교과서 질문
A long, straight conducting wire of radius R has a nonuniform current density J = J₀r/R, where J₀ is a constant. The wire carries total current I. Find an expression for the magnetic field strength inside the wire at radius r.
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Ch 29: The Magnetic Field
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
29장, 문제 83b
An infinitely wide flat sheet of charge flows out of the figure in FIGURE CP29.83. The current per unit width along the sheet (amps per meter) is given by the linear current density Js. Find the magnetic field strength at distance d above or below the current sheet.
검증된 단계별 안내1
Step 1: Recognize that the problem involves calculating the magnetic field due to an infinitely wide current sheet. The current sheet has a linear current density Js (current per unit width), and we need to find the magnetic field at a distance d above or below the sheet.
Step 2: Use Ampere's Law, which states that the line integral of the magnetic field around a closed loop is equal to μ₀ times the current enclosed by the loop. The formula is: ∮B⋅dl = μ₀I_enclosed.
Step 3: Choose an appropriate Amperian loop. For this problem, a rectangular loop is ideal, with one side parallel to the current sheet above it at distance d, and the other side parallel to the sheet below it at distance d. The other two sides of the loop are perpendicular to the sheet.
Step 4: Analyze the symmetry of the problem. The magnetic field will be uniform and parallel to the current sheet at distances d above and below the sheet. The contributions to the integral from the sides perpendicular to the sheet cancel out due to symmetry.
Step 5: Solve for the magnetic field strength. The total current enclosed by the Amperian loop is Js (linear current density) multiplied by the width of the loop. Using Ampere's Law, the magnetic field strength B at distance d is given by: B = (μ₀Js)/2, where μ₀ is the permeability of free space. The factor of 1/2 arises because the field splits equally above and below the sheet.
비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Linear Current Density (Js)
Linear current density, denoted as Js, represents the amount of electric current flowing per unit width of a current-carrying conductor. It is measured in amperes per meter (A/m) and is crucial for understanding how current is distributed across a surface. In the context of a current sheet, Js helps determine the magnetic field generated by the current flowing through the sheet.
추천 영상:
가이드 코스
8:13
Intro to Density
Magnetic Field Due to Current
The magnetic field generated by a current-carrying conductor can be described by Ampère's Law, which relates the magnetic field around a conductor to the current flowing through it. For an infinitely wide current sheet, the magnetic field strength is uniform and perpendicular to the direction of the current. This concept is essential for calculating the magnetic field strength at a distance from the current sheet.
추천 영상:
가이드 코스
12:33Magnetic Field Produced by Straight Currents
Biot-Savart Law
The Biot-Savart Law provides a method to calculate the magnetic field produced by a small segment of current-carrying wire. It states that the magnetic field at a point in space is proportional to the current, the length of the wire segment, and the sine of the angle between the wire and the line connecting the wire to the point. This law is foundational for understanding how current distributions, like those in a current sheet, generate magnetic fields.
추천 영상:
가이드 코스
04:53Biot-Savart Law with Calculus
관련 실천
교과서 질문
FIGURE CP29.79 is an edge view of a 2.0 kg square loop, 2.5 m on each side, with its lower edge resting on a frictionless, horizontal surface. A 25 A current is flowing around the loop in the direction shown. What is the strength of a uniform, horizontal magnetic field for which the loop is in static equilibrium at the angle shown?
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교과서 질문
Determine the field strength at the center of a current-carrying square loop having sides of length 2R.
251
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