Textbook Question
An L-R-C series circuit has L = 0.450 H, C = 2.50 × 10-5 F, and resistance R. What is the angular frequency of the circuit when R = 0?
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Ch 30: Inductance
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors37
- Ch 02: Motion Along a Straight Line57
- Ch 03: Motion in Two or Three Dimensions45
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws54
- Ch 06: Work & Kinetic Energy11
- Ch 07: Potential Energy & Conservation6
- Ch 08: Momentum, Impulse, and Collisions15
- Ch 09: Rotation of Rigid Bodies37
- Ch 10: Dynamics of Rotational Motion44
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics27
- Ch 13: Gravitation34
- Ch 14: Periodic Motion26
- Ch 15: Mechanical Waves22
- Ch 16: Sound & Hearing28
- Ch 17: Temperature and Heat28
- Ch 18: Thermal Properties of Matter35
- Ch 19: The First Law of Thermodynamics29
- Ch 20: The Second Law of Thermodynamics18
- Ch 21: Electric Charge and Electric Field28
- Ch 22: Gauss' Law21
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF24
- Ch 26: Direct-Current Circuits29
- Ch 27: Magnetic Field and Magnetic Forces16
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction22
- Ch 30: Inductance14
- Ch 31: Alternating Current20
- Ch 33: The Nature and Propagation of Light17
- Ch 34: Geometric Optics29
- Ch 35: Interference13
- Ch 36: Diffraction19
- Ch 37: Special Relativity19
- Ch 38: Photons: Light Waves Behaving as Particles12
- Ch 39: Particles Behaving as Waves25
- Ch 40: Quantum Mechanics I: Wave Functions20
- Ch 41: Quantum Mechanics II: Atomic Structure21
- Ch 42: Molecules and Condensed Matter17
- Ch 43: Nuclear Physics13
- Ch 44: Particle Physics and Cosmology17
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 30, Problem 17c
A solenoid 25.0 cm long and with a cross-sectional area of 0.500 cm2 contains 400 turns of wire and carries a current of 80.0 A. Calculate: the total energy contained in the coil's magnetic field (assume the field is uniform);
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First, understand that the energy stored in the magnetic field of a solenoid can be calculated using the formula for the energy density of a magnetic field: \( u = \frac{1}{2} \frac{B^2}{\mu_0} \), where \( B \) is the magnetic field and \( \mu_0 \) is the permeability of free space.
Calculate the magnetic field \( B \) inside the solenoid using the formula \( B = \mu_0 n I \), where \( n \) is the number of turns per unit length (\( n = \frac{N}{L} \)), \( N \) is the total number of turns, \( L \) is the length of the solenoid, and \( I \) is the current.
Substitute the values into the formula for \( n \): \( n = \frac{400}{0.25} \) turns/m. Then calculate \( B \) using \( B = \mu_0 n I \).
Once \( B \) is calculated, find the energy density \( u \) using \( u = \frac{1}{2} \frac{B^2}{\mu_0} \).
Finally, calculate the total energy stored in the solenoid by multiplying the energy density \( u \) by the volume of the solenoid: \( E = u \times \text{Volume} \), where the volume is \( A \times L \) and \( A \) is the cross-sectional area.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Magnetic Field in a Solenoid
The magnetic field inside a long solenoid is uniform and can be calculated using the formula B = μ₀nI, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. This concept is crucial for understanding how the solenoid generates a magnetic field.
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Magnetic Field Produced by Loops and Solenoids
Magnetic Energy Density
Magnetic energy density is the energy stored per unit volume in a magnetic field, given by the formula u = B²/(2μ₀), where u is the energy density, B is the magnetic field, and μ₀ is the permeability of free space. This concept helps in determining how much energy is stored in the magnetic field of the solenoid.
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Total Magnetic Energy in a Solenoid
The total energy stored in the magnetic field of a solenoid can be calculated by multiplying the magnetic energy density by the volume of the solenoid. The volume is found by multiplying the cross-sectional area by the length of the solenoid. This concept is essential for calculating the total energy contained in the solenoid's magnetic field.
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Related Practice
Textbook Question
It has been proposed to use large inductors as energy storage devices. How much electrical energy is converted to light and thermal energy by a 150 W light bulb in one day?
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Textbook Question
An air-filled toroidal solenoid has a mean radius of 15.0 cm and a cross-sectional area of 5.00 cm2. When the current is 12.0 A, the energy stored is 0.390 J. How many turns does the winding have?
Textbook Question
A long, straight solenoid has 800 turns. When the current in the solenoid is 2.90 A, the average flux through each turn of the solenoid is 3.25 × 10-3 Wb. What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.20 mV?
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Textbook Question
Inductance of a Solenoid. A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?
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Textbook Question
It has been proposed to use large inductors as energy storage devices. If the amount of energy calculated in part is stored in an inductor in which the current is 80.0 A, what is the inductance?
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