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Chapter 25: Problem 27

What is the current density in an aluminum wire having a radius of \(1.00 \mathrm{~mm}\) and carrying a current of \(1.00 \mathrm{~mA}\) ? What is the drift speed of the electrons carrying this current? The density of aluminum is \(2.70 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3},\) and 1 mole of aluminum has a mass of \(26.98 \mathrm{~g}\). There is one conduction electron per atom in aluminum.

Short Answer

Expert verified
Based on the given information in the problem, the current density in the aluminum wire is approximately \(3.18\cdot10^2\, \text{A/m}^2\), and the drift speed of the electrons carrying this current is approximately \(3.29\cdot10^{-4}\, \text{m/s}\).

Step by step solution

01

Calculate the cross-sectional area of the wire

First, let's calculate the cross-sectional area \(A\) of the aluminum wire using its given radius \(r=1.00\,\text{mm}=1.00\cdot 10^{-3}\, \text{m}\). Since the wire is cylindrical, we can calculate its area using the formula \(A= \pi r^2\): $$ A = \pi r^2 = \pi (1.00\cdot10^{-3}\, \text{m})^2 = 3.14\cdot10^{-6}\, \text{m}^2. $$
02

Calculate the current density

Now that we have the cross-sectional area of the wire, we can calculate the current density \(J\) using the formula \(J=I/A\), where \(I=1.00\, \text{mA}=1.00\cdot10^{-3}\, \text{A}\): $$ J = \frac{I}{A} = \frac{1.00\cdot10^{-3}\, \text{A}}{3.14\cdot10^{-6}\, \text{m}^2} = 3.18\cdot10^2\, \frac{\text{A}}{\text{m}^2}. $$
03

Calculate the number of moles in a unit volume of aluminum

The next step is to find the charge density \(n\) of the aluminum wire. First, we need to calculate the number of moles in a unit volume of aluminum. We know the density of aluminum is \(\rho = 2.70\cdot10^3\, \text{kg/m}^3\), and 1 mole of aluminum has a mass of \(M = 26.98\, \text{g} = 26.98\cdot10^{-3}\, \text{kg}\): $$ \text{Number of moles per unit volume} = \frac{\rho}{M} = \frac{2.70\cdot10^3\, \text{kg/m}^3}{26.98\cdot10^{-3}\, \text{kg/mol}} = 1.00\cdot10^{5}\, \text{mol/m}^3. $$
04

Calculate the charge density

Since there is one conduction electron per atom in aluminum, the charge density \(n\) is equal to the number of moles per unit volume multiplied by Avogadro's number \(N_A = 6.022\cdot10^{23}\, \text{mol}^{-1}\): $$ n = (1.00\cdot10^{5}\, \text{mol/m}^3)\cdot(6.022\cdot10^{23}\, \text{mol}^{-1}) = 6.022\cdot10^{28}\, \text{m}^{-3}. $$
05

Calculate the drift speed of electrons

Finally, we can calculate the drift speed of the electrons \(v_d\) using the formula \(v_d = I/(neA)\), where \(e\) is the elementary charge \(e = 1.60\cdot10^{-19}\, \text{C}\): $$ v_d = \frac{I}{neA} = \frac{1.00\cdot10^{-3}\, \text{A}}{(6.022\cdot10^{28}\, \text{m}^{-3})(1.60\cdot10^{-19}\, \text{C})(3.14\cdot10^{-6}\, \text{m}^2)} = 3.29\cdot10^{-4}\, \frac{\text{m}}{\text{s}}. $$ So, the current density in the aluminum wire is \(3.18\cdot10^2\, \text{A/m}^2\), and the drift speed of the electrons carrying this current is \(3.29\cdot10^{-4}\, \text{m/s}\).

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Most popular questions from this chapter

When a \(40.0-V\) emf device is placed across two resistors in series, a current of \(10.0 \mathrm{~A}\) is flowing in each of the resistors. When the same emf device is placed across the same two resistors in parallel, the current through the emf device is \(50.0 \mathrm{~A}\). What is the magnitude of the larger of the two resistances?

The Stanford Linear Accelerator accelerated a beam consisting of \(2.0 \cdot 10^{14}\) electrons per second through a potential difference of \(2.0 \cdot 10^{10} \mathrm{~V}\) a) Calculate the current in the beam. b) Calculate the power of the beam. c) Calculate the effective ohmic resistance of the accelerator.

Why do light bulbs typically burn out just as they are turned on rather than while they are lit?

A modern house is wired for \(115 \mathrm{~V}\), and the current is limited by circuit breakers to a maximum of \(200 .\) A. (For the purpose of this problem, treat these as DC quantities.) a) Calculate the minimum total resistance the circuitry in the house can have at any time. b) Calculate the maximum electrical power the house can consume.

What is (a) the conductance and (b) the radius of a \(3.5-\mathrm{m}\) -long iron heating element for a \(110-\mathrm{V}, 1500-\mathrm{W}\) heater?
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