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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

We can model some of the electrical properties of an unmyelinated axon as an electric cable covered with defective insulation so that current leaks out of the axon to the surrounding fluid. We assume the axon consists of a cylindrical membrane filled with conducting fluid. A current of ions can travel along the axon in this fluid and can also leak out through the membrane. The inner radius of the cylinder is \(5.0 \mu \mathrm{m} ;\) the membrane thickness is \(8.0 \mathrm{nm} .\) (a) If the resistivity of the axon fluid is \(2.0 \Omega \cdot \mathrm{m},\) calculate the resistance of a 1.0 -cm length of axon to current flow along its length. (b) If the resistivity of the porous membrane is \(2.5 \times 10^{7} \Omega\).m, calculate the resistance of the wall of a 1.0 -cm length of axon to current flow across the membrane. (c) Find the length of axon for which the two resistances are equal. This length is a rough measure of the distance a signal can travel without amplification.
Near Earth's surface the air contains both negative and positive ions, due to radioactivity in the soil and cosmic rays from space. As a simplified model, assume there are 600.0 singly charged positive ions per \(\mathrm{cm}^{3}\) and 500.0 singly charged negative ions per \(\mathrm{cm}^{3} ;\) ignore the presence of multiply charged ions. The electric field is $100.0 \mathrm{V} / \mathrm{m},$ directed downward. (a) In which direction do the positive ions move? The negative ions? (b) What is the direction of the current due to these ions? (c) The measured resistivity of the air in the region is $4.0 \times 10^{13} \Omega \cdot \mathrm{m} .$ Calculate the drift speed of the ions, assuming it to be the same for positive and negative ions. [Hint: Consider a vertical tube of air of length \(L\) and cross-sectional area \(A\) How is the potential difference across the tube related to the electric field strength?] (d) If these conditions existed over the entire surface of the Earth, what is the total current due to the movement of ions in the air?
A 1.5 -horsepower motor operates on \(120 \mathrm{V}\). Ignoring \(l^{2} R\) losses, how much current does it draw?
In the physics laboratory, Oscar measured the resistance between his hands to be \(2.0 \mathrm{k} \Omega .\) Being curious by nature, he then took hold of two conducting wires that were connected to the terminals of an emf with a terminal voltage of \(100.0 \mathrm{V} .\) (a) What current passes through Oscar? (b) If one of the conducting wires is grounded and the other has an alternate path to ground through a \(15-\Omega\) resistor (so that Oscar and the resistor are in parallel), how much current would pass through Oscar if the maximum current that can be drawn from the emf is \(1.00 \mathrm{A} ?\)
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