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A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

Short Answer

Expert verified
(a) \(3.125 \times 10^{19}\) electrons/s, (b) Current density remains the same, (c) Drift speed decreases, (d) Current density and drift speed decrease with larger diameter.

Step by step solution

01

Calculate the charge of electrons passing through the bulb

The current is given as 5.00 A. Current \( I \) is the rate of charge flow, \( I = \frac{Q}{t} \), where \( Q \) is the total charge and \( t \) is the time. Each electron carries a charge of \( 1.6 \times 10^{-19} \) coulombs. Thus, the number of electrons \( n \) passing through in one second is given by \( n = \frac{I}{e} = \frac{5.00}{1.6 \times 10^{-19}} \approx 3.125 \times 10^{19} \) electrons.
02

Calculate the current density in the wire

Current density \( J \) is defined as the current \( I \) per unit cross-sectional area \( A \). The cross-sectional area of the wire, using the diameter \( d = 2.05 \ mm \) or \( 2.05 \times 10^{-3} \ m \), is \( A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{2.05 \times 10^{-3}}{2}\right)^2 \). The current density \( J = \frac{I}{A} \).
03

Compute the speed of electrons

The drift velocity \( v_d \) can be found using \( I = nAqv_d \), where \( n \) is the number density of electrons, \( A \) is the cross-sectional area, \( q \) is the charge of an electron, and \( v_d \) is the drift velocity. Rearranging gives \( v_d = \frac{I}{nAq} \), where \( n = 8.5 \times 10^{28} \) m⁻³. Substitute the values to find \( v_d \).
04

Analyze effect of resizing the wire

If the wire's diameter is doubled, the cross-sectional area increases by a factor of four since \( A \propto d^2 \). This means the current density would decrease by a factor of four. The number of electrons per second (part a) will stay the same because the current is unchanged, but the drift speed will decrease by a factor of four.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drift Velocity
Drift velocity is the average velocity that a charge carrier, such as an electron, attains due to an electric field. In a copper wire carrying current, these charge carriers are predominantly free electrons. The flow of electricity in the wire can be thought of as these electrons drifting through the metallic structure of the wire. Although individual electrons move at tremendous speeds randomly, their average net motion is slow and their drift velocity can be calculated using the formula:\[v_d = \frac{I}{nAq}\]where:
  • \(I\) is the current through the wire, which is given as 5.00 A.
  • \(n\) represents the number density of electrons, for copper it is given as \(8.5 \times 10^{28}\) electrons per cubic meter.
  • \(A\) is the cross-sectional area of the wire.
  • \(q\) is the charge of an electron, which is \(1.6 \times 10^{-19}\) coulombs.
Substituting these values calculates how fast electrons drift through the wire, highlighting that the motion responsible for electric current is far slower than one might assume.
Copper Wire
Copper wire is a common conductor in electrical circuits, revered for its excellent conductivity. Conductors allow electrons to move freely through their structure, making them ideal for carrying electric current. Copper’s ability to conduct electricity is tied to its atomic structure, where free electrons can move without significant resistance.For this exercise, the copper wire is specified as a 12-gauge wire with a diameter of 2.05 mm. The wire's cross-sectional area, crucial for calculating current density and drift velocity, can be found using \[A = \pi \left(\frac{d}{2}\right)^2\]where \(d\) is the diameter. This structural property, an integral aspect of its design, ensures the wire can efficiently carry the defined current. If the diameter doubles, the area increases by a factor of four, affecting how current flows through the wire.
Electron Flow
Electron flow refers to the movement of electrons through a conductor, like a copper wire, under the influence of an electric field. Each electron acts as a tiny charge carrier, contributing to the overall electric current. In this setting, electron flow is continuous and quantified as how many electrons pass a point per second.The formula to calculate the number of electrons passing through the light bulb every second is:\[n = \frac{I}{e}\]where:
  • \(I\) represents the electric current (5.00 A),
  • \(e\) is the elementary charge of an electron \(1.6 \times 10^{-19}\) C.
Substituting these values gives the electron flow for any cross-section of the wire, immobilizing the abstract concept of current into an understandable count of electrons.
Current Calculation
Current calculation in a wire involves understanding how current, charge, and time are interconnected. Current, denoted as \(I\), measures how much charge passes a given point in the wire per second.Rowing through the relationship \(I = \frac{Q}{t}\), where \(Q\) is total charge and \(t\) is time, we can measure the flow of electrons in a tangible manner. In the context of this exercise, understanding that current is essentially a count of electrons passing a wire's cross section per second, reifies its definition.The concept of current density, defined as the current per unit area \(J = \frac{I}{A}\), builds on this foundation. For a given wire's cross-sectional area calculated using its diameter, this value indicates how densely packed the current flow is within the cable's structure, informing both theoretical calculations and practical applications, especially in circuit design.

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