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Chapter 30: Problem 5

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400 turns. When the current in solenoid 1 is 6.52 \(\mathrm{A}\) , the average flux through each turn of solenoid 2 is 0.0320 \(\mathrm{Wb}\) . (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is 2.54 \(\mathrm{A}\) , what is the average flux through each turn of solenoid 1 1 ?

Short Answer

Expert verified
(a) \( M \approx 1.963 \) H; (b) \( \Phi_1 \approx 0.00712 \) Wb.

Step by step solution

01

Understanding Mutual Inductance

Mutual inductance occurs when the magnetic field created by one solenoid affects another. It is defined by the formula \( M = \frac{N_2 \Phi_2}{I_1} \), where \( N_2 \) is the number of turns in solenoid 2, \( \Phi_2 \) is the average magnetic flux through each turn of solenoid 2, and \( I_1 \) is the current in solenoid 1.
02

Calculating Mutual Inductance

Using the values given: \( N_2 = 400 \) turns, \( \Phi_2 = 0.0320 \) Wb, and \( I_1 = 6.52 \) A, plug these into the mutual inductance formula: \[ M = \frac{(400 \times 0.0320)}{6.52} \]. Calculate this to find the mutual inductance, \( M \).
03

Computing the Result for Mutual Inductance

Perform the calculation: \[ M = \frac{12.8}{6.52} \approx 1.963 \ ext{H} \].Thus, the mutual inductance \( M \) is approximately 1.963 Henry.
04

Understanding Average Flux Through Solenoid

To find the average flux through each turn of solenoid 1 when the current is in solenoid 2, use the formula \( \Phi_1 = \frac{M I_2}{N_1} \), where \( I_2 \) is the current in solenoid 2 and \( N_1 \) is the number of turns in solenoid 1.
05

Calculating Average Magnetic Flux for Solenoid 1

Given \( I_2 = 2.54 \) A, \( N_1 = 700 \) turns, and previously calculated \( M = 1.963 \) H, substitute into the formula: \[ \Phi_1 = \frac{1.963 \times 2.54}{700} \].Calculate this to find the average flux \( \Phi_1 \).
06

Computing the Result for Average Flux

Perform the calculation: \[ \Phi_1 = \frac{4.98652}{700} \approx 0.00712 \ ext{Wb} \].Thus, the average flux through each turn of solenoid 1 is approximately 0.00712 Weber.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Toroidal Solenoid
A toroidal solenoid is a coil of wire, wound into a shape resembling a donut or torus. The beauty of a toroidal solenoid is how it confines the magnetic field it generates entirely within its coils. This confinement is crucial for minimizing magnetic interference with surrounding objects.
A toroidal solenoid is effective because it:
  • Allows the magnetic field lines to circulate within its closed loop, thus enhancing the solenoid's efficiency.
  • Is configured so that the magnetic field it produces can be continuous and stable, especially important in electrical circuits.
The uniformity of the field within the toroid is why toroidal solenoids are favored in sensitive applications like transformers and inductors.
Magnetic Flux
Magnetic Flux ( \( \Phi \) ) is a vital concept in understanding electromagnetism. It quantifies the total magnetic field passing through a given area, measuring how much field is "flowing" through a surface.
The amount of magnetic flux depends on:
  • The strength of the magnetic field (denoted by \( B \) - Magnetic Field).
  • The size of the area the field penetrates.
  • The angle between the magnetic field lines and the perpendicular (normal) to the surface.
The formula for magnetic flux is given by \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( A \) is the area and \( \theta \) is the angle. In toroidal solenoids, the flux through the coils is influenced by the geometry of the coil and the current running through it, which helps us calculate inductance.
Inductance Formula
Inductance is a measure of how effectively a solenoid or inductor can "store" magnetic energy. In the context of mutual inductance between two solenoids, we use a specific formula: \( M = \frac{N_2 \Phi_2}{I_1} \).In this relationship:
  • \( M \) is the mutual inductance.
  • \( N_2 \) is the number of turns in the second solenoid.
  • \( \Phi_2 \) is the magnetic flux through solenoid 2.
  • \( I_1 \) is the current flowing in solenoid 1.
Using this formula allows us to calculate how efficiently the magnetic field from one solenoid affects the second solenoid. This is pivotal in designing systems where energy transfer through inductance is required, like in transformers.
Magnetic Field
The magnetic field is a fundamental concept when analyzing solenoids. It is essentially the region around a magnet or current-carrying conductor where magnetic forces can be detected. Within a toroidal solenoid, the magnetic field is very much contained, not only supporting efficient inductance but also preventing external field interference.
Key characteristics of a magnetic field include:
  • Its strength, often represented by \( B \) and measured in Teslas (T).
  • How its direction is depicted with magnetic field lines.
  • Its ability to influence other magnetic materials within its range.
The behavior and impact of the magnetic field are integral to solutions involving mutual inductance, as they directly affect how power and signals are transferred between solenoids.
Turns of Coil
The number of turns in a coil dramatically affects the inductance and magnetic effects within a solenoid system. With more turns, the coil can produce a stronger magnetic field, which in turn enhances the inductance.
The actual number of turns impacts:
  • The potential energy storage, with more turns enabling greater energy within the magnetic field.
  • Magnetic coupling efficiency between two solenoids, which is crucial in systems designed for mutual inductance.
In the case of the original exercise's solenoids, solenoid 1 with 700 turns, and solenoid 2 with 400 turns, are used to quantify the magnetic flux and resulting inductance. More turns in a coil mean a higher ability to affect or be affected by the magnetic field within the system.

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Most popular questions from this chapter

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1\() .\) The inner solenoid is 25.0 \(\mathrm{cm}\) long and has a diameter of 2.00 \(\mathrm{cm} .\) At a certain time, the current in the inner solenoid is 0.120 \(\mathrm{A}\) and is increasing at a rate of \(1.75 \times 10^{3} \mathrm{A} / \mathrm{s}\) . For this time, calculate: (a)the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

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

An inductor is connected to the terminals of a battery that has an emf of 12.0 \(\mathrm{V}\) and negligible internal resistance. The current is 4.86 \(\mathrm{mA}\) at 0.940 \(\mathrm{ms}\) after the connection is completed. After a long time the current is 6.45 \(\mathrm{mA} .\) What are (a) the resistance \(R\) of the inductor and (b) the inductance \(L\) of the inductor?

An Electromagnetic Car Alarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3500 Hz. To do this, the car-alarm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be 12.0 \(\mathrm{V}\) (the same voltage as the car battery. To produce a sufficiently loud sound, the capacitor must store 0.0160 \(\mathrm{J}\) of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

A toroidal solenoid with mean radius \(r\) and cross-sectional area \(A\) is wound uniformly with \(N_{1}\) turns. A second toroidal solenoid with \(N_{2}\) turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. (a) What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. (b) If \(N_{1}=500\) turns, \(N_{2}=300\) turns, \(r=10.0 \mathrm{cm},\) and \(A=0.800 \mathrm{cm}^{2},\) what is the value of the mutual inductance?
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