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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

A water heater consisting of a metal coil that is connected across the terminals of a 15 -V power supply is able to heat \(250 \mathrm{~mL}\) of water from room temperature to boiling point in \(45 \mathrm{~s}\). What is the resistance of the coil?

You make a parallel combination of resistors consisting of resistor A having a very large resistance and resistor B having a very small resistance. The equivalent resistance for this combination will be: a) slightly greater than the resistance of the resistor A. b) slightly less than the resistance of the resistor \(\mathrm{A}\). c) slightly greater than the resistance of the resistor B. d) slightly less than the resistance of the resistor B.

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Two resistors with resistances \(200 . \Omega\) and \(400 . \Omega\) are connected (a) in series and (b) in parallel with an ideal 9.00-V battery. Compare the power delivered to the \(200 .-\Omega\) resistor.
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