/*! 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 32 Consider a solenoid of length \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a solenoid of length \(\ell\) and radius \(R,\) containing \(N\) closely spaced turns and carrying a steady current I. (a) In terms of these parameters, find the magnetic field at a point along the axis as a function of distance \(a\) from the end of the solenoid. (b) Show that as \(\ell\) becomes very long, \(B\) approaches \(\mu_{0} N I / 2 \ell\) at each end of the solenoid.

Short Answer

Expert verified
For a point along the axis, a distance a from the end of the solenoid, the magnetic field B can be found using \(B(a) = \mu_0 * N * I / 2 * ℓ * (1 - a/√(R^2 + a^2))\). As the length ℓ becomes very long compared to the solenoid's radius R, the magnetic field B at each end of the solenoid simplifies to \(\mu_0 * N * I / 2 * ℓ\).

Step by step solution

01

Expression for Magnetic Field Inside a Solenoid

The magnetic field produced within a closely wound solenoid carrying a current I, having n turns per unit length is given by \(B = \mu_0 n I\). Knowing that n = N/â„“ (N is the total number of turns and â„“ is the length of the solenoid), the magnetic field inside the solenoid is given by \(B = \mu_0 * N * I / \â„“\).
02

Integrating to Find the Total Magnetic Field

To find the total magnetic field, we integrate over the length of the solenoid. Let's consider a point along the axis that is a distance a from the end. We divide the solenoid into two halves, each of length a/2. For the magnetic field B(a), we will include counts from both halves from the centerline to the points that make an angle θ with the centerline. The contribution dB from a small angle dθ is given by \(dB = \mu_0 * n * I * R^2 * sin(θ) * cos(θ) * dθ / (R^2 + a^2)\).
03

Evaluating the Integration

We'll carry out the integration from θ=0 to θ=π/2 to get the total magnetic field on each side of the solenoid, which means integrating \(∫ dB = ∫_{0}^{π/2} (\mu_0 * n * I * R^2 * sin(θ) * cos(θ) * dθ / (R^2 + a^2)\). The result of this integration gives \(B(a) = \mu_0 * N * I / 2 * ℓ * (1 - a/√(R^2 + a^2))\).
04

Long Solenoid Limit

The second part of the question asks to find the limit when the length of the solenoid is much longer than the radius (i.e. \(ℓ >> R\)). In this case, we have that \(a / √(R^2 + a^2)) ≈ 0\) for \(a << ℓ\). So, from step 3, we have that, when \(ℓ\) becomes very long, \(B\) approaches \(\mu_0 * N * I / 2 * ℓ\) at each end of the solenoid.

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.

Magnetic Field Inside Solenoid
Understanding the magnetic field inside a solenoid is crucial for many applications in electromagnetism. A solenoid is a coil of wire, and when an electric current passes through it, it generates a magnetic field. The field is remarkably uniform in the central region away from the ends, which makes solenoids useful in a variety of applications, such as electromagnets and inductors in electrical circuits.

The magnetic field inside a long solenoid can be derived using the principle that the field inside a closely wound solenoid is proportional to the number of turns per unit length, and the current. The formula for the field is expressed as \(B = \frac{\text{{}}}{} I\) where \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(I\) is the current passing through the solenoid. When the solenoid is very long compared to its radius, the field inside is nearly uniform and is a direct multiplication of these factors.
Ampere's Law
Ampere's Law is a fundamental law of electromagnetism that relates the magnetic field around a closed loop to the electric current passing through the loop. The law is given by the formula \(\oint B \cdot dl = \mu_0 I_{enc}\), where \(B\) is the magnetic field, \(dl\) is an infinitesimal element of the loop, \(I_{enc}\) is the current enclosed by the loop, and \(\mu_0\) is the magnetic constant. In the context of a solenoid, Ampere's Law can be applied to a loop that runs inside and outside the solenoid to determine the magnetic field generated by the current. The law illustrates that the magnetic field inside the solenoid is proportional to the current passing through it and the density of the turns in the coil.

By using Ampere's Law, we can understand the factors contributing to the strength of the magnetic field inside a solenoid and gain insights into how manipulating these factors will affect the field. For example, increasing the current or the number of turns per unit length will increase the magnetic field inside the solenoid.
Solenoid Inductance
The inductance of a solenoid is a measure of its ability to store energy in the magnetic field when a current is passed through it. Inductance is a key parameter in the design of circuits involving solenoids because it affects how the magnetic field develops over time. The formula for the inductance \(L\) of a solenoid can be derived from the basic principles and is given by \(L = \dfrac{N^2 \mu_0 A}{}\), where \(N\) is the number of turns, \(\mu_0\) is the permeability of free space, \(A\) is the cross-sectional area of the solenoid, and \(\ell\) is the length of the solenoid.

It's essential to understand that the inductance of a solenoid is directly proportional to the square of the number of turns, meaning that a solenoid with more turns can store more energy for a given current. As students work through problems related to solenoid inductance, they should pay attention to the connection between physical dimensions, the number of turns, and their impact on inductance, as this relationship has practical implications in the design of circuits and electrical devices.

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

Table \(P 30.70\) contains data taken for a ferromagnetic material. (a) Construct a magnetization curve from the data. Remember that \(\mathbf{B}=\mathbf{B}_{0}+\mu_{0} \mathbf{M} .\) (b) Determine the ratio \(B / B_{0}\) for each pair of values of \(B\) and \(B_{0},\) and construct a graph of \(B / B_{0}\) versus \(B_{0}\). (The fraction \(B / B_{0}\) is called the relative permeability, and it is a measure of the induced magnetic field.) $$\begin{array}{cc}B(T) & B_{0}(T) \\\\\hline 0.2 & 4.8 \times 10^{-5} \\\0.4 & 7.0 \times 10^{-5} \\\0.6 & 8.8 \times 10^{-5} \\\0.8 & 1.2 \times 10^{-4} \\\1.0 & 1.8 \times 10^{-4} \\\1.2 & 3.1 \times 10^{-4} \\\1.4 & 8.7 \times 10^{-4} \\\1.6 & 3.4 \times 10^{-3} \\\1.8 & 1.2 \times 10^{-1} \\\\\hline\end{array}$$

A circular coil of 5 turns and a diameter of \(30.0 \mathrm{cm}\) is oriented in a vertical plane with its axis perpendicular to the horizontal component of the Earth's magnetic field. A horizontal compass placed at the center of the coil is made to deflect \(45.0^{\circ}\) from magnetic north by a current of \(0.600 \mathrm{A}\) in the coil. (a) What is the horizontal component of the Earth's magnetic field? (b) The current in the coil is switched off. A "dip needle" is a magnetic compass mounted so that it can rotate in a vertical north-south plane. At this location a dip needle makes an angle of \(13.0^{\circ}\) from the vertical. What is the total magnitude of the Earth's magnetic field at this location?

Consider a flat circular current loop of radius \(R\) carrying current \(I\). Choose the \(x\) axis to be along the axis of the loop, with the origin at the center of the loop. Plot a graph of the ratio of the magnitude of the magnetic field at coordinate \(x\) to that at the origin, for \(x=0\) to \(x=5 R .\) It may be useful to use a programmable calculator or a computer to solve this problem.

A magnetic field of \(1.30 \mathrm{T}\) is to be set up in an iron-core toroid. The toroid has a mean radius of \(10.0 \mathrm{cm},\) and magnetic permeability of \(5000 \mu_{0}\). What current is required if the winding has 470 turns of wire? The thickness of the iron ring is small compared to \(10 \mathrm{cm},\) so the field in the material is nearly uniform.

A particular paramagnetic substance achieves \(10.0 \%\) of its saturation magnetization when placed in a magnetic field of \(5.00 \mathrm{T}\) at a temperature of \(4.00 \mathrm{K}\). The density of magnetic atoms in the sample is \(8.00 \times 10^{27}\) atoms \(/ \mathrm{m}^{3},\) and the magnetic moment per atom is 5.00 Bohr magnetons. Calculate the Curie constant for this substance.

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.