In an experiment to determine the resistivity of the material of a wire, a student measured the length, diameter and resistance of a sample of nichrome wire - Leaving Cert Physics - Question 4 - 2010

The student ensured the wire was straight and securely fixed, then used a meter stick or ruler to measure the distance between the crocodile clips attached to both ends of the wire. This measured length corresponds to the effective length of the wire in the circuit.

The student used a micrometer or caliper to measure the diameter of the wire. This instrument provides precise measurements by using adjustable jaws that can grip the wire and give a direct reading on the scale.

The diameter of the wire may vary along its length due to manufacturing inconsistencies or physical deformities. By measuring at different points, the student could obtain an average diameter, which is crucial for accurate calculations of the cross-sectional area and subsequently the resistivity.

To calculate the cross-sectional area (A) of the wire using the diameter measurements, the average diameter is first determined. For the given readings (0.21 mm, 0.20 mm, and 0.18 mm), the average diameter is:

$d_{avg} = \frac{0.21 + 0.20 + 0.18}{3} = 0.1967 \text{ mm}$

Now converting this to meters: 0.1967 mm = 0.1967 x 10^{-3} m.

The radius (r) is half of the diameter:

$r = \frac{d_{avg}}{2} = \frac{0.1967 \times 10^{-3}}{2} = 0.09835 \times 10^{-3} m$

Using the formula for the area of a circle, the cross-sectional area is:

$A = \pi r^2 = \pi (0.09835 \times 10^{-3})^2 \approx 3.03 \times 10^{-9} m^2$

The resistivity (ρ) can be calculated using the formula:

$\rho = \frac{R A}{l}$

Where:

• R = resistance = 20.2 Ω
• A = cross-sectional area ≈ 3.03 x 10^{-9} m²
• l = length = 48.8 cm = 0.488 m

Substituting the values into the equation:

$\rho = \frac{20.2 \times 3.03 \times 10^{-9}}{0.488} \approx 1.25 \times 10^{-8} Ω \, m$