What do you understand by the term drift velocity as applied to any current carriers in a wire?

$$n = \frac{\rho}{M} \times N_A \times (\text{electrons per atom})$$ $$n = \frac{1.05 \times 10^4 \text{ kg/m}^3}{0.108 \text{ kg/mol}} \times 6.022 \times 10^{23} \text{ mol}^{-1} \times 1$$ $$n = 9.7222 \times 10^4 \times 6.022 \times 10^{23} \text{ m}^{-3}$$ $$n = 5.855 \times 10^{28} \text{ m}^{-3}$$

$$v_d = \frac{15 \text{ A}}{(5.855 \times 10^{28} \text{ m}^{-3}) \times (4.5 \times 10^{-6} \text{ m}^2) \times (1.602 \times 10^{-19} \text{ C})}$$ $$v_d = \frac{15}{42.19 \times 10^3}$$ $$v_d = \frac{15}{42190}$$ $$v_d = 3.555 \times 10^{-4} \text{ m/s}$$

The drift velocity of electrons is $\boxed{\text{3.56} \times 10^{-4} \text{ m/s}}$.

3. An unknown wire of $1 \text{ mm}$ diameter is found to carry and passes a total charge of $90 \text{ C}$ in $1 \text{ hour}$ and $15 \text{ min}$. If the wire has $5.8 \times 10^{28}$ free electrons per $m^3$, find:

Given:
Diameter $d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m}$
Radius $r = d/2 = 0.5 \times 10^{-3} \text{ m}$
Total charge $Q = 90 \text{ C}$
Time $t = 1 \text{ hour } 15 \text{ min}$
Number of free electrons per $m^3$, $n = 5.8 \times 10^{28} \text{ m}^{-3}$
Charge of an electron $e = 1.602 \times 10^{-19} \text{ C}$

a) The current in the wire.

$$I = \frac{90 \text{ C}}{4500 \text{ s}}$$ $$I = 0.02 \text{ A}$$

The current in the wire is $\boxed{\text{0.02 A}}$.

b) The drift velocity of the electrons in m s⁻¹.

$$A = \pi (0.5 \times 10^{-3} \text{ m})^2$$ $$A = \pi \times 0.25 \times 10^{-6} \text{ m}^2$$ $$A = 0.7854 \times 10^{-6} \text{ m}^2$$

$$v_d = \frac{0.02 \text{ A}}{(5.8 \times 10^{28} \text{ m}^{-3}) \times (0.7854 \times 10^{-6} \text{ m}^2) \times (1.602 \times 10^{-19} \text{ C})}$$ $$v_d = \frac{0.02}{7.30 \times 10^3}$$ $$v_d = \frac{0.02}{7300}$$ $$v_d = 2.74 \times 10^{-6} \text{ m/s}$$

The drift velocity of the electrons is $\boxed{\text{2.74} \times 10^{-6} \text{ m/s}}$.

4. The current of $12 \text{ A}$ is made to pass through an aluminium wire of radius $1.5 \text{ mm}$ which is joined in series with a copper wire of radius $0.8 \text{ mm}$. Determine:

Given:
Current $I = 12 \text{ A}$
Aluminium wire radius $r_{Al} = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m}$
Copper wire radius $r_{Cu} = 0.8 \text{ mm} = 0.8 \times 10^{-3} \text{ m}$
Number of free electrons per unit volume in copper wire $n_{Cu} = 10^{29} \text{ m}^{-3}$
Charge of an electron $e = 1.602 \times 10^{-19} \text{ C}$

a) The current density in an aluminium wire.

$$A_{Al} = \pi (1.5 \times 10^{-3} \text{ m})^2$$ $$A_{Al} = \pi \times 2.25 \times 10^{-6} \text{ m}^2$$ $$A_{Al} = 7.069 \times 10^{-6} \text{ m}^2$$

$$J_{Al} = \frac{12 \text{ A}}{7.069 \times 10^{-6} \text{ m}^2}$$ $$J_{Al} = 1.698 \times 10^6 \text{ A/m}^2$$

The current density in the aluminium wire is $\boxed{\text{1.70} \times 10^6 \text{ A/m}^2}$.

b) The drift velocity of the electron in the copper wire.

$$A_{Cu} = \pi (0.8 \times 10^{-3} \text{ m})^2$$ $$A_{Cu} = \pi \times 0.64 \times 10^{-6} \text{ m}^2$$ $$A_{Cu} = 2.011 \times 10^{-6} \text{ m}^2$$

$$v_{d,Cu} = \frac{12 \text{ A}}{(10^{29} \text{ m}^{-3}) \times (2.011 \times 10^{-6} \text{ m}^2) \times (1.602 \times 10^{-19} \text{ C})}$$ $$v_{d,Cu} = \frac{12}{3.221 \times 10^4}$$ $$v_{d,Cu} = \frac{12}{32210}$$ $$v_{d,Cu} = 3.725 \times 10^{-4} \text{ m/s}$$

The drift velocity of the electron in the copper wire is $\boxed{\text{3.73} \times 10^{-4} \text{ m/s}}$.