Calculate the conductivity σ of the material of which this wire is made

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SUMMARY

The discussion focuses on calculating the conductivity (σ) of a wire with a length of 1.0 m and a cross-sectional area of 1.1 mm², carrying a current of 5.4 A under a potential difference of 2.1 V. The resistance (R) is determined using the formula R = V/I, leading to R = 2.1 V / 5.4 A = 0.3889 ohms. The resistivity (ρ) is then calculated using the formula R = ρL/A, which allows for the determination of conductivity as σ = 1/ρ.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of resistance and resistivity formulas (R = ρL/A)
  • Familiarity with units of measurement for electrical properties (ohms, volts, amperes)
  • Basic geometry for calculating cross-sectional area (A = πr²)
NEXT STEPS
  • Calculate the resistivity (ρ) of the wire using the derived resistance value.
  • Explore the relationship between conductivity (σ) and resistivity (ρ) in different materials.
  • Investigate the impact of temperature on the conductivity of various materials.
  • Learn about practical applications of conductivity in electrical engineering and materials science.
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Students in physics or electrical engineering, professionals working with electrical materials, and anyone interested in understanding the properties of conductive materials.

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A wire is 1.0 m long and 1.1 mm2 in cross-sectional area. It carries a current of 5.4 A when a 2.1 V potential difference is applied between its ends. Calculate the conductivity σ of the material of which this wire is made.

Known data: I hate Wileyplus.

Ok, REAL known data: I know that I have to use 2pi(r), and pi(r)^2

R= (roe)(L)/(A)


I understand that if I were to use a specific material, I could find the resistance. However, this does not ask for that. Please help!
 
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Welcome to PF.

You know the resistance from V/I.

You know the A.

You know the Length.

So what is σ?

σ = 1/ρ
and
R = ρL/A
 

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