A bird lands on a bare copper wire carrying a current of 32 A. Its cross-sectional area is 0.13 cm 2 . (a) Find the difference in potential between the bird's feet, assuming they are separated by a distance of 6.0 cm. (b) Will your answer to Part a increase or decrease if the separation between the bird's feet increases? Explain. 

To solve this problem, we will use the formula for the **potential difference** (\( V \)) along a conductor due to the flow of current:

\[
V = I \cdot R
\]
where:
- \( I \) is the current (in Amperes),
- \( R \) is the resistance of the portion of the wire between the bird’s feet (in Ohms).

The resistance \( R \) of a wire can be calculated using the formula:

\[
R = \rho \cdot \frac{L}{A}
\]
where:
- \( \rho \) is the resistivity of the material (in Ohm meters),
- \( L \) is the length of the wire (the distance between the bird's feet, in meters),
- \( A \) is the cross-sectional area of the wire (in square meters).

### Given:
- Current \( I = 32 \, \text{A} \),
- Cross-sectional area \( A = 0.13 \, \text{cm}^2 = 0.13 \times 10^{-4} \, \text{m}^2 \),
- Distance between the bird's feet \( L = 6.0 \, \text{cm} = 6.0 \times 10^{-2} \, \text{m} \),
- Resistivity of copper \( \rho = 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \) (standard value).

### (a) Potential difference between the bird’s feet:
First, we calculate the resistance of the wire between the bird's feet:

\[
R = \rho \cdot \frac{L}{A} = 1.68 \times 10^{-8} \cdot \frac{6.0 \times 10^{-2}}{0.13 \times 10^{-4}}
\]

Now, we can find the potential difference using \( V = I \cdot R \).

Let's calculate it.The potential difference between the bird's feet is approximately \( 0.00248 \, \text{V} \) (or 2.48 millivolts).

### (b) Will the potential difference increase or decrease if the separation between the bird's feet increases?

The potential difference is directly proportional to the distance between the bird’s feet, as shown in the resistance formula \( R = \rho \frac{L}{A} \). If the separation between the bird's feet \( L \) increases, the resistance \( R \) increases, and hence the potential difference \( V = I \cdot R \) will also increase.

### Summary:
- (a) The potential difference between the bird's feet is approximately 2.48 millivolts.
- (b) The potential difference will **increase** if the separation between the bird’s feet increases, because resistance is directly proportional to the length of the wire segment.

---

Would you like further clarification on any part of the solution? Here are some related questions to expand your understanding:

1. How does the resistivity of a material affect the potential difference?
2. What would happen to the potential difference if the current were doubled?
3. Why do birds not get electrocuted when sitting on power lines?
4. How would the potential difference change if the wire were made of a different material, like aluminum?
5. What happens to the potential difference if the cross-sectional area of the wire increases?

**Tip:** The current through a conductor causes a small voltage drop along its length, and this voltage depends on the material’s resistivity, length, and cross-sectional area.

A bird lands on a bare copper wire carrying a current of 32 A. Its cross-sectional area is 0.13 cm². (a) Find the difference in potential between the bird's feet, assuming they are separated by a distance of 6.0 cm. (b) Will your answer to Part a increase or decrease if the separation between the bird's feet increases? Explain.

Learn how to calculate the potential difference between a bird's feet on a copper wire carrying 32 A current. This problem covers Ohm's Law, resistivity, and how changes in the separation distance affect the potential difference. Suitable for students in Grades 10-12.

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