/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A long, current-carrying solenoi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A long, current-carrying solenoid with an air core has 1750 turns per meter of length and a radius of \(0.0180 \mathrm{m}\). A coil of 125 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system?

Short Answer

Expert verified
The mutual inductance is approximately 0.501 mH.

Step by step solution

01

Understanding Mutual Inductance

Mutual inductance is a measure of the inductance between two coils that are wound together. It is defined as the ability of one coil to induce an electromotive force in another coil when the current in the first coil changes. The mutual inductance, \(M\), is calculated as \(M = \frac{N_2 imes ext{Flux through one turn of the secondary}}{I_1}\), where \(N_2\) is the number of turns on the secondary coil and \(I_1\) is the current in the primary coil.
02

Calculate the Magnetic Field Inside the Solenoid

The magnetic field inside a long solenoid is given by the formula \(B = \mu_0 imes n imes I\), where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T}\, \text{m/A}\) is the permeability of free space, \(n\) is the number of turns per unit length and \(I\) is the current.
03

Calculate the Flux Through Coil Turn

Magnetic flux through one turn of the coil is given by \(\Phi = B \times A\), where \(A\) is the cross-sectional area of the coil. The cross-sectional area \(A\) can be calculated using \(A = \pi r^2\), where \(r\) is the radius of the solenoid.
04

Determine the Flux Linkage for One Turn

Since the flux through one turn of the coil is \(\Phi\), the total flux linkage for one turn is simply \(\Phi\) because the flux from the solenoid is uniform across the turns of the coil.
05

Calculate Total Flux Linkage

The total flux linkage for the coil is \(N_2 \times \Phi\), where \(N_2 = 125\) is the number of turns in the coil.
06

Derive the Formula for Mutual Inductance

By replacing \(B\) in the flux expression, the mutual inductance \(M\) can be calculated by combining all expressions: \(M = N_2 \times \frac{(\mu_0 \times n \times I) \times \pi r^2}{I} = \mu_0 \times N_2 \times n \times \pi r^2\).
07

Input the Given Values and Calculate "M"

Substitute \(\mu_0 = 4\pi \times 10^{-7}\), \(N_2 = 125\), \(n = 1750\), and \(r = 0.0180\, \text{m}\) into the mutual inductance formula: \[ M = (4\pi \times 10^{-7}) \times 125 \times 1750 \times \pi \times (0.0180)^2 \] Simplify and solve this expression to find the mutual inductance \(M\).
08

Compute the Numerical Result

Calculate the expression from Step 7 using a calculator: \[ M \approx (1.26 \times 10^{-6}) \times 125 \times 1750 \times \pi \times 3.24 \times 10^{-4} \] \[ M \approx 5.01 \times 10^{-4} \, \text{H} \] Thus, the mutual inductance of the system is approximately \(0.501 \, \text{mH}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Solenoid
A solenoid is essentially a coil of wire, usually in a cylindrical shape, that generates a magnetic field when an electric current passes through it. Solenoids are widely used in various applications, such as electromagnets, inductors, and valves. Their design allows them to produce uniform magnetic fields, which are useful in controlling electromotive force and mutual inductance.

The strength of the magnetic field inside a solenoid depends on several factors:
  • The number of loops or turns of the coil, expressed as turns per unit length.
  • The current running through the coil.
  • The core material inside the solenoid, which is often air in basic setups.
The central position of a solenoid exhibits a highly constant magnetic field compared to the ends, making it a valuable component in experiments and devices requiring precise control over electromagnetic fields.
Magnetic Flux
Magnetic flux is a measure of the quantity of magnetism, represented by the total number of magnetic field lines passing through an area. It forms a core part of electromagnetic theory, often associated with the operation of solenoids in generating or inducing electromotive force (emf).

To understand magnetic flux, consider two key aspects:
  • The magnetic field (
  • The area the field lines pass through, usually the cross-section of the solenoid in electromagnetic contexts.
Formulaically, magnetic flux (\(\Phi\)) is calculated as:\[\Phi = B \times A\]where \(\Phi\) is the flux, \(B\) is the magnetic field intensity, and \(A\) is the area. A changing magnetic flux is what drives induction, bringing us to the concept of mutual induction, which applies when a solenoid and a wrapped coil interact. The complete integration of flux interactions is critical in determining the behavior and efficiency of electrical circuits.
Permeability of Free Space
The permeability of free space, often denoted by the symbol \(\mu_0\), is a fundamental physical constant. This constant describes how a magnetic field interacts with an unbounded medium, like air or vacuum. The standard value for the permeability of free space is \(4\pi \times 10^{-7} \, \text{T m/A}\). It plays a critical role in the calculations involving magnetic fields and inductance.

In the context of a solenoid, \(\mu_0\) is used to determine the magnetic field inside the solenoid, which directly relates to how much flux it can produce:\[B = \mu_0 \times n \times I\]where \(B\) is the magnetic field, \(n\) is the number of turns per unit length of the solenoid, and \(I\) is the current. This constant establishes the whole backbone for the derivation of formulas related to mutual inductance, connecting microscopic magnetic field interactions with macroscopic electromagnetic behavior.
Electromotive Force
Electromotive force (emf) is the voltage developed by any source of electrical energy such as a battery or generator. In the context of mutual inductance observed with solenoids and coils, emf is induced when there's a change in magnetic flux through the coil.

The key principles underlying this concept include:
  • Faraday's Law of Induction - This law states that the magnitude of the electromotive force induced in a circuit is proportional to the rate of change of magnetic flux through the circuit.
  • Lenz's Law - This principle indicates that the direction of the induced emf is such that it opposes the change in flux that produces it.
This induced emf is central to understanding mutual inductance, where one coil can influence another through varying currents and magnetic fields. Induced currents and emf due to changing magnetic fields make electrical transformers and electromagnetic devices possible, showcasing the practical significance of these principles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A conducting coil of 1850 turns is connected to a galvanometer, and the total resistance of the circuit is \(45.0 \Omega .\) The area of each turn is \(4.70 \times 10^{-4} \mathrm{m}^{2} .\) This coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. The amount of charge that is induced to flow around the circuit is measured to be \(8.87 \times 10^{-3} \mathrm{C} .\) Find the magnitude of the magnetic field.

A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius \(r=0.20 \mathrm{m} .\) The normal to the plane of the loop is parallel to a constant magnetic field \(\left(\phi=0^{\circ}\right)\) of magnitude 0.75 T. What is the change \(\Delta \Phi\) in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?

Parts \(a\) and \(b\) of the drawing show the same uniform and constant (in time) magnetic field \(\overrightarrow{\mathbf{B}}\) directed perpendicularly into the paper over a rectangular region. Outside this region, there is no field. Also shown is a rectangular coil (one turn), which lies in the plane of the paper. In part \(a\) the long side of the coil (length \(=L\) ) is just at the edge of the field region, while in part \(b\) the short side (width \(=W\) ) is just at the edge. It is known that \(L / W=3.0 .\) In both parts of the drawing the coil is pushed into the field with the same velocity \(\overrightarrow{\mathbf{v}}\) until it is completely within the field region. The magnitude of the average emf induced in the coil in part \(a\) is 0.15 V. What is its magnitude in part \(b ?\)

Suppose you wish to make a solenoid whose self-inductance is 1.4 \(\mathrm{mH}\). The inductor is to have a cross-sectional area of \(1.2 \times 10^{-3} \mathrm{m}^{2}\) and a length of \(0.052 \mathrm{m}\). How many turns of wire are needed?

A magnetic field has a magnitude of \(0.078 \mathrm{T}\) and is uniform over a circular surface whose radius is \(0.10 \mathrm{m}\). The field is oriented at an angle of \(\phi=25^{\circ}\) with respect to the normal to the surface. What is the magnetic flux through the surface?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.