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1. Question: A wire with a circular cross-section of 2.5 mm in diameter carries a weak current of $1.2 \times 10^{-10} \, \text{A}$. Assuming a uniform current distribution, calculate:
   • (a) the current density in the wire.
   • (b) the drift velocity of the electrons in the wire.

Solution:

(a) Current Density ($J$):

The formula for current density is:

$J = \frac{I}{A}$

Where:

• $I = 1.2 \times 10^{-10} \, \text{A}$ (current),
• $A = \pi r^2$ (cross-sectional area of the wire),
• $r$ is the radius of the wire.

The diameter of the wire is $2.5 \, \text{mm}$, so the radius is:

$r = \frac{2.5}{2} = 1.25 \, \text{mm} = 1.25 \times 10^{-3} \, \text{m}.$

Now calculate the area $A$:

$A = \pi r^2 = \pi (1.25 \times 10^{-3})^2 = \pi (1.5625 \times 10^{-6}) \approx 4.91 \times 10^{-6} \, \text{m}^2.$

Substitute the values of $I$ and $A$ into the current density formula:

$J = \frac{1.2 \times 10^{-10}}{4.91 \times 10^{-6}} \approx 2.44 \times 10^{-5} \, \text{A/m}^2.$

So, the current density is:

$J \approx 2.44 \times 10^{-5} \, \text{A/m}^2.$

(b) Drift Velocity ($v_d$):

The drift velocity of electrons is given by:

$v_d = \frac{J}{n e}$

Where:

• $J = 2.44 \times 10^{-5} \, \text{A/m}^2$ (current density),
• $n$ is the number density of electrons,
• $e = 1.6 \times 10^{-19} \, \text{C}$ (charge of an electron).

The number density of electrons $n$ for a typical conductor (like copper) is approximately:

$n = 8.5 \times 10^{28} \, \text{electrons/m}^3.$

Substitute these values into the drift velocity formula:

$v_d = \frac{2.44 \times 10^{-5}}{(8.5 \times 10^{28})(1.6 \times 10^{-19})}.$

First, calculate the denominator:

$n e = (8.5 \times 10^{28})(1.6 \times 10^{-19}) \approx 1.36 \times 10^{10}.$

Now calculate $v_d$:

$v_d = \frac{2.44 \times 10^{-5}}{1.36 \times 10^{10}} \approx 1.79 \times 10^{-15} \, \text{m/s}.$

So, the drift velocity is:

$v_d \approx 1.79 \times 10^{-15} \, \text{m/s}.$

Final Answers:

• (a) Current density: $J \approx 2.44 \times 10^{-5} \, \text{A/m}^2$,
• (b) Drift velocity: $v_d \approx 1.79 \times 10^{-15} \, \text{m/s}$.

Do you want me to elaborate on any step or explain related concepts in more depth?

Related Questions:

1. What is the physical significance of current density and drift velocity?
2. How does the diameter of the wire affect the drift velocity?
3. Why is the drift velocity of electrons so small in conductors?
4. How does the material's electron density ($n$) influence current flow?
5. What happens to $J$ and $v_d$ if the current $I$ is doubled?

Tip:

In conductors, the drift velocity is typically very small because the high electron density compensates for the weak current.