Domanda del libro di testo
The current in a 2.0 mm x 2.0 mm square aluminum wire is 2.5 A. What are (a) the current density and (b) the electron drift speed?
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Ch 27: Current and Resistance
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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TOC.tableOfContents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 27, Problema 19
What electric field strength is needed to create a 5.0 A current in a 2.0-mm-diameter iron wire?
Guida verificata passo dopo passo1
Determine the resistivity of iron (ρ). This is a material property and can be found in a reference table. For iron, ρ ≈ 1.0 × 10⁻⁷ Ω·m.
Calculate the cross-sectional area (A) of the wire. The diameter is given as 2.0 mm, so the radius (r) is 1.0 mm = 1.0 × 10⁻³ m. Use the formula for the area of a circle: A = πr².
Use Ohm's Law in the form E = Jρ, where E is the electric field strength, J is the current density, and ρ is the resistivity. First, calculate the current density (J) using J = I/A, where I is the current (5.0 A) and A is the cross-sectional area calculated in the previous step.
Substitute the values of J and ρ into the formula E = Jρ to calculate the electric field strength. Ensure all units are consistent (e.g., meters, amperes, ohms).
Simplify the expression to find the electric field strength (E). This will give you the magnitude of the electric field required to create the specified current in the wire.
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Concetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Ohm's Law
Ohm's Law states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. It is mathematically expressed as V = I × R. This law is fundamental in understanding how electric fields and currents interact in conductive materials.
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Percorso guidato
03:07
Resistance and Ohm's Law
Electric Field Strength
Electric field strength (E) is defined as the force (F) per unit charge (q) experienced by a positive test charge placed in the field. It is measured in volts per meter (V/m). The relationship between electric field strength and current can be explored through the concept of drift velocity in conductors, where a stronger electric field can lead to a higher current.
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Percorso guidato
03:16Intro to Electric Fields
Resistivity
Resistivity is a material property that quantifies how strongly a given material opposes the flow of electric current. It is denoted by the symbol ρ (rho) and is influenced by factors such as temperature and material composition. The resistivity of iron, for example, will affect the resistance of the wire, which in turn influences the electric field strength required to achieve a specific current.
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Percorso guidato
05:25Resistivity & Resistors in Circuits
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