A copper wire has a square cross-section, 2.0 mm on a side. It carries a current of 8 A and the density of free electrons is `8 xx 10^(28)m^(-3)`. The drift speed of electrons is equal to

Options

A. `0.156 xx 10^(-3) ms`
B. `0.156 xx 10^(-2)ms`
C. `3.12 xx 10^(-3)ms`
D. `3.12 xx 10^(-2)ms`

Answer

To find the drift speed of electrons in a copper wire, we can use the formula that relates current (I), number density of electrons (n), charge of an electron (e), drift velocity (vd), and cross-sectional area (A):

\[ I = n \cdot e \cdot v_d \cdot A \]

Where:
- \( I \) = current (in Amperes)
- \( n \) = number density of free electrons (in \( m^{-3} \))
- \( e \) = charge of an electron (approximately \( 1.6 \times 10^{-19} \) Coulombs)
- \( v_d \) = drift velocity (in \( m/s \))
- \( A \) = cross-sectional area of the wire (in \( m^2 \))

### Step-by-Step Solution:

**Step 1: Calculate the cross-sectional area (A) of the wire.**

The wire has a square cross-section with each side measuring 2.0 mm. First, convert this measurement to meters:

\[ 2.0 \, \text{mm} = 2.0 \times 10^{-3} \, \text{m} \]

Now, calculate the area:

\[ A = \text{side}^2 = (2.0 \times 10^{-3} \, \text{m})^2 = 4.0 \times 10^{-6} \, \text{m}^2 \]

**Step 2: Substitute known values into the formula.**

We know:
- \( I = 8 \, \text{A} \)
- \( n = 8 \times 10^{28} \, \text{m}^{-3} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( A = 4.0 \times 10^{-6} \, \text{m}^2 \)

Now, rearranging the formula to solve for drift velocity (\( v_d \)):

\[ v_d = \frac{I}{n \cdot e \cdot A} \]

**Step 3: Plug in the values and calculate \( v_d \).**

\[ v_d = \frac{8 \, \text{A}}{(8 \times 10^{28} \, \text{m}^{-3}) \cdot (1.6 \times 10^{-19} \, \text{C}) \cdot (4.0 \times 10^{-6} \, \text{m}^2)} \]

Calculating the denominator:

\[ n \cdot e \cdot A = (8 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (4.0 \times 10^{-6}) \]

Calculating this step-by-step:

1. \( 8 \times 1.6 = 12.8 \)
2. \( 12.8 \times 4.0 = 51.2 \)
3. \( 51.2 \times 10^{28 - 19 - 6} = 51.2 \times 10^{3} \)

So, the denominator becomes:

\[ 51.2 \times 10^{3} \]

Now, we can calculate \( v_d \):

\[ v_d = \frac{8}{51.2 \times 10^{3}} \]

Calculating this gives:

\[ v_d = \frac{8}{51200} = 0.15625 \times 10^{-3} \, \text{m/s} \]

Thus, the drift speed of electrons is approximately:

\[ v_d \approx 0.156 \times 10^{-3} \, \text{m/s} \]

### Final Answer:
The drift speed of electrons is approximately \( 0.156 \times 10^{-3} \, \text{m/s} \).

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