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# Question:The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5 \times 10^{28} m^{-3}$. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of the cross-section of the wire is $2.0 \times 10^{-6} m^2$ and it carries a current of 3.0 A.

Last updated date: 12th Sep 2024
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Hint: To find the time taken by an electron to drift from one end of the wire to the other, use the formula $I = nAeV_d$, where I is the current, n is the number density of free electrons, A is the cross-sectional area, e is the electric charge, and $V_d$ is the drift velocity.

Step-by-Step Solution:

Given information:

Number density of free electrons $n = 8.5 \times 10^{28} \text{m}^{-3}\$

Length of the copper wire $l = 3.0 \text{m}$
Area of the cross-section of the wire $A = 2.0 \times 10^{-6} \text{m}^2$

Current carried by the wire $I = 3.0 \text{A}$
Electric charge $e = 1.6 \times 10^{-19} \text{C}$
Use the formula $I = nAeV_d$​ to calculate the drift velocity $V_d$:

$V_d = \dfrac{I}{nAe}$

Plug in the given values to calculate $V_d$:

$\ V_d = \dfrac{3.0 \text{A}}{8.5 \times 10^{28} \text{m}^{-3} \times 2.0 \times 10^{-6} \text{m}^2 \times 1.6 \times 10^{-19} \text{C}}\$

Calculate the value of $V_d$ .

Now, to find the time taken by an electron to drift from one end of the wire to the other, use the formula $\ t = \dfrac{l}{V_d}\$, where ${l}$ is the length of the wire and $V_d$ is the drift velocity.

Plug in the values to calculate ${t}$:

$t = \dfrac{3.0 \text{m}}{1.103 \times 10^{-4} \text{m/s}}$

The time taken by an electron to drift from one end of the wire to the other is approximately $\ 2.7 \times 10^4\$  seconds.