If a ferromagnetic substance is subjected to an external magnetic field, the external field exerts a torque on the magnetic dipole of each domain. The torque causes many of the domains to rotate and become aligned with the external field, just as a compass needle aligns with a magnetic field, although internal forces between the domains generally prevent the alignment from being perfect. In addition, atomic-level forces between the spins can cause the domain boundaries to move. Domains that are aligned along the external field become larger at the expense of domains that are opposed to the field. These changes in the size and orientation of the domains cause the material to develop a net magnetic dipole that is aligned with the external field. This magnetic dipole has been induced by the external field, so it is called an induced magnetic dipole.

FIGURE 29.55 shows a ferromagnetic material near the end of a solenoid. The magnetic dipole moments of the domains align with the solenoid’s field, creating an induced magnetic dipole whose south pole faces the solenoid’s north pole. Consequently, the magnetic force between the poles pulls the ferromagnetic object to the electromagnet.

FIGURE 29.55 The magnetic field of the solenoid creates an induced magnetic dipole moment in the iron.

The fact that a magnet attracts and picks up ferromagnetic objects was one of the basic observations about magnetism with which we started the chapter. Now we have an explanation of how it works, based on three ideas:

1. Electrons are microscopic magnets due to their spin.

2. A ferromagnetic material in which the spins are aligned is organized into magnetic domains.

3. The individual domains align with an external magnetic field to produce an induced magnetic dipole moment for the entire object.

The object’s magnetic dipole may not return to zero when the external field is removed because some domains remain “frozen” in the alignment they had in the external field. Thus a ferromagnetic object that has been in an external field may be left with a net magnetic dipole moment after the field is removed. In other words, the object has become a permanent magnet. A permanent magnet is simply a ferromagnetic material in which a majority of the magnetic domains are aligned with each other to produce a net magnetic dipole moment.

Whether or not a ferromagnetic material can be made into a permanent magnet depends on the internal crystalline structure of the material. Steel is an alloy of iron with other elements. An alloy of mostly iron with the right percentages of chromium and nickel produces stainless steel, which has virtually no magnetic properties at all because its particular crystalline structure is not conducive to the formation of domains. A very different steel alloy called Alnico V is made with 51% iron, 24% cobalt, 14% nickel, 8% aluminum, and 3% copper. It has extremely prominent magnetic properties and is used to make high-quality permanent magnets.

So we’ve come full circle. One of our initial observations about magnetism was that a permanent magnet can exert forces on some materials but not others. The theory of magnetism that we then proceeded to develop was about the interactions between moving charges. What moving charges had to do with permanent magnets was not obvious. But finally, by considering magnetic effects at the atomic level, we found that properties of permanent magnets and magnetic materials can be traced to the interactions of vast numbers of electron spins.

CHAPTER 29 CHALLENGE EXAMPLE Designing a loudspeaker

A loudspeaker consists of a paper cone wrapped at the bottom with several turns of fine wire. As FIGURE 29.56 shows, this coil sits in a narrow gap between the poles of a circular magnet. To produce sound, the amplifier drives a current through the coil. The magnetic field then exerts a force on this current, pushing the cone and thus pushing the air to create a sound wave. An ideal speaker would experience only forces from the magnetic field, thus responding only to the current from the amplifier. Real speakers are balanced so as to come close to this ideal unless driven very hard.

FIGURE 29.56 The coil and magnet of a loudspeaker.

Full Alternative Text

Consider a 5.5 g loudspeaker cone with a 5.0-cm-diameter, 20-turn coil having a resistance of $8.0\text{\hspace{0.17em}Ω}$. There is a 0.18 T field in the gap between the poles. These values are typical of the loudspeakers found in car stereo systems. What is the oscillation amplitude of this speaker if driven by a 100 Hz oscillatory voltage from the amplifier with a peak value of 12 V?

MODEL Model the loudspeaker as ideal, responding only to magnetic forces. These forces cause the cone to accelerate. We’ll use kinematics to relate the acceleration to the displacement.

VISUALIZE FIGURE 29.57 shows the coil in the gap between the magnet poles. Magnetic fields go from north to south poles, so the field is radially outward. Consequently, the field at all points is perpendicular to the circular current. According to the right-hand rule, the magnetic force on the current is into or out of the figure, depending on whether the current is counterclockwise or clockwise, respectively.

FIGURE 29.57 The magnetic field in the gap, from north to south, is perpendicular to the current.

FIGURE 29.57 Full Alternative Text

SOLVE We can write the output voltage of the amplifier as $\text{Δ}V=V_0\text{cos}\omega t$, where $V_0=12\text{V}$ is the peak voltage and $\omega =2\pi f=628\text{\hspace{0.17em}rad/s}$ is the angular frequency at 100 Hz. The voltage drives current

$$I={\displaystyle \frac{\text{Δ}V}{R}}={\displaystyle \frac{V_0\text{cos}\omega t}{R}}$$

through the coil, where $R$ is the coil’s resistance. This causes the oscillating in-and-out force that drives the speaker cone back and forth. Even though the coil isn’t a straight wire, the fact that the magnetic field is everywhere perpendicular to the current means that we can calculate the magnetic force as $F=IlB$ where l is the total length of the wire in the coil. The circumference of the coil is $\pi (0.050\text{m})=0.157\text{m}$ so 20 turns gives $l=3.1\text{m}$. The cone responds to the force by accelerating with $a=F\text{/}m$. Combining these pieces, we find the cone’s acceleration is

$$a={\displaystyle \frac{IlB}{m}}={\displaystyle \frac{V_{0}lB\text{cos}\omega t}{mR}}=a_{\text{max}}\text{cos}\omega t$$

It is straightforward to evaluate $a_{\text{max}}=152\text{m}\text{/}{\text{s}}^2$.

From kinematics, $a=dv\text{/}dt$ and $v=dx\text{/}dt$. We need to integrate twice to find the displacement. First,

$$v={{\displaystyle \int }}^{\text{}}adt=a_{\text{max}}{{\displaystyle \int }}^{\text{}}\cos \omega tdt=\frac{a_{\text{max}}}{\omega }\text{sin}\omega t$$

The integration constant is zero because we know, from simple harmonic motion, that the average velocity is zero. Integrating again, we get

$$x={{\displaystyle \int }}^{\text{}}vdt=\frac{a_{\text{max}}}{\omega }{{\displaystyle \int }}^{\text{}}\text{sin }\omega tdt=-\frac{a_{\text{max}}}{{\omega }^2}\text{cos }\omega t$$

where the integration constant is again zero if we assume the oscillation takes place around the origin. The minus sign tells us that the displacement and acceleration are out of phase. The amplitude of the oscillation, which we seek, is

$$A={\displaystyle \frac{a_{\text{max}}}{{\omega }^2}}={\displaystyle \frac{152{\text{m/s}}^2}{(628{\text{rad/s})}^2}}=3.8\times {10}^{-4}\text{m}=0.38\text{mm}$$

REVIEW If you’ve ever placed your hand on a loudspeaker cone, you know that you can feel a slight vibration. An amplitude of 0.38 mm is consistent with this observation. The fact that the amplitude increases with the inverse square of the frequency explains why you can sometimes see the cone vibrating with an amplitude of several millimeters for low-frequency bass notes.