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Ch 30: Inductance
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 30: InductanceProblem 13a
Chapter 30, Problem 13a

A toroidal solenoid has mean radius 12.0 cm and crosssectional area 0.600 cm2. How many turns does the solenoid have if its inductance is 0.100 mH?

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Start by recalling the formula for the inductance of a toroidal solenoid: \( L = \frac{\mu_0 N^2 A}{2\pi r} \), where \( L \) is the inductance, \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \text{ Tm/A}) \), \( N \) is the number of turns, \( A \) is the cross-sectional area, and \( r \) is the mean radius.
Convert the given measurements to meters: the mean radius \( r = 12.0 \text{ cm} = 0.12 \text{ m} \) and the cross-sectional area \( A = 0.600 \text{ cm}^2 = 0.0000600 \text{ m}^2 \).
Rearrange the formula to solve for the number of turns \( N \): \( N = \sqrt{\frac{2\pi r L}{\mu_0 A}} \).
Substitute the known values into the rearranged formula: \( L = 0.100 \text{ mH} = 0.0001 \text{ H} \), \( r = 0.12 \text{ m} \), \( A = 0.0000600 \text{ m}^2 \), and \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \).
Calculate the number of turns \( N \) using the substituted values in the formula: \( N = \sqrt{\frac{2\pi \times 0.12 \times 0.0001}{4\pi \times 10^{-7} \times 0.0000600}} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inductance

Inductance is a property of an electrical conductor that quantifies its ability to induce an electromotive force (EMF) when the current flowing through it changes. It is measured in henries (H) and is a crucial factor in the design of coils and solenoids, as it determines how effectively they can store magnetic energy.
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Toroidal Solenoid

A toroidal solenoid is a coil of wire shaped like a doughnut, with the wire wound around a circular core. This configuration confines the magnetic field within the core, minimizing external magnetic interference. The inductance of a toroidal solenoid depends on its number of turns, cross-sectional area, and the core's permeability.
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Magnetic Permeability

Magnetic permeability is a measure of how easily a material can support the formation of a magnetic field within itself. It is a key factor in determining the inductance of a solenoid, as materials with higher permeability allow for stronger magnetic fields, thus increasing the solenoid's inductance. The permeability of free space (vacuum) is a constant used in calculations involving inductance.
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Related Practice
Textbook Question

The inductor shown in Fig. E30.11 has inductance 0.260 H and carries a current in the direction shown. The current is changing at a constant rate. The potential between points a and b is Vab = 1.04 V, with point a at higher potential. Is the current increasing or decreasing?

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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

At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 V. If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

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Textbook Question

When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

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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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