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The cross-sectional area of a wire is $1.0\times 10^{-7} m^2$ and the density of free electrons is $2.0\times 10^{28} m^{-3}$. What will be the drift velocity of the free electrons for a current of 3.2A in the wire?

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Last updated date: 20th Jun 2024
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Answer
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Hint: Drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an external electric field, caused by applying potential difference. In general, an electron in a conductor will propagate randomly here and there such that the average velocity could be assumed to be zero. But due to motion, electrons collide with one another which transfers electrical energy through the wire.
Formula used: $I = ev_dAn$

Complete answer:
Current is caused by the transfer of energy through collisions. $I = ev_dAn$, here $n=2.0\times 10^{28} m^{-3}$is the electron density inside the wire. ‘e’ is the charge of an electron and ‘A’ is the area of cross section. $v_d$ is the drift velocity of the electrons causing the current ‘I’.
Hence putting the values in;
$I = ev_dAn$
$v_d = \dfrac{I}{eAn}$
$v_d = \dfrac{3.2}{1.6\times 10^{-19}\times 10^{-7}\times 2\times 10^{28}} = 0.01 = 1 cm/s$

Hence the drift velocity of current (electrons) is 1cm/s.

Note:
Drift velocity in this problem is of order of cm. But in reality, it is even lesser (of order $10^{-4} m/s$). One might wonder, if the drift velocity of electrons in a conducting wire is this much less, then how come the tube light or fan turns on immediately when the switch is on? This doubt arises because we think the flow of charge in a wrong way. Actually, electrons don’t actually flow through the wires. It just vibrates (or travels a very short distance) with this velocity. After travelling a shorter distance, one electron gets collided with another electron and energy gets transferred. This process continues and energy is passed from one terminal to the battery to the other and since the electron density inside a conductor is very high, hence rate of energy transfer is also higher.