A potential difference of 6.00 nV is set up across a 1.50 cm length of copper wire that has a radius of 2.00 mm. How much charge drifts through a cross section in 4.70 ms?

Answer

To solve the problem step by step, we will follow the outlined approach to find the charge that drifts through a cross-section of the copper wire.

### Step 1: Convert Given Values to Standard Units
- **Potential Difference (V)**: Given as 6.00 nV.
\[ V = 6.00 \, \text{nV} = 6.00 \times 10^{-9} \, \text{V} \]

- **Length of the Wire (L)**: Given as 1.50 cm.
\[ L = 1.50 \, \text{cm} = 1.50 \times 10^{-2} \, \text{m} \]

- **Radius of the Wire (r)**: Given as 2.00 mm.
\[ r = 2.00 \, \text{mm} = 2.00 \times 10^{-3} \, \text{m} \]

- **Time (t)**: Given as 4.70 ms.
\[ t = 4.70 \, \text{ms} = 4.70 \times 10^{-3} \, \text{s} \]

### Step 2: Calculate the Cross-Sectional Area (A) of the Wire
The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (2.00 \times 10^{-3})^2 = \pi (4.00 \times 10^{-6}) \approx 1.25664 \times 10^{-5} \, \text{m}^2 \]

### Step 3: Calculate the Resistance (R) of the Wire
Using the formula for resistance \( R \):
\[ R = \frac{\rho L}{A} \]
Where \( \rho \) (resistivity of copper) is given as \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \).
Substituting the values:
\[ R = \frac{1.68 \times 10^{-8} \times 1.50 \times 10^{-2}}{1.25664 \times 10^{-5}} \approx 2.01 \times 10^{-3} \, \Omega \]

### Step 4: Calculate the Current (I) Using Ohm's Law
Using Ohm's Law:
\[ I = \frac{V}{R} \]
Substituting the values we calculated:
\[ I = \frac{6.00 \times 10^{-9}}{2.01 \times 10^{-3}} \approx 2.98 \times 10^{-6} \, \text{A} \]

### Step 5: Calculate the Charge (Q) That Drifts Through the Wire
Using the formula for charge:
\[ Q = I \cdot t \]
Substituting the values:
\[ Q = (2.98 \times 10^{-6}) \cdot (4.70 \times 10^{-3}) \approx 1.40 \times 10^{-6} \, \text{C} \]

### Final Answer
The charge that drifts through a cross-section of the copper wire in 4.70 ms is approximately:
\[ Q \approx 1.40 \times 10^{-6} \, \text{C} \]