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Chapter 18: Problem 121

A gold wire and an aluminum wire have the same dimensions and carry the same current. The electron density (in electrons/cm \(^{3}\) ) in aluminum is three times larger than the density in gold. How do the drift speeds of the electrons in the two wires, \(v_{\mathrm{Au}}\) and \(v_{\mathrm{Al}},\) compare?

Short Answer

Expert verified
Answer: Under the same conditions, the drift speed of electrons in the gold wire is one-third of the drift speed of electrons in the aluminum wire.

Step by step solution

01

Recall the equation for drift speed

The drift speed, \(v\), of electrons in a metallic wire is given by the formula: \(v = \frac{I}{nqA}\), where \(I\) is the current, \(n\) is the electron density, \(q\) is the charge of an electron, and \(A\) is the cross-sectional area of the wire.
02

Set up the ratios

Since the gold and aluminum wires have the same dimensions and carry the same current, we can set up the following ratios: \(\frac{v_{Au}}{v_{Al}} = \frac{I_{Au}/(n_{Au}qA_{Au})}{I_{Al}/(n_{Al}qA_{Al})}\) Since \(I_{Au}=I_{Al}\), \(A_{Au} = A_{Al}\), and \(n_{Al}=3n_{Au}\), we can simplify this expression:
03

Simplify the ratios

Our expression simplifies as follows: \(\frac{v_{Au}}{v_{Al}} = \frac{I_{Au}/(n_{Au}qA_{Au})}{I_{Al}/(3n_{Au}qA_{Au})}\) Note that \(I_{Au}\), \(q\), and \(A_{Au}\) will cancel out, leaving: \(\frac{v_{Au}}{v_{Al}} = \frac{1}{3}\)
04

Interpret the result

We found that the drift speed of electrons in the gold wire, \(v_{Au}\), is one-third of the drift speed of electrons in the aluminum wire, \(v_{Al}\). Therefore, electrons in the gold wire drift with a slower speed as compared to electrons in the aluminum wire under the same conditions. In conclusion, the drift speeds of electrons in gold and aluminum wires compare as: \(v_{Au} = \frac{1}{3}v_{Al}\).

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Most popular questions from this chapter

A 20 - \(\mu\) F capacitor is discharged through a 5 -k\Omega resistor. The initial charge on the capacitor is \(200 \mu \mathrm{C}\) (a) Sketch a graph of the current through the resistor as a function of time. Label both axes with numbers and units. (b) What is the initial power dissipated in the resistor? (c) What is the total energy dissipated?
A \(12-\Omega\) resistor has a potential difference of \(16 \mathrm{V}\) across it. What current flows through the resistor?
(a) Given two identical, ideal batteries (emf \(=\mathscr{E}\) ) and two identical light-bulbs (resistance \(=R\) assumed constant), design a circuit to make both bulbs glow as brightly as possible. (b) What is the power dissipated by each bulb? (c) Design a circuit to make both bulbs glow, but one more brightly than the other. Identify the brighter bulb.
About \(5.0 \times 10^{4} \mathrm{m}\) above Earth's surface, the atmosphere is sufficiently ionized that it behaves as a conductor. The Earth and the ionosphere form a giant spherical capacitor, with the lower atmosphere acting as a leaky dielectric. (a) Find the capacitance \(C\) of the Earth-ionosphere system by treating it as a parallel plate capacitor. Why is it OK to do that? [Hint: Com- pare Earth's radius to the distance between the "plates." (b) The fair-weather electric field is about \(150 \mathrm{V} / \mathrm{m},\) downward. How much energy is stored in this capacitor? (c) Due to radioactivity and cosmic rays, some air molecules are ionized even in fair weather. The resistivity of air is roughly \(3.0 \times 10^{14} \Omega \cdot \mathrm{m}\) Find the resistance of the lower atmosphere and the total current that flows between Earth's surface and the ionosphere. [Hint: since we treat the system as a parallel plate capacitor, treat the atmosphere as a dielectric of uniform thickness between the plates.] (d) If there were no lightning, the capacitor would discharge. In this model, how much time would elapse before Earth's charge were reduced to \(1 \%\) of its normal value? (Thunderstorms are the sources of emf that maintain the charge on this leaky capacitor.)
During a "brownout," which occurs when the power companies cannot keep up with high demand, the voltage of the household circuits drops below its normal $120 \mathrm{V} .\( (a) If the voltage drops to \)108 \mathrm{V},$ what would be the power consumed by a "100-W" light-bulb (that is, a light-bulb that consumes \(100.0 \mathrm{W}\) when connected to \(120 \mathrm{V}\) )? Ignore (for now) changes in the resistance of the light-bulb filament. (b) More realistically, the light-bulb filament will not be as hot as usual during the brownout. Does this make the power drop more or less than that you calculated in part (a)? Explain.
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