Question

In the given figure, a battery of emf \(E\) is connected across a conductor \(\text{PQ}\) of length \('l'\) and different area of cross-sections having radii \(r_1\) and \(r_2~(r_2<r_1)\).


Choose the correct option as one moves from \(\text P\) to \(\text Q\) :

• A. Electron current decreases
• B. Electric field decreases.
• C. Drift velocity of electron increases.
• D. All of these

Answer 1

The correct option is C Drift velocity of electron increases.


Take an elementary disc of radius $r$ and thickness $dx$ in conductor.

Resistance of element, $dR=\frac{\rho dx}{\pi r^2}$

Current is constant in conductor,

$\therefore i=$ constant

So, the potential drop across thickness $dx$ will be,

$dV=idR=\frac{i\rho dx}{\pi r^2}$

Electric field in the conductor will be,

$E_{field}=\frac{dV}{dx}=\frac{i\rho }{\pi r^2}$

$\therefore E_{field}\propto \frac{1}{r^2}$

And drift velocity will be,

$v_d=\frac{e{E}_{field}\tau }{m}$

$\therefore v_d\propto E_{field}\propto \frac{1}{r^2}$

From above expression, we can say that, If $r$ decreases, $E_{field}\&v_d$ will increase.

$\therefore$ option $\left(\text{A}\right)$ is correct.