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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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