Domanda del libro di testo
FIGURE EX30.13 shows a 10-cm-diameter loop in three different magnetic fields. The loop's resistance is 0.20 Ω. For each, what are the size and direction of the induced current?
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Ch 30: Electromagnetic Induction
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 30, Problema 6
An equilateral triangle 8.0 cm on a side is in a 5.0 mT uniform magnetic field. The magnetic flux through the triangle is 6.0 μWb. What is the angle between the magnetic field and an axis perpendicular to the plane of the triangle?
Guida verificata passo dopo passo1
Step 1: Recall the formula for magnetic flux, which is given by Φ = B × A × cos(θ), where Φ is the magnetic flux, B is the magnetic field strength, A is the area of the surface, and θ is the angle between the magnetic field and the normal (perpendicular) to the surface.
Step 2: Calculate the area of the equilateral triangle. The formula for the area of an equilateral triangle is A = (√3 / 4) × s², where s is the length of a side. Substitute s = 8.0 cm (convert to meters: 0.08 m) into the formula to find the area.
Step 3: Rearrange the magnetic flux formula to solve for cos(θ): cos(θ) = Φ / (B × A). Substitute the given values: Φ = 6.0 μWb (convert to webers: 6.0 × 10⁻⁶ Wb), B = 5.0 mT (convert to teslas: 5.0 × 10⁻³ T), and the calculated area A from Step 2.
Step 4: Use the value of cos(θ) obtained in Step 3 to find the angle θ. Take the inverse cosine (arccos) of cos(θ) to determine θ in radians or degrees, depending on the desired unit.
Step 5: Verify the result by checking the units and ensuring the calculated angle is consistent with the physical setup of the problem (e.g., the angle should be between 0° and 90° since the flux is positive).
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Concetti chiave
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Magnetic Flux
Magnetic flux is a measure of the quantity of magnetism, taking into account the strength and the extent of a magnetic field. It is defined as the product of the magnetic field strength (B) and the area (A) through which the field lines pass, adjusted for the angle (θ) between the field and the normal to the surface. The formula is given by Φ = B * A * cos(θ), where Φ is the magnetic flux.
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04:52
Magnetic Flux
Uniform Magnetic Field
A uniform magnetic field is one in which the magnetic field strength is constant in both magnitude and direction throughout a given region. This means that any magnetic field lines are parallel and evenly spaced, indicating that the force experienced by a charged particle moving through the field will be consistent. In this problem, the uniform magnetic field is specified as 5.0 mT.
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05:30Magnetic Fields and Magnetic Dipoles
Angle Between Magnetic Field and Normal
The angle between the magnetic field and the normal to a surface is crucial for calculating magnetic flux. The normal is an imaginary line perpendicular to the surface, and the angle θ affects how much of the magnetic field passes through the area. In the context of the problem, determining this angle is essential to relate the given magnetic flux to the magnetic field and the area of the triangle.
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05:30Magnetic Fields and Magnetic Dipoles
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INT A 10-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.20 Ω. Pulling the wire at a steady speed of 4.0 m/s causes 4.0 W of power to be dissipated in the circuit. How big is the pulling force?
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