Ch 29: Electromagnetic Induction

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 29: Electromagnetic Induction

Problem 37

Chapter 29, Problem 37

A long, straight solenoid with a cross-sectional area of 8.00 cm² is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?

Verified step by step guidance

First, understand that the solenoid creates a magnetic field when current flows through it. The magnetic field inside a long solenoid is given by the formula:

B = μ n I

, where

μ

is the permeability of free space,

n

is the number of turns per unit length, and

I

is the current.

Calculate the initial magnetic field inside the solenoid using the given values:

n = 90 / 1 cm

(convert to meters),

I = 0.350 A

, and

μ = 4 π \times 10 ⁻ 7 T m / A

Determine the change in magnetic flux through the second winding. The magnetic flux

Φ

is given by

Φ = B A

, where

A

is the cross-sectional area of the solenoid. Initially,

Φ

is non-zero, and finally, it becomes zero when the current is turned off.

Use Faraday's law of electromagnetic induction to find the average induced emf in the second winding. Faraday's law states that the induced emf

ε

is equal to the negative rate of change of magnetic flux:

ε = - \frac{Δ Φ}{Δ t}

. Calculate

Δ Φ

and divide by the time interval

Δ t = 0.0400 s

Finally, multiply the induced emf by the number of turns in the second winding to find the total average induced emf. The second winding has

12

turns, so the total induced emf is

ε \times 12

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Key Concepts

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

Faraday's Law of Electromagnetic Induction

Faraday's Law states that a change in magnetic flux through a circuit induces an electromotive force (emf) in the circuit. The induced emf is proportional to the rate of change of the magnetic flux. In this problem, the solenoid's magnetic field changes, inducing an emf in the second winding.

Magnetic Flux

Magnetic flux is the product of the magnetic field and the area it penetrates, perpendicular to the field lines. It is a measure of the quantity of magnetism, considering the strength and extent of a magnetic field. The change in magnetic flux through the second winding is crucial for calculating the induced emf.

Solenoid Magnetic Field

A solenoid generates a uniform magnetic field inside its coils when a current passes through it. The field strength is determined by the number of turns per unit length and the current. Understanding how the solenoid's field changes when the current is turned off helps in determining the change in magnetic flux and the resulting induced emf.

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Related Practice

Textbook Question

A metal ring 4.50 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 T/s. What is the magnitude of the electric field induced in the ring?

Textbook Question

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.520 A. What is the rate at which the electric field between the plates is changing?

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Textbook Question

A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the center of the solenoid and 3.50 cm from its axis is 8.00 × 10^-6 V/m. Calculate di/dt.

Textbook Question

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.520 A. (a) What is the displacement current density j_D in the air space between the plates?

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Textbook Question

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.520 A. (c) What is the induced magnetic field between the plates at a distance of 2.00 cm from the axis? (d) At 1.00 cm from the axis?

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Textbook Question

A long, thin solenoid has 900 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s. What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?

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