Question

2) An aluminum wire (ρ = 3.15 × 10-8 Ω.m at 20.0°C) with a diameter of 0.100 mm has a uniform electric field of 0.200 V/m imposed along its entire length. The temperature of the wire is 50.0°C. Assume n = 6.02 × 1028 e−/m3. (a) Determine the resistivity of aluminum at this temperature. (b) What is the current density in the wire? (c) What is the total current in the wire? (d) What is the drift speed of the conduction electrons? (e) What potential difference must exist between the ends of a 2.00-m length of the wire to produce the stated electric field? (f) What is the resistance of a 2.00-m length of the wire at 50.0°C?

Answer 1

Given that,

Length of wire = 2.00 m

Diameter =0.100 mm

Electric field = 0.200 V/m

Resistivity $\rho=3.15\times10^{-8}\ \Omega m$

Temperature = 20.0°C

(a). We need to calculate the resistivity of aluminum at this temperature

Using formula of resistivity of aluminum

$\rho=\rho_{0}(1+\alpha(T-T_{0}))$

Put the value into the formula

$\rho=3.15\times10^{-8}(1+3.9\times10^{-3}(50-20))$

$\rho= 3.51\times10^{-8}\ \Omega-m$

(b). We need to calculate the current density in the wire

Using formula of current density

$J=\sigma E$

$J=\dfrac{E}{\rho}$

Put the value into the formula

$J=\dfrac{0.200}{3.51\times10^{-8}}$

$J=5698005.69\ A/m^2$

$J=5.6\times10^{6}\ A/m^2$

(c). We need to calculate the total current in the wire

Using formula of current density

$J=\dfrac{I}{A}$

$I=J\times A$

$I=J\times\pi r^2$

$I=5.6\times10^{6}\times\pi\times(0.05\times10^{-3})^2$

$I=0.04398\ A$

(d). We need to calculate the drift speed of the conduction electron

Using formula of drift speed

$v_{d}=\dfrac{IM}{N\rho_{Al}qA}$

Put the value into the formula

$v_{d}=\dfrac{0.04398\times0.027}{6.02\times10^{23}\times2.7\times10^3\times1.6\times10^{-19}\times\pi\times(0.05\times10^{-3})^2}$

$v_{d}=0.000581\ m/s$

$v_{d}=5.81\times10^{-4}\ m/s$

(e). We need to calculate the potential difference

Using formula of potential difference

$\Delta V=El$

Put the value into the formula

$\Delta V=0.200\times2.00$

$\Delta V=0.4\ V$

(f). We need to calculate the resistance

Using formula of resistance

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

Put the value into the formula

$R=\dfrac{3.15\times10^{-8}\times2}{\pi\times(0.05\times10^{-3})^2}$

$R=8.02\ \Omega$

Hence, (a). The resistivity of aluminum at this temperature is $3.51\times10^{-8}\ \Omega-m$

(b). The current density in the wire  is $5.6\times10^{6}\ A/m^2$

(c). The total current in the wire is 0.04398 A.

(d). The drift speed of the conduction electrons is $5.81\times10^{-4}\ m/s$

(e). The potential difference is 0.4 V.

(f). The resistance of the wire is 8.02 Ω.

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Topic : resistivity

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