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

Current (I) is defined as $nAv_dq_e$, with n representing electron density, A signifying cross-sectional area, $v_d$ denoting drift velocity, q indicating the charge of an electron, and e being the electronic charge.

Given that $A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}$, and consequently $A \propto d^2$, the current can be expressed as $I \propto d^2 v_d$.

The equation $\frac{100}{200} = \frac{d'^2 v'}{\left(\frac{d}{2}\right)^2 v'}$ simplifies to $v' = 2 \times 2^2 v$, resulting in $v' = 8v$.