Textbook Question
A 100 A current circulates around a 2.0-mm-diameter superconducting ring. What is the ring's magnetic dipole moment?
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Ch 29: The Magnetic Field
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- 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: 9780137344796Not the one you use?Change textbook
Chapter 29, Problem 16b
A 100 A current circulates around a 2.0-mm-diameter superconducting ring. What is the on-axis magnetic field strength 5.0 cm from the ring?
Verified step by step guidance1
Determine the radius of the superconducting ring. Since the diameter is given as 2.0 mm, the radius \( r \) is half of the diameter: \( r = \frac{2.0 \text{ mm}}{2} = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m} \).
Use the Biot-Savart law to calculate the magnetic field on the axis of a current loop. The formula for the magnetic field at a point along the axis of a circular current loop is: \( B = \frac{\mu_0 I r^2}{2 (r^2 + z^2)^{3/2}} \), where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \ \text{T·m/A} \)), \( I \) is the current, \( r \) is the radius of the loop, and \( z \) is the distance from the center of the loop along the axis.
Substitute the known values into the formula: \( I = 100 \ \text{A} \), \( r = 1.0 \times 10^{-3} \ \text{m} \), and \( z = 5.0 \ \text{cm} = 0.05 \ \text{m} \).
Simplify the expression for \( B \) by calculating the terms \( r^2 \), \( z^2 \), and \( (r^2 + z^2)^{3/2} \). Then multiply by the constants \( \mu_0 \), \( I \), and \( r^2 \) as per the formula.
After simplifying, the result will give the magnetic field strength \( B \) at a distance of 5.0 cm from the ring along its axis.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Superconductivity
Superconductivity is a phenomenon where certain materials exhibit zero electrical resistance and expel magnetic fields when cooled below a critical temperature. In superconductors, electric current can flow indefinitely without energy loss, making them ideal for applications like magnetic levitation and powerful electromagnets.
Magnetic Field of a Current Loop
The magnetic field generated by a current-carrying loop can be calculated using the Biot-Savart law or Ampère's law. For a circular loop, the magnetic field strength at a point along the axis can be determined by considering the contributions from each segment of the loop, which combine to create a net magnetic field.
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Biot-Savart Law
The Biot-Savart law describes how electric currents produce magnetic fields. It states that the magnetic field dB at a point in space due to a small segment of current-carrying wire is directly proportional to the current and inversely proportional to the square of the distance from the wire segment to the point, factoring in the angle between the current direction and the line connecting the wire to the point.
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Related Practice
Textbook Question
What is the line integral of between points i and f in FIGURE EX29.19?
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Textbook Question
The on-axis magnetic field strength 10 cm from a small bar magnet is 500 μT. What is the on-axis field strength 15 cm from the magnet?
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Textbook Question
The value of the line integral of around the closed path in FIGURE EX29.21 is 1.38 x 10-5 T m. What are the direction (into or out of the figure) and magnitude of I3?
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Textbook Question
What are the magnetic fields at points a to c in FIGURE EX29.13? Give your answers as vectors.
Textbook Question
The element niobium, which is a metal, is a superconductor (i.e., no electrical resistance) at temperatures below 9 K. However, the superconductivity is destroyed if the magnetic field at the surface of the metal reaches or exceeds 0.10 T. What is the maximum current in a straight, 3.0-mm-diameter superconducting niobium wire?
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