First slide

Drift velocity

Question

The mass and charge of an electron are m and e respectively. The momentum acquired by the electrons in l cm of wire when a current i starts to flow is

Easy

A

$\frac{mil}{e}$
B

$\frac{ml}{ie}$
C

$\frac{mie}{l}$
D

$\frac{e}{mil}$

Solution

If $α$ be the area of cross-section of wire, then its volume is a l. Let n be the number of electrons per unit volume. Now, the number of electrons in the wire is n a L The mass of the electron is m and if its drift velocity be v_d, then momentum acquired by the electrons is the momentum acquired by the wire. So, momentum acquired by the wire
$p = {nalmv}_{d}$
We know that $v_{d} = i / nae$
$∴ p = nalm (i / nae)$
or $p = \frac{mil}{e}$

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