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Chapter 6: Problem 19

Solenoid field, inside and outside *** Consider an infinite solenoid with circular cross section. The current is \(I\), and there are \(n\) turns per unit length. Show that the magnetic field is zero outside and \(B=\mu_{0} n I\) (in the longitudinal direction) everywhere inside. Do this in three steps as follows. (a) Show that the field has only a longitudinal component. Hint: Consider the contributions to the field from rings that are symmetrically located with respect to a given point. (b) Use Amp猫re's law to show that the field has a uniform value outside and a uniform value inside, and that these two values differ by \(\mu_{0} n I\). (c) Show that \(B \rightarrow 0\) as \(r \rightarrow \infty .\) There are various ways to do this. One is to obtain an upper bound on the field contribution due to a given ring by unwrapping the ring into a straight wire segment, and then finding the field due to this straight segment.

Short Answer

Expert verified
The magnetic field outside the solenoid is zero and the magnetic field inside the solenoid is \(B=\mu_{0} n I\) in the longitudinal direction.

Step by step solution

01

Apply the Symmetry Principle

Since the solenoid is infinite and has circular symmetry, there is no reason for the magnetic field direction to steer towards any particular radial direction. Therefore, only the longitudinal component can exist.
02

Apply Amp猫re's law

Amp猫re's law states that the line integral of the magnetic field around a closed loop is equal to \(\mu_0\) times the total current enclosed by the loop. Consider selecting Amp猫re's loop as a rectangle, one side inside the solenoid parallel to the axis, and the other side outside. Because there exists no magnetic field outside the solenoid and due to its consistency inside the solenoid, the integral of \(B\cdot dl\) over the loop equals to \(B l\). Moreover, the total current enclosed by the loop is \(n\cdot I\cdot l\), where \(l\) is the length of the loop along the solenoid. Equating the results from Amp猫re's law: \(\mu_0 n I l = B l\). After simplifying, one obtains \(B = \mu_0 n I\) inside the solenoid.
03

Look at the field at infinity

Consider a ring of the solenoid at a distance \(r\) from the observation point. If you unwrap the ring into a straight wire segment, its field contribution at the point will not exceed the field that would have been produced by the same current in a straight wire of length \(2\pi r\). This field at a distance \(r\) would be \(B_{wire} = \frac{\mu_0 I}{2\pi r}\), and considering the field contribution from all such rings: \(B = nB_{wire} = \frac{\mu_0 In}{2\pi r}\). This tends to zero as \(r\) approaches infinity. Therefore the field outside the solenoid must be zero.

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Key Concepts

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

Amp猫re's Law
Understanding Amp猫re's Law is pivotal when studying the relationship between electric currents and magnetic fields. Amp猫re's Law, a fundamental principle in electromagnetism, is expressed mathematically as
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \]
This law states that the integral of the magnetic field \( \vec{B} \) along a closed loop is proportional to the electric current \( I_{\text{enc}} \) that passes through the loop. The constant \( \mu_0 \) is the permeability of free space and is a measure of the medium's ability to support the formation of a magnetic field.

In simple terms, Amp猫re's Law links the magnetic field traversing around a loop to the current flowing within it, serving as a bridge between magnetism and electricity. This is what allows us to analyze magnetic fields due to current-carrying conductors, like in the case of a solenoid.
Magnetic Field
A magnetic field is a vector field that surrounds magnets and electric currents, and exerts a force on other magnets and currents in the field. This field can be visualized using magnetic field lines, where the direction of the field at any point is tangent to the field line, and the density of the field lines corresponds to the strength of the magnetic field.

The magnetic field is typically denoted by the symbol \( \vec{B} \) and is measured in teslas (T) in the International System of Units. It's crucial for students to understand that the magnetic field is a result of moving electric charges and it changes with the velocity and direction of the electric current.
Electric Current
Electric current is the flow of electric charge in a specific direction. It's a key concept in electricity and is directly related to the generation of a magnetic field. It is quantified as the rate at which charge passes through a surface, measured in amperes (A). When current flows through wires or conductors, it creates a magnetic field around them, the nature of which is determined by the right-hand rule. The strength and direction of this magnetic field can be influenced by factors such as the shape of the conductor and the current magnitude.

It's important to grasp that not only does electric current produce a magnetic field, but the field itself interacts with other currents and magnetic moments, which can lead to complex behaviors especially in devices like solenoids or inductors.
Magnetic Field due to a Solenoid
A solenoid is a coil of wire designed to create a nearly uniform magnetic field inside when electric current flows through it. The magnetic field due to a solenoid with \( n \) turns per unit length and carrying a current \( I \) is given by the formula:
\[ B = \mu_0 n I \]
This applies within the interior of the solenoid, assuming infinitely long solenoid and no fringing effects at the ends. Outside of the solenoid, however, the idealization of having zero field holds true only for an infinite solenoid; in practical circumstances, there will be some leakage of the field.

When discussing a solenoid's magnetic field, it's useful to consider its physical characteristics: the number of turns (which intensifies the field) and the current (directly proportional to the field strength). Understanding this topic is vital in electromagnetism, as solenoids are fundamental components in many electrical devices, such as relays, electromagnets, and inductors.

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Most popular questions from this chapter

The retarded potential \(* * *\) A point charge \(q\) moves with speed \(v\) along the line \(y=r\) in the \(x y\) plane. We want to find the magnetic field at the origin at the moment the charge crosses the \(y\) axis. (a) Starting with the electric field in the charge's frame, use the Lorentz transformation to show that, in the lab frame, the magnitude of the magnetic field at the origin (at the moment the charge crosses the \(y\) axis) equals \(B=\left(\mu_{0} / 4 \pi\right)\left(\gamma q v / r^{2}\right)\) (b) Use the Biot-Savart law to calculate the magnetic field at the origin. For the purposes of obtaining the current, you may assume that the "point" charge takes the shape of a very short stick. You should obtain an incorrect answer, lacking the \(\gamma\) factor in the above correct answer. (c) The Biot-Savart method is invalid because the Biot-Savart law holds for steady currents (or slowly changing ones, but see Footnote 8 ). But the current due to the point charge is certainly not steady. At a given location along the line of the charge's motion, the current is zero, then nonzero, then zero again. For non-steady currents, the validity of the Biot-Savart law can be restored if we use the so-called "retarded time." \(^{\prime 11}\) The basic idea with the retarded time is that, since information can travel no faster than the speed of light, the magnetic field at the origin, at the moment the charge crosses the \(y\) axis, must be related to what the charge was doing at an earlier time. More precisely, this earlier time (the "retarded time") is the time such that if a light signal were emitted from the charge at this time, then it would reach the origin at the same instant the charge crosses the \(y\) axis. Said in another way, if someone standing at the origin takes a photograph of the surroundings at the moment the charge crosses the \(y\) axis, then the position of the charge in the photograph (which will not be on the \(y\) axis) is the charge's location we are concerned with. \(^{12}\)

Field at different radii \(*\) A current of 8000 amperes flows through an aluminum rod \(4 \mathrm{~cm}\) in diameter. Assuming the current density is uniform through the cross section, find the strength of the magnetic field at \(1 \mathrm{~cm}\), at \(2 \mathrm{~cm}\), and at \(3 \mathrm{~cm}\) from the axis of the rod.

Integral of \(A\), flux of \(B\) Show that the line integral of the vector potential \(\mathbf{A}\) around a closed curve \(C\) equals the magnetic flux \(\Phi\) through a surface \(S\) bounded by the curve. This result is very similar to Amp猫re's law, which says that the line integral of the magnetic field \(\mathbf{B}\) around a closed curve \(C\) equals (up to a factor of \(\mu_{0}\) ) the current flux \(I\) through a surface \(S\) bounded by the curve.

Field at the center of an orbit \(*\) An electron is moving at a speed \(0.01 c\) on a circular orbit of radius \(10^{-10} \mathrm{~m}\). What is the strength of the resulting magnetic field at the center of the orbit? (The numbers given are typical, in order of magnitude, for an electron in an atom.)

Far field from a square loop ** Consider a square loop with current \(I\) and side length \(a\). The goal of this problem is to determine the magnetic field at a point a large, distance \(r\) (with \(r \gg a\) ) from the loop. (a) At the distant point \(P\) in Fig. 6.36, the two vertical sides give essentially zero Biot-Savart contributions to the field, because they are essentially parallel to the radius vector to \(P\). What are the Biot-Savart contributions from the two horizontal sides? These are easy to calculate because every little interval in these sides is essentially perpendicular to the radius vector to \(P\). Show that the sum (or difference) of these contributions equals \(\mu_{0} I a^{2} / 2 \pi r^{3}\), to leading order in \(a\). (b) This result of \(\mu_{0} I a^{2} / 2 \pi r^{3}\) is not the correct field from the loop at point \(P\). The correct field is half of this, or \(\mu_{0} I a^{2} / 4 \pi r^{3} .\) We will eventually derive this in Chapter 11, where we will show that the general result is \(\mu_{0} I A / 4 \pi r^{3}\), where \(A\) is the area of a loop with arbitrary shape. But we should be able to calculate it via the Biot-Savart law. Where is the error in the reasoning in part (a), and how do you go about fixing it? This is a nice one - don't peek at the answer too soon!
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