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Chapter 5: Problem 16

Two long coaxial solenoids each carry current \(I\), but in opposite directions, as shown in Fig. \(5.42\). The inner solenoid (radius \(a\) ) has \(n_{1}\) turns per unit length, and the outer one (radius \(b\) ) has \(n_{2}\). Find \(\mathbf{B}\) in cach of the three regions: (i) inside the inner solenoid, (ii) between them, and (iii) outside both.

Short Answer

Expert verified
Inside the inner solenoid: \( \mathbf{B} = \mu_0 n_1 I \); Between the solenoids: \( \mathbf{B} = \mu_0 (n_1 - n_2) I \); Outside both: \( \mathbf{B} \approx 0 \).

Step by step solution

01

Understanding the Problem

We are given two coaxial solenoids carrying currents in opposite directions. We need to find the magnetic field \( \mathbf{B} \) in three different regions: inside the inner solenoid (radius \( a \)), between the solenoids (radius between \( a \) and \( b \)), and outside both solenoids (radius greater than \( b \)).
02

Applying Ampere's Law

For each region, we will use Ampere's Law: \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \), where \( I_{enc} \) is the enclosed current. We assume the solenoids are very long, allowing us to consider the magnetic field as uniform inside the solenoids and approximately zero outside in practice.
03

Finding \( \mathbf{B} \) Inside the Inner Solenoid

Inside the inner solenoid, only its own current contributes to the magnetic field. The magnetic field is uniform and given by \( \mathbf{B} = \mu_0 n_1 I \) inside the solenoid. Here, \( n_1 \) is the number of turns per unit length of the inner solenoid.
04

Finding \( \mathbf{B} \) Between the Solenoids

In the region between the solenoids, the fields from the two solenoids oppose each other. The net magnetic field \( \mathbf{B} \) is given by the difference between the fields of the individual solenoids: \( \mathbf{B} = \mu_0 n_1 I - \mu_0 n_2 I \). This simplifies to \( \mathbf{B} = \mu_0 (n_1 - n_2) I \).
05

Finding \( \mathbf{B} \) Outside Both Solenoids

Outside both solenoids, the current in one direction is cancelled by the current in the opposite direction in the inner solenoid. As long as the solenoids are sufficiently aligned and very long, the magnetic field outside is approximately zero: \( \mathbf{B} \approx 0 \).

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Key Concepts

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

Ampere's Law
Ampere's Law is a fundamental principle in electromagnetism that relates the integrated magnetic field around a closed loop to the electric current passing through the loop. Mathematically, it is expressed as:\[\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}\]Here, \( \mathbf{B} \) is the magnetic field, \( d\mathbf{l} \) is an infinitesimal segment of the loop, \( \mu_0 \) is the permeability of free space, and \( I_{enc} \) is the current enclosed by the loop.
  • The law is particularly useful for calculating magnetic fields in symmetrical situations.
  • It simplifies problems involving long, straight wires, solenoids, and toroids.
  • The direction of the magnetic field is determined by the right-hand rule.
Ampere's Law assumes a steady current and a static field, making it ideal for analyzing the magnetic fields of solenoids as it provides a straightforward way to calculate the field based on enclosed current.
Magnetic Field
A magnetic field is a vector field surrounding magnets and electric currents, representing the magnetic influence exerted by these entities. The strength and direction of the field can be depicted by magnetic field lines, which help visualize how magnetic forces act at different points in space.
  • Magnitude: The strength of the magnetic field, often measured in teslas (T).
  • Direction: Given by the orientation of the magnetic field lines; it shows the path a north pole would take.
  • Sources: Generated by moving electric charges (current) or inherent magnetic materials.
In the context of solenoids, the magnetic field appears inside along the coil due to the flow of current and is typically uniform, assuming the solenoid is long enough. Outside a solenoid, the magnetic fields tend to cancel each other out if the currents in nearby solenoids are opposite, leaving an approximate field of zero.
Solenoids
Solenoids are coil-shaped wires that generate magnetic fields when electric current passes through them. They are often used in devices like electromechanical relays and are characterized by a series of loops creating a uniform magnetic field inside them.
  • Structure: Consists of multiple turns of wire often wrapped cylindrically.
  • Function: Converts electrical energy into magnetic energy, creating a uniform magnetic field inside the coil.
  • Applications: Used in motors, inductors, transformers, and electromagnets.
The number of coils per unit length, denoted by \( n \), directly impacts the strength of the magnetic field inside. This number, coupled with the current passing through the solenoid, determines the magnitude of the field according to the equation \( \mathbf{B} = \mu_0 n I \). The direction of the magnetic field inside a solenoid can be determined using the right-hand grip rule: if you curl your fingers in the direction of current flow, your thumb points in the direction of the magnetic field.
Coaxial Systems
Coaxial systems include two or more tubes or structures sharing a common axis, typically with one nested inside another. In electromagnetism, this configuration is often adapted to optimize the magnetic or electric fields within a given area.
  • Configuration: Two or more solenoids or cylindrical structures, often used to create designed field patterns.
  • Advantages: Allows for multiple layers of winding or shielding which can enhance or manipulate field distributions.
  • Applications: Utilized in RF transmission lines, magnetic field studies, and cable shielding.
In the exercise of coaxial solenoids carrying current in opposite directions, the cancellation effect is significant. The system's configuration ensures that the net magnetic field outside the coaxial structure remains nearly zero, as opposite currents negate each other's effects, maintaining functionality in applications requiring minimal external interference.

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Most popular questions from this chapter

A circularly symmetrical magnetic field (B depends only on the distance from the axis), pointing perpendicular to the page, occupies the shaded region in Fig. 5.58. If the total flux \((f \mathrm{~B} \cdot d a)\) is zero, show that a charged particle that starts out at the center will emerge from the field region on a radial path (provided

Consider a plane loop of wire that carries a steady current \(I ;\) we want to calculate the magnetic field at a point in the plane. We might as well take that point to be the origin (it could be inside or outside the loop). The shape of the wire is given, in polar coordinates, by a specified function \(r(\theta)\) (Fig. \(5.62\) ). (a) Show that the magnitude of the field is \(^{28}\) $$ B=\frac{\mu_{0} I}{4 \pi} \oint \frac{d \theta}{r} . $$ [Hint: Start with the Biot-Savart law; note that \(r=-\mathbf{r}\), and \(d \mathbf{l} \times \hat{\mathbf{r}}\) points perpendicular to the plane; show that \(|d \mathbf{l} \times \hat{\mathbf{r}}|=d l \sin \phi=r d \theta\).] (b) Test this formula by calculating the field at the center of a circular loop. (c) The "lituus spiral" is defined by $$ r(\theta)=\frac{a}{\sqrt{\theta}}, \quad(0<\theta \leq 2 \pi) $$ (for some constant \(a\) ). Sketch this figure, and complete the loop with a straight segment along the \(x\) axis. What is the magnetic field at the origin?

For a configuration of charges and currents confined within a volume \(\mathcal{V}\), show that $$ \int_{\mathcal{V}} \mathbf{J} d \tau=d \mathbf{p} / d t $$ where \(p\) is the total dipole moment. [Hint: evaluate \(\int_{\nu} \nabla \cdot(x \mathbf{J}) d r\).]

(a) A current \(I\) is uniformly distributed over a wire of circular cross section, with radius \(a\) (Fig. 5.15). Find the volume current density \(J\). Solution The area (perpendicular to the flow) is \(\pi a^{2}\), so $$ J=\frac{I}{\pi a^{2}} $$ This was trivial because the current density was uniform. (b) Suppose the current density in the wire is proportional to the distance from the axis, $$ J=k s $$ (for some constant \(k\) ). Find the total current in the wire.

A steady current \(I\) flows down a long cylindrical wire of radius \(a\) (Fig. 5.40). Find the magnetic field, both inside and outside the wire, if (a) The current is uniformly distributed over the outside surface of the wire. (b) The current is distributed in such a way that \(J\) is proportional to \(s\), the distance from the axis.
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