Question

A uniform wire of diameter \(d\) carries a current of \(100 \, \text{mA}\) when the mean drift velocity of electrons in the wire is \(v\). For a wire of diameter \(\frac{d}{2}\) of the same material to carry a current of \(200 \, \text{mA}\), the mean drift velocity of electrons in the wire is

• A. 4v
• B. 8v
• C. v
• D. 2v

Answer 1

The Correct Option is B

1. Current in terms of drift velocity

$$ I = nAv_d q $$  

Where:

• $n$ = electron density
• $A$ = cross-sectional area of wire
• $v_d$ = drift velocity
• $q$ = charge of an electron ($e$)

2. Area relation with diameter $d$

$$ A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} $$

So, $A \propto d^2$

Therefore:

$$ I \propto d^2 v_d $$

For same material, $n$ and $q$ are constant.

3. Comparing two cases

Let initial: diameter = $d$, current = $I$, drift velocity = $v_d$

New: diameter = $d' = \frac{d}{2}$, current = $I' = 2I$, drift velocity = $v_d'$

$$ \frac{I'}{I} = \frac{d'^2 v_d'}{d^2 v_d} $$

$$ \frac{2I}{I} = \frac{\left(\frac{d}{2}\right)^2 v_d'}{d^2 v_d} $$

$$ 2 = \frac{\frac{d^2}{4} \cdot v_d'}{d^2 v_d} = \frac{v_d'}{4 v_d} $$

$$ v_d' = 2 \times 4 \, v_d = 8 v_d $$

$$ \therefore v_d' = 8v $$