$, having free electrons per unit volume $2n$ and $n$, respectively, are joined in series. A current $I$ flows through them in steady state. Then the ratio of drift velocity of free electron in left rod to drift velocity of electron in the right rod is $\left(\frac{v_L}{v_R}\right)$ is $\left(\frac{x}{y}\right)$. Find $(x+y)$.

Answer 1

Step 1: Calculation of drift velocity

For rod having area $A$, carrier density $=2n$

For rod having area $2A$, carrier density $=n$

Since, they are connected in series, equal current flows through them i.e. current $=I$

We know $I=neA{v}_d$

$v_d=\frac{I}{neA}$

$n\to$ Carrier density

$A\to$ Cross-sectional area

$v_d\to$ Drift velocity

Now, for rod having area $A$, drift velocity

$v_L=\frac{I}{2neA}$

For rod having area $2A$, drift velocity

$v_R=\frac{I}{ne2A}$

Step 2: Calculating ratio of drift velocities

So, $\frac{v_L}{v_R}=\frac{I/2neA}{I/2neA}=1/1$

$\therefore x=1,y=1$

So, $x+y=1+1=2$

Hence, 2 is the correct answer.