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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

A rotating cylinder \(*\) An infinite cylinder with radius \(R\) and surface charge density \(\sigma\) spins around its symmetry axis with angular frequency \(\omega\). Find the magnetic field inside the cylinder.

A rotating solid cylinder ** (a) A very long cylinder with radius \(R\) and uniform volume charge density \(\rho\) spins with frequency \(\omega\) around its axis. What is the magnetic field at a point on the axis? (b) How would your answer change if all the charge were concentrated on the surface?

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 in the plane of a ring ** A ring with radius \(R\) carries a current \(I\). Show that the magnetic field due to the ring, at a point in the plane of the ring, a distance \(a\) from the center (either inside or outside the ring), is given by $$ B=2 \cdot \frac{\mu_{0} I}{4 \pi} \int_{0}^{\pi} \frac{(R-a \cos \theta) R d \theta}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}} $$ Hint: The easiest way to handle the cross product in the BiotSavart law is to write the Cartesian coordinates of \(d l\) and \(\mathbf{r}\) in terms of an angle \(\theta\) in the ring. This integral can't be evaluated in closed form (except in terms of elliptic functions), but it can always be evaluated numerically if desired. For the special case of \(a=0\) at the center of the ring, the integral is easy to do; verify that it yields the result given in

Hall voltage \(* *\) A Hall probe for measuring magnetic fields is made from arsenic-doped silicon, which has \(2 \cdot 10^{21}\) conduction electrons per \(\mathrm{m}^{3}\) and a resistivity of \(0.016 \mathrm{ohm}-\mathrm{m}\). The Hall voltage is measured across a ribbon of this \(n\)-type silicon that is \(0.2 \mathrm{~cm}\) wide, \(0.005\) \(\mathrm{cm}\) thick, and \(0.5 \mathrm{~cm}\) long between thicker ends at which it is connected into a \(1 \mathrm{~V}\) battery circuit. What voltage will be measured across the \(0.2 \mathrm{~cm}\) dimension of the ribbon when the probe is inserted into a field of 1 kilogauss?
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