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What property of a material determines how strongly it resists current?

Resistivity and the equation R = rho L / A, the effect of length and cross-sectional area on resistance, how resistivity varies with temperature for a metal, and superconductivity and its uses.

A focused answer to AQA A-Level Physics 3.5.1.3, covering resistivity and the equation R = rho L / A, the effect of length and cross-sectional area on resistance, how the resistivity of a metal varies with temperature, and superconductivity with its applications.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. Resistivity and the resistance equation
  3. Effect of dimensions
  4. Temperature dependence in a metal
  5. Superconductivity
  6. Try this

What this dot point is asking

AQA specification point 3.5.1.3 wants you to define resistivity, use R=ρLAR = \dfrac{\rho L}{A}, explain how resistance depends on length and area, describe how a metal's resistivity changes with temperature, and explain superconductivity and its uses.

Resistivity and the resistance equation

Resistivity is a property of the material, while resistance also depends on the dimensions of the sample. A short, fat copper bar and a long, thin copper wire are made of the same material (same ρ\rho) but have very different resistances. Typical values span an enormous range: copper has ρ≈1.7×10−8 Ω m\rho \approx 1.7 \times 10^{-8} \text{ }\Omega \text{ m}, while a good insulator can exceed 1012 Ω m10^{12} \text{ }\Omega \text{ m}, a difference of about twenty orders of magnitude, which reflects the huge difference in free-carrier number density.

Effect of dimensions

This is why high-current wiring uses thick cables (large AA keeps the resistance and hence the power loss I2RI^2 R low) and why a long extension lead has measurably more resistance than a short one of the same gauge.

Temperature dependence in a metal

As a metal heats up, its positive ions vibrate with greater amplitude about their lattice sites, so the drifting conduction electrons collide with them more frequently. Each collision transfers energy from the electrons to the lattice, so a larger field is needed to maintain a given drift velocity. The resistivity, and therefore the resistance, increases roughly linearly with temperature over a wide range. This is the microscopic reason a filament lamp is non-ohmic: as it draws current it heats up and its resistance rises.

Superconductivity

Conventional superconductors must be cooled with liquid helium to a few kelvin, while high-temperature superconductors work above the boiling point of cheaper liquid nitrogen (77 K77 \text{ K}). Applications include the powerful electromagnets in MRI scanners and particle accelerators such as the Large Hadron Collider, magnetically levitated (maglev) trains, and the prospect of lossless power transmission cables that waste no energy as heat.

Try this

Q1. State how the resistance of a wire changes if its length is doubled. [1 mark]

  • Cue. The resistance doubles.

Q2. Explain why the resistivity of a metal increases with temperature. [2 marks]

  • Cue. The ions vibrate more, so electrons collide with them more frequently, increasing resistance.

Q3. State one application of superconductors. [1 mark]

  • Cue. The electromagnets in an MRI scanner (or a particle accelerator, or maglev trains).

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20184 marksA wire of length 1.5 m1.5 \text{ m} and diameter 0.40 mm0.40 \text{ mm} is made of metal of resistivity 4.9×10−7 Ω m4.9 \times 10^{-7} \text{ }\Omega \text{ m}. Calculate its resistance.
Show worked answer →

First find the cross-sectional area, then apply R=ρLAR = \dfrac{\rho L}{A}.

The radius is 0.20 mm=2.0×10−4 m0.20 \text{ mm} = 2.0 \times 10^{-4} \text{ m}, so A=πr2=π(2.0×10−4)2=1.26×10−7 m2A = \pi r^2 = \pi (2.0 \times 10^{-4})^2 = 1.26 \times 10^{-7} \text{ m}^2.

R=ρLA=(4.9×10−7)(1.5)1.26×10−7=5.8 ΩR = \dfrac{\rho L}{A} = \dfrac{(4.9 \times 10^{-7})(1.5)}{1.26 \times 10^{-7}} = 5.8 \text{ }\Omega.

Markers reward converting the diameter to a radius in metres, finding the area with πr2\pi r^2, and correct substitution into the resistivity equation.

AQA 20224 marksDescribe an experiment to determine the resistivity of the metal of a uniform wire, stating the measurements taken and how the result is calculated.
Show worked answer →

Measure the diameter of the wire at several points along its length with a micrometer and average, then calculate the cross-sectional area A=πr2A = \pi r^2. Connect a known length of the wire into a circuit with an ammeter and voltmeter, and vary the length using a crocodile clip.

For each length LL, record the current and voltage and find the resistance R=VIR = \dfrac{V}{I}. Plot RR against LL; the graph is a straight line through the origin with gradient ρA\dfrac{\rho}{A}.

The resistivity is then ρ=gradient×A\rho = \text{gradient} \times A. Using a graph reduces the effect of random errors and avoids relying on a single length.

Markers reward measuring the diameter (micrometer) and length, using R=VIR = \dfrac{V}{I}, plotting RR against LL, and extracting ρ\rho from the gradient times the area.

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