#set document(title: "12.7 Magnetism in Matter", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 12.7#h(0.6em)Magnetism in Matter Why are certain materials magnetic and others not? And why do certain substances become magnetized by a field, whereas others are unaffected? To answer such questions, we need an understanding of magnetism on a microscopic level. Within an atom, every electron travels in an orbit and spins on an internal axis. Both types of motion produce current loops and therefore magnetic dipoles. For a particular atom, the net magnetic dipole moment is the vector sum of the magnetic dipole moments. Values of #math.equation(block: false, alt: "μ")[$μ$] for several types of atoms are given in . Notice that some atoms have a zero net dipole moment and that the magnitudes of the nonvanishing moments are typically #math.equation(block: false, alt: "10 to the power −23 A times m squared .")[$10^(−23) "A" · "m"^(2) .$] #figure(table( columns: 2, align: left, inset: 6pt, table.header([Atom], [Magnetic Moment #math.equation(block: false, alt: "open parenthesis 10 to the power −24 A times m squared close parenthesis")[$( 10^(−24) #h(0.2em) "A" · "m"^(2) )$]]), [H], [9.27], [He], [0], [Li], [9.27], [O], [13.9], [Na], [9.27], [S], [13.9], )) A handful of matter has approximately #math.equation(block: false, alt: "10 to the power 26")[$10^(26)$] atoms and ions, each with its magnetic dipole moment. If no external magnetic field is present, the magnetic dipoles are randomly oriented—as many are pointed up as down, as many are pointed east as west, and so on. Consequently, the net magnetic dipole moment of the sample is zero. However, if the sample is placed in a magnetic field, these dipoles tend to align with the field, and this alignment determines how the sample responds to the field. On the basis of this response, a material is said to be either paramagnetic, ferromagnetic, or diamagnetic. In a #strong[paramagnetic material], only a small fraction (roughly one-third) of the magnetic dipoles are aligned with the applied field. Since each dipole produces its own magnetic field, this alignment contributes an extra magnetic field, which enhances the applied field. When a #strong[ferromagnetic material] is placed in a magnetic field, its magnetic dipoles also become aligned; furthermore, they become locked together so that a permanent magnetization results, even when the field is turned off or reversed. This permanent magnetization happens in ferromagnetic materials but not paramagnetic materials. #strong[Diamagnetic materials] are composed of atoms that have no net magnetic dipole moment. However, when a diamagnetic material is placed in a magnetic field, a magnetic dipole moment is directed opposite to the applied field and therefore produces a magnetic field that opposes the applied field. We now consider each type of material in greater detail. === Paramagnetic Materials For simplicity, we assume our sample is a long, cylindrical piece that completely fills the interior of a long, tightly wound solenoid. When there is no current in the solenoid, the magnetic dipoles in the sample are randomly oriented and produce no net magnetic field. With a solenoid current, the magnetic field due to the solenoid exerts a torque on the dipoles that tends to align them with the field. In competition with the aligning torque are thermal collisions that tend to randomize the orientations of the dipoles. The relative importance of these two competing processes can be estimated by comparing the energies involved. From , the energy difference between a magnetic dipole aligned with and against a magnetic field is #math.equation(block: false, alt: "U sub B equals 2 μ B .")[$U_(B) = 2 μ B .$] If #math.equation(block: false, alt: "μ equals 9.3 times 10 to the power −24 A times m squared")[$μ = 9.3 #h(0.2em) × #h(0.2em) 10^(−24) "A" · "m"^(2)$] (the value of atomic hydrogen) and #emph[B] = 1.0 T, then #math.equation(block: true, alt: "U sub B equals 1.9 times 10 to the power −23 J .")[$U_(B) = 1.9 #h(0.2em) × #h(0.2em) 10^(−23) "J" .$] At a room temperature of #math.equation(block: false, alt: "27 °C ,")[$27 #h(0.2em) "°C" ,$] the thermal energy per atom is #math.equation(block: true, alt: "U sub T approximately equals k T equals open parenthesis 1.38 times 10 to the power −23 J/K close parenthesis open parenthesis 300 K close parenthesis equals 4.1 times 10 to the power −21 J ,")[$U_(T) ≈ k T = ( 1.38 #h(0.2em) × #h(0.2em) 10^(−23) "J/K" ) ( 300 #h(0.2em) "K" ) = 4.1 #h(0.2em) × #h(0.2em) 10^(−21) "J" ,$] which is about 220 times greater than #math.equation(block: false, alt: "U sub B .")[$U_(B) .$] Clearly, energy exchanges in thermal collisions can seriously interfere with the alignment of the magnetic dipoles. As a result, only a small fraction of the dipoles is aligned at any instant. The four sketches of furnish a simple model of this alignment process. In part (a), before the field of the solenoid (not shown) containing the paramagnetic sample is applied, the magnetic dipoles are randomly oriented and there is no net magnetic dipole moment associated with the material. With the introduction of the field, a partial alignment of the dipoles takes place, as depicted in part (b). The component of the net magnetic dipole moment that is perpendicular to the field vanishes. We may then represent the sample by part (c), which shows a collection of magnetic dipoles completely aligned with the field. By treating these dipoles as current loops, we can picture the dipole alignment as equivalent to a current around the surface of the material, as in part (d). This fictitious surface current produces its own magnetic field, which enhances the field of the solenoid. #figure(figph[Figure a shows a rod with randomly oriented magnetic dipoles. Figure b shows domains that got partially oriented after the magnetic field was applied along the axis of the rod. Figure c shows fully oriented domains. Figure d shows that the dipoles are aligned within the individual domains and are equivalent to a current around the surface of the material. This surface current produces its own magnetic field which enhances the field of the solenoid.], alt: "Figure a shows a rod with randomly oriented magnetic dipoles. Figure b shows domains that got partially oriented after the magnetic field was applied along the axis of the rod. Figure c shows fully oriented domains. Figure d shows that the dipoles are aligned within the individual domains and are equivalent to a current around the surface of the material. This surface current produces its own magnetic field which enhances the field of the solenoid.", caption: [The alignment process in a paramagnetic material filling a solenoid (not shown). (a) Without an applied field, the magnetic dipoles are randomly oriented. (b) With a field, partial alignment occurs. (c) An equivalent representation of part (b). (d) The internal currents cancel, leaving an effective surface current that produces a magnetic field similar to that of a finite solenoid.]) We can express the total magnetic field #math.equation(block: false, alt: "B →")[$arrow(B)$] in the material as #math.equation(block: true, alt: "B → equals B → sub 0 plus B → sub m ,")[$arrow(B) = arrow(B)_(0) + arrow(B)_(m) ,$] where #math.equation(block: false, alt: "B → sub 0")[$arrow(B)_(0)$] is the field due to the current #math.equation(block: false, alt: "I sub 0")[$I_(0)$] in the solenoid and #math.equation(block: false, alt: "B → sub m")[$arrow(B)_(m)$] is the field due to the surface current #math.equation(block: false, alt: "I sub m")[$I_(m)$] around the sample. Now #math.equation(block: false, alt: "B → sub m")[$arrow(B)_(m)$] is usually proportional to #math.equation(block: false, alt: "B → sub 0 ,")[$arrow(B)_(0) ,$] a fact we express by #math.equation(block: true, alt: "B → sub m equals χ B → sub 0 ,")[$arrow(B)_(m) = χ arrow(B)_(0) ,$] where #math.equation(block: false, alt: "χ")[$χ$] is a dimensionless quantity called the #strong[magnetic susceptibility]. Values of #math.equation(block: false, alt: "χ")[$χ$] for some paramagnetic materials are given in . Since the alignment of magnetic dipoles is so weak, #math.equation(block: false, alt: "χ")[$χ$] is very small for paramagnetic materials. By combining and , we obtain: #math.equation(block: true, alt: "B → equals B → sub 0 plus χ B → sub 0 equals open parenthesis 1 plus χ close parenthesis B → sub 0 .")[$arrow(B) = arrow(B)_(0) + χ arrow(B)_(0) = ( 1 + χ ) arrow(B)_(0) .$] For a sample within an infinite solenoid, this becomes #math.equation(block: true, alt: "B equals open parenthesis 1 plus χ close parenthesis μ sub 0 n I .")[$B = ( 1 + χ ) μ_(0) n I .$] This expression tells us that the insertion of a paramagnetic material into a solenoid increases the field by a factor of #math.equation(block: false, alt: "open parenthesis 1 plus χ close parenthesis .")[$( 1 + χ ) .$] However, since #math.equation(block: false, alt: "χ")[$χ$] is so small, the field isn’t enhanced very much. The quantity #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "μ equals open parenthesis 1 plus χ close parenthesis μ sub 0 .")[$μ = ( 1 + χ ) μ_(0) .$] ] is called the magnetic permeability of a material. In terms of #math.equation(block: false, alt: "μ ,")[$μ ,$] can be written as #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "B equals μ n I")[$B = μ n I$] ] for the filled solenoid. #figure(table( columns: 4, align: left, inset: 6pt, table.header([Paramagnetic Materials], [#math.equation(block: false, alt: "χ")[$χ$]], [Diamagnetic Materials], [#math.equation(block: false, alt: "χ")[$χ$]]), [Aluminum], [#math.equation(block: false, alt: "2.2 times 10 to the power −5")[$2.2 #h(0.2em) × #h(0.2em) 10^(−5)$]], [Bismuth], [#math.equation(block: false, alt: "−1.7 times 10 to the power −5")[$−1.7 #h(0.2em) × #h(0.2em) 10^(−5)$]], [Calcium], [#math.equation(block: false, alt: "1.4 times 10 to the power −5")[$1.4 #h(0.2em) × #h(0.2em) 10^(−5)$]], [Carbon (diamond)], [#math.equation(block: false, alt: "−2.2 times 10 to the power −5")[$−2.2 #h(0.2em) × #h(0.2em) 10^(−5)$]], [Chromium], [#math.equation(block: false, alt: "3.1 times 10 to the power −4")[$3.1 #h(0.2em) × #h(0.2em) 10^(−4)$]], [Copper], [#math.equation(block: false, alt: "−9.7 times 10 to the power −6")[$−9.7 #h(0.2em) × #h(0.2em) 10^(−6)$]], [Magnesium], [#math.equation(block: false, alt: "1.2 times 10 to the power −5")[$1.2 #h(0.2em) × #h(0.2em) 10^(−5)$]], [Lead], [#math.equation(block: false, alt: "−1.8 times 10 to the power −5")[$−1.8 #h(0.2em) × #h(0.2em) 10^(−5)$]], [Oxygen gas (1 atm)], [#math.equation(block: false, alt: "1.8 times 10 to the power −6")[$1.8 #h(0.2em) × #h(0.2em) 10^(−6)$]], [Mercury], [#math.equation(block: false, alt: "−2.8 times 10 to the power −5")[$−2.8 #h(0.2em) × #h(0.2em) 10^(−5)$]], [Oxygen liquid (90 K)], [#math.equation(block: false, alt: "3.5 times 10 to the power −3")[$3.5 #h(0.2em) × #h(0.2em) 10^(−3)$]], [Hydrogen gas (1 atm)], [#math.equation(block: false, alt: "−2.2 times 10 to the power −9")[$−2.2 #h(0.2em) × #h(0.2em) 10^(−9)$]], [Tungsten], [#math.equation(block: false, alt: "6.8 times 10 to the power −5")[$6.8 #h(0.2em) × #h(0.2em) 10^(−5)$]], [Nitrogen gas (1 atm)], [#math.equation(block: false, alt: "−6.7 times 10 to the power −9")[$−6.7 #h(0.2em) × #h(0.2em) 10^(−9)$]], [Air (1 atm)], [#math.equation(block: false, alt: "3.6 times 10 to the power −7")[$3.6 #h(0.2em) × #h(0.2em) 10^(−7)$]], [Water], [#math.equation(block: false, alt: "−9.1 times 10 to the power −6")[$−9.1 #h(0.2em) × #h(0.2em) 10^(−6)$]], )) === Diamagnetic Materials A magnetic field always induces a magnetic dipole in an atom. This induced dipole points opposite to the applied field, so its magnetic field is also directed opposite to the applied field. In paramagnetic and ferromagnetic materials, the induced magnetic dipole is masked by much stronger permanent magnetic dipoles of the atoms. However, in diamagnetic materials, whose atoms have no permanent magnetic dipole moments, the effect of the induced dipole is observable. We can now describe the magnetic effects of diamagnetic materials with the same model developed for paramagnetic materials. In this case, however, the fictitious surface current flows opposite to the solenoid current, and the magnetic susceptibility #math.equation(block: false, alt: "χ")[$χ$] is negative. Values of #math.equation(block: false, alt: "χ")[$χ$] for some diamagnetic materials are also given in . #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Water is a common diamagnetic material. Animals are mostly composed of water. Experiments have been performed on #link("https://openstax.org/l/21frogs")[frogs] and #link("https://openstax.org/l/21mice")[mice] in diverging magnetic fields. The water molecules are repelled from the applied magnetic field against gravity until the animal reaches an equilibrium. The result is that the animal is levitated by the magnetic field. ] === Ferromagnetic Materials Common magnets are made of a ferromagnetic material such as iron or one of its alloys. Experiments reveal that a ferromagnetic material consists of tiny regions known as #strong[magnetic domains]. Their volumes typically range from #math.equation(block: false, alt: "10 to the power −12")[$10^(−12)$] to #math.equation(block: false, alt: "10 to the power −8 m cubed ,")[$10^(−8) "m"^(3) ,$] and they contain about #math.equation(block: false, alt: "10 to the power 17")[$10^(17)$] to #math.equation(block: false, alt: "10 to the power 21")[$10^(21)$] atoms. Within a domain, the magnetic dipoles are rigidly aligned in the same direction by coupling among the atoms. This coupling, which is due to quantum mechanical effects, is so strong that even thermal agitation at room temperature cannot break it. The result is that each domain has a net dipole moment. Some materials have weaker coupling and are ferromagnetic only at lower temperatures. If the domains in a ferromagnetic sample are randomly oriented, as shown in , the sample has no net magnetic dipole moment and is said to be unmagnetized. Suppose that we fill the volume of a solenoid with an unmagnetized ferromagnetic sample. When the magnetic field #math.equation(block: false, alt: "B → sub 0")[$arrow(B)_(0)$] of the solenoid is turned on, the dipole moments of the domains rotate so that they align somewhat with the field, as depicted in . In addition, the aligned domains tend to increase in size at the expense of unaligned ones. The net effect of these two processes is the creation of a net magnetic dipole moment for the ferromagnet that is directed along the applied magnetic field. This net magnetic dipole moment is much larger than that of a paramagnetic sample, and the domains, with their large numbers of atoms, do not become misaligned by thermal agitation. Consequently, the field due to the alignment of the domains is quite large. #figure(figph[Picture a shows small randomly oriented domains in the unmagnetized piece of the ferromagnetic sample. Picture b shows small partially aligned domains upon the application of a magnetic field. Figure c shows domains of a single crystal of nickel. Clear domain boundaries are visible.], alt: "Picture a shows small randomly oriented domains in the unmagnetized piece of the ferromagnetic sample. Picture b shows small partially aligned domains upon the application of a magnetic field. Figure c shows domains of a single crystal of nickel. Clear domain boundaries are visible.", caption: [(a) Domains are randomly oriented in an unmagnetized ferromagnetic sample such as iron. The arrows represent the orientations of the magnetic dipoles within the domains. (b) In an applied magnetic field, the domains align somewhat with the field. (c) The domains of a single crystal of nickel. The white lines show the boundaries of the domains. These lines are produced by iron oxide powder sprinkled on the crystal.]) Besides iron, only four elements contain the magnetic domains needed to exhibit ferromagnetic behavior: cobalt, nickel, gadolinium, and dysprosium. Many alloys of these elements are also ferromagnetic. Ferromagnetic materials can be described using through , the paramagnetic equations. However, the value of #math.equation(block: false, alt: "χ")[$χ$] for ferromagnetic material is usually on the order of #math.equation(block: false, alt: "10 cubed")[$10^(3)$] to #math.equation(block: false, alt: "10 to the power 4 ,")[$10^(4) ,$] and it also depends on the history of the magnetic field to which the material has been subject. A typical plot of #emph[B] (the total field in the material) versus #math.equation(block: false, alt: "B sub 0")[$B_(0)$] (the applied field) for an initially unmagnetized piece of iron is shown. Some sample numbers are (1) for #math.equation(block: false, alt: "B sub 0 equals 1.0 times 10 to the power −4 T , B equals 0.60 T ,")[$B_(0) = 1.0 #h(0.2em) × #h(0.2em) 10^(−4) "T" , B = 0.60 #h(0.2em) "T" ,$] and #math.equation(block: false, alt: "χ equals open parenthesis the fraction 0.60 over 1.0 times 10 to the power −4 close parenthesis minus 1 approximately equals 6.0 times 10 cubed")[$χ = ( frac(0.60, 1.0 #h(0.2em) × #h(0.2em) 10^(−4)) ) − 1 ≈ 6.0 #h(0.2em) × #h(0.2em) 10^(3)$]; (2) for #math.equation(block: false, alt: "B sub 0 equals 6.0 times 10 to the power −4 T , B equals 1.5 T ,")[$B_(0) = 6.0 #h(0.2em) × #h(0.2em) 10^(−4) "T" , B = 1.5 #h(0.2em) "T" ,$] and #math.equation(block: false, alt: "χ equals open parenthesis the fraction 1.5 over 6.0 times 10 to the power −4 close parenthesis minus 1 approximately equals 2.5 times 10 cubed .")[$χ = ( frac(1.5, 6.0 #h(0.2em) × #h(0.2em) 10^(−4)) ) − 1 ≈ 2.5 #h(0.2em) × #h(0.2em) 10^(3) .$] #figure(figph[This picture shows a plot of the total field in the material versus the applied field for an initially unmagnetized piece of iron. The initial increase in the total field is followed by the saturation.], alt: "This picture shows a plot of the total field in the material versus the applied field for an initially unmagnetized piece of iron. The initial increase in the total field is followed by the saturation.", caption: [The magnetic field #emph[B] in annealed iron as a function of the applied field #math.equation(block: false, alt: "B sub 0 .")[$B_(0) .$]]) When #math.equation(block: false, alt: "B sub 0")[$B_(0)$] is varied over a range of positive and negative values, #emph[B] is found to behave as shown. Note that the same #math.equation(block: false, alt: "B sub 0")[$B_(0)$] (corresponding to the same current in the solenoid) can produce different values of #emph[B] in the material. The magnetic field #emph[B] produced in a ferromagnetic material by an applied field #math.equation(block: false, alt: "B sub 0")[$B_(0)$] depends on the magnetic history of the material. This effect is called #strong[hysteresis], and the curve of is called a hysteresis loop. Notice that #emph[B] does not disappear when #math.equation(block: false, alt: "B sub 0 equals 0")[$B_(0) = 0$] (i.e., when the current in the solenoid is turned off). The iron stays magnetized, which means that it has become a permanent magnet. #figure(figph[This picture shows a typical hysteresis loop for a ferromagnet. It starts at the origin with the upward curve that is the initial magnetization curve to the saturation point a, followed by the downward curve to point b after the saturation, along with the lower return curve back to the point a.], alt: "This picture shows a typical hysteresis loop for a ferromagnet. It starts at the origin with the upward curve that is the initial magnetization curve to the saturation point a, followed by the downward curve to point b after the saturation, along with the lower return curve back to the point a.", caption: [A typical hysteresis loop for a ferromagnet. When the material is first magnetized, it follows a curve from 0 to #emph[a]. When #math.equation(block: false, alt: "B sub 0")[$B_(0)$] is reversed, it takes the path shown from #emph[a] to #emph[b]. If #math.equation(block: false, alt: "B sub 0")[$B_(0)$] is reversed again, the material follows the curve from #emph[b] to #emph[a].]) Like the paramagnetic sample of , the partial alignment of the domains in a ferromagnet is equivalent to a current flowing around the surface. A bar magnet can therefore be pictured as a tightly wound solenoid with a large current circulating through its coils (the surface current). You can see in that this model fits quite well. The fields of the bar magnet and the finite solenoid are strikingly similar. The figure also shows how the poles of the bar magnet are identified. To form closed loops, the field lines outside the magnet leave the north (N) pole and enter the south (S) pole, whereas inside the magnet, they leave S and enter N. #figure(figph[The left picture shows magnetic fields of a finite solenoid; the right picture shows magnetic fields of a bar magnet. The fields are strikingly similar and form closed loops in both situations.], alt: "The left picture shows magnetic fields of a finite solenoid; the right picture shows magnetic fields of a bar magnet. The fields are strikingly similar and form closed loops in both situations.", caption: [Comparison of the magnetic fields of a finite solenoid and a bar magnet.]) Ferromagnetic materials are found in computer hard disk drives and permanent data storage devices . A material used in your hard disk drives is called a spin valve, which has alternating layers of ferromagnetic (aligning with the external magnetic field) and antiferromagnetic (each atom is aligned opposite to the next) metals. It was observed that a significant change in resistance was discovered based on whether an applied magnetic field was on the spin valve or not. This large change in resistance creates a quick and consistent way for recording or reading information by an applied current. #figure(figph[Photo shows the inside of a hard disk drive. The silver disk contains the information, whereas the thin stylus on top of the disk reads and writes information to the disk.], alt: "Photo shows the inside of a hard disk drive. The silver disk contains the information, whereas the thin stylus on top of the disk reads and writes information to the disk.", caption: [The inside of a hard disk drive. The silver disk contains the information, whereas the thin stylus on top of the disk reads and writes information to the disk.]) #examplebox("Example 1")[Iron Core in a Coil][ A long coil is tightly wound around an iron cylinder whose magnetization curve is shown. (a) If #math.equation(block: false, alt: "n equals 20")[$n = 20$] turns per centimeter, what is the applied field #math.equation(block: false, alt: "B sub 0")[$B_(0)$] when #math.equation(block: false, alt: "I sub 0 equals 0.20 A ?")[$I_(0) = 0.20 #h(0.2em) "A" ?$] (b) What is the net magnetic field for this same current? (c) What is the magnetic susceptibility in this case? Strategy (a) The magnetic field of a solenoid is calculated using . (b) The graph is read to determine the net magnetic field for this same current. (c) The magnetic susceptibility is calculated using . Solution + The applied field #math.equation(block: false, alt: "B sub 0")[$B_(0)$] of the coil is #math.equation(block: true, alt: "B sub 0, equals, μ sub 0 n I sub 0 equals open parenthesis 4 π times 10 to the power −7 T times m/A close parenthesis open parenthesis 2000 / m close parenthesis open parenthesis 0.20 A close parenthesis; B sub 0, equals, 5.0 times 10 to the power −4 T .")[$B_(0) & = & μ_(0) n I_(0) = ( 4 π #h(0.2em) × #h(0.2em) 10^(−7) "T" · "m/A" ) ( 2000 #h(0.1em) "/" #h(0.1em) "m" ) ( 0.20 #h(0.2em) "A" ) \ B_(0) & = & 5.0 #h(0.2em) × #h(0.2em) 10^(−4) "T" .$] + From inspection of the magnetization curve of , we see that, for this value of #math.equation(block: false, alt: "B sub 0 ,")[$B_(0) ,$] #math.equation(block: false, alt: "B equals 1.4 T .")[$B = 1.4 #h(0.2em) "T" .$] Notice that the internal field of the aligned atoms is much larger than the externally applied field. + The magnetic susceptibility is calculated to be #math.equation(block: true, alt: "χ equals the fraction B over B sub 0 minus 1 equals the fraction 1.4 T over 5.0 times 10 to the power −4 T −1 equals 2.8 times 10 cubed .")[$χ = frac(B, B_(0)) − 1 = frac(1.4 #h(0.2em) "T", 5.0 #h(0.2em) × #h(0.2em) 10^(−4) "T") "−1" = 2.8 #h(0.2em) × #h(0.2em) 10^(3) .$] Significance Ferromagnetic materials have susceptibilities in the range of #math.equation(block: false, alt: "10 cubed")[$10^(3)$] which compares well to our results here. Paramagnetic materials have fractional susceptibilities, so their applied field of the coil is much greater than the magnetic field generated by the material. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Repeat the calculations from the previous example for #math.equation(block: false, alt: "I sub 0 equals 0.040 A .")[$I_(0) = 0.040 #h(0.2em) "A" .$] #solutionbox[ a. #math.equation(block: false, alt: "1.0 times 10 to the power −4 T")[$1.0 #h(0.2em) × #h(0.2em) 10^(−4) "T"$]; b. 0.60 T; c. #math.equation(block: false, alt: "6.0 times 10 cubed")[$6.0 #h(0.2em) × #h(0.2em) 10^(3)$] ] ] === Summary - Materials are classified as paramagnetic, diamagnetic, or ferromagnetic, depending on how they behave in an applied magnetic field. - Paramagnetic materials have partial alignment of their magnetic dipoles with an applied magnetic field. This is a positive magnetic susceptibility. Only a surface current remains, creating a solenoid-like magnetic field. - Diamagnetic materials exhibit induced dipoles opposite to an applied magnetic field. This is a negative magnetic susceptibility. - Ferromagnetic materials have groups of dipoles, called domains, which align with the applied magnetic field. However, when the field is removed, the ferromagnetic material remains magnetized, unlike paramagnetic materials. This magnetization of the material versus the applied field effect is called hysteresis. === Key Equations #figure(table( columns: 2, align: left, inset: 6pt, table.header([Permeability of free space], [#math.equation(block: false, alt: "μ sub 0 equals 4 π times 10 to the power −7 T times m/A")[$μ_(0) = 4 π #h(0.2em) × #h(0.2em) 10^(−7) "T" ⋅ "m/A"$]]), [Contribution to magnetic field #linebreak() from a current element], [#math.equation(block: false, alt: "d B equals the fraction μ sub 0 over 4 π the fraction I d l sin θ over r squared")[$d B = frac(μ_(0), 4 π) #h(0.2em) frac(I #h(0.2em) d l #h(0.2em) "sin" #h(0.1em) θ, r^(2))$]], [Biot–Savart law], [#math.equation(block: false, alt: "B → equals the fraction μ sub 0 over 4 π ∫ wire the fraction I d l → times r ^ over r squared")[$arrow(B) = frac(μ_(0), 4 π) display(limits(∫)_("wire") frac(I d arrow(l) #h(0.2em) × #h(0.2em) hat(r), r^(2)))$]], [Magnetic field due to a #linebreak() long straight wire], [#math.equation(block: false, alt: "B equals the fraction μ sub 0 I over 2 π R")[$B = frac(μ_(0) I, 2 π R)$]], [Force between two parallel currents], [#math.equation(block: false, alt: "the fraction F over l equals the fraction μ sub 0 I sub 1 I sub 2 over 2 π r")[$frac(F, l) = frac(μ_(0) I_(1) I_(2), 2 π r)$]], [Magnetic field of a current loop], [#math.equation(block: false, alt: "B equals the fraction μ sub 0 I over 2 R (at center of loop)")[$B = frac(μ_(0) I, 2 R) #h(0.2em) "(at center of loop)"$]], [Ampère’s law], [#math.equation(block: false, alt: "∮ B → times d l → equals μ sub 0 I")[$display(∮ arrow(B)) · d arrow(l) = μ_(0) I$]], [Magnetic field strength #linebreak() inside a solenoid], [#math.equation(block: false, alt: "B equals μ sub 0 n I")[$B = μ_(0) n I$]], [Magnetic field strength inside a toroid], [#math.equation(block: false, alt: "B equals the fraction μ sub o N I over 2 π r")[$B = frac(μ_(o) N I, 2 π r)$]], [Magnetic permeability], [#math.equation(block: false, alt: "μ equals open parenthesis 1 plus χ close parenthesis μ sub 0")[$μ = ( 1 + χ ) μ_(0)$]], [Magnetic field of a solenoid #linebreak() filled with paramagnetic material], [#math.equation(block: false, alt: "B equals μ n I")[$B = μ n I$]], )) === Conceptual Questions A diamagnetic material is brought close to a permanent magnet. What happens to the material? If you cut a bar magnet into two pieces, will you end up with one magnet with an isolated north pole and another magnet with an isolated south pole? Explain your answer. #solutionbox[ The bar magnet will then become two magnets, each with their own north and south poles. There are no magnetic monopoles or single pole magnets. ] === Problems The magnetic field in the core of an air-filled solenoid is 1.50 T. By how much will this magnetic field decrease if the air is pumped out of the core while the current is held constant? A solenoid has a ferromagnetic core, #emph[n] = 1000 turns per meter, and #emph[I] = 5.0 A. If #emph[B] inside the solenoid is 2.0 T, what is #math.equation(block: false, alt: "χ")[$χ$] for the core material? #solutionbox[ 317.31 ] A 20-A current flows through a solenoid with 2000 turns per meter. What is the magnetic field inside the solenoid if its core is (a) a vacuum and (b) filled with liquid oxygen at 90 K? The magnetic dipole moment of the iron atom is about #math.equation(block: false, alt: "2.1 times 10 to the power −23 A times m squared .")[$2.1 #h(0.2em) × #h(0.2em) 10^(−23) "A" · "m"^(2) .$] (a) Calculate the maximum magnetic dipole moment of a domain consisting of #math.equation(block: false, alt: "10 to the power 19")[$10^(19)$] iron atoms. (b) What current would have to flow through a single circular loop of wire of diameter 1.0 cm to produce this magnetic dipole moment? #solutionbox[ #math.equation(block: true, alt: "2.1 times 10 to the power −4 A times m squared")[$2.1 #h(0.2em) × #h(0.2em) 10^(−4) "A" · "m"^(2)$] #linebreak() #math.equation(block: true, alt: "2.7 A")[$2.7 #h(0.2em) "A"$] ] Suppose you wish to produce a 1.2-T magnetic field in a toroid with an iron core for which #math.equation(block: false, alt: "χ equals 4.0 times 10 cubed .")[$χ = 4.0 #h(0.2em) × #h(0.2em) 10^(3) .$] The toroid has a mean radius of 15 cm and is wound with 500 turns. What current is required? A current of 1.5 A flows through the windings of a large, thin toroid with 200 turns per meter and a radius of 1 meter. If the toroid is filled with iron for which #math.equation(block: false, alt: "χ equals 3.0 times 10 cubed ,")[$χ = 3.0 #h(0.2em) × #h(0.2em) 10^(3) ,$] what is the magnetic field within it? #solutionbox[ 0.18 T ] A solenoid with an iron core is 25 cm long and is wrapped with 100 turns of wire. When the current through the solenoid is 10 A, the magnetic field inside it is 2.0 T. For this current, what is the permeability of the iron? If the current is turned off and then restored to 10 A, will the magnetic field necessarily return to 2.0 T? === Additional Problems Three long, straight, parallel wires, all carrying 20 A, are positioned as shown in the accompanying figure. What is the magnitude of the magnetic field at the point #emph[P]? #figure(figph[This figure shows three long, straight, parallel wires. Each wire forms a vertex of an equilateral triangle with 10 centimeter sides. Point P is the center of a triangle.], alt: "This figure shows three long, straight, parallel wires. Each wire forms a vertex of an equilateral triangle with 10 centimeter sides. Point P is the center of a triangle.", caption: none) #solutionbox[ #math.equation(block: true, alt: "B equals 1.4 times 10 to the power −4 T")[$B = 1.4 #h(0.2em) × #h(0.2em) 10^(−4) "T"$] ] A current #emph[I] flows around a wire bent into the shape of a square of side #emph[a]. What is the magnetic field at the point P that is a distance #emph[z] above the center of the square (see the accompanying figure)? #figure(figph[This figure shows a wire bent into the shape of a rhombus of side a. Point P that is a distance z above the center of the rhombus.], alt: "This figure shows a wire bent into the shape of a rhombus of side a. Point P that is a distance z above the center of the rhombus.", caption: none) The accompanying figure shows a long, straight wire carrying a current of 10 A. What is the magnetic force on an electron at the instant it is 20 cm from the wire, traveling parallel to the wire with a speed of #math.equation(block: false, alt: "2.0 times 10 to the power 5 m/s?")[$2.0 #h(0.2em) × #h(0.2em) 10^(5) "m/s?"$] Describe qualitatively the subsequent motion of the electron. #figure(figph[Figure shows a long, straight wire carrying a current. An electron is located 20 cm from the wire and travels parallel to it.], alt: "Figure shows a long, straight wire carrying a current. An electron is located 20 cm from the wire and travels parallel to it.", caption: none) #solutionbox[ #math.equation(block: false, alt: "3.2 times 10 to the power −19 N")[$3.2 #h(0.2em) × #h(0.2em) 10^("−19") N$] in an arc away from the wire ] Current flows along a thin, infinite sheet as shown in the accompanying figure. The current per unit length along the sheet is #emph[J] in amperes per meter. (a) Use the Biot-Savart law to show that #math.equation(block: false, alt: "B equals μ sub 0 J / 2")[$B = μ_(0) J / 2$] on either side of the sheet. What is the direction of #math.equation(block: false, alt: "B →")[$arrow(B)$] on each side? (b) Now use Ampère’s law to calculate the field. #figure(figph[Figure shows current flowing along a thin, infinite sheet.], alt: "Figure shows current flowing along a thin, infinite sheet.", caption: none) (a) Use the result of the previous problem to calculate the magnetic field between, above, and below the pair of infinite sheets shown in the accompanying figure. (b) Repeat your calculations if the direction of the current in the lower sheet is reversed. #figure(figph[Figure shows currents flowing along two thin, infinite sheets. Sheets are located in the parallel planes and current flows in the same direction.], alt: "Figure shows currents flowing along two thin, infinite sheets. Sheets are located in the parallel planes and current flows in the same direction.", caption: none) #solutionbox[ a. above and below #math.equation(block: false, alt: "B equals μ sub 0 j ,")[$B = μ_(0) j ,$] in the middle #math.equation(block: false, alt: "B equals 0 ;")[$B = 0 ;$] b. above and below #math.equation(block: false, alt: "B equals 0 ,")[$B = 0 ,$] in the middle #math.equation(block: false, alt: "B equals μ sub 0 j")[$B = μ_(0) j$] ] We often assume that the magnetic field is uniform in a region and zero everywhere else. Show that in reality it is impossible for a magnetic field to drop abruptly to zero, as illustrated in the accompanying figure. (#emph[Hint]: Apply Ampère’s law over the path shown.) #figure(figph[Figure shows the magnetic field that is perpendicular to the rectangular current path and intersects it.], alt: "Figure shows the magnetic field that is perpendicular to the rectangular current path and intersects it.", caption: none) How is the fractional change in the strength of the magnetic field across the face of the toroid related to the fractional change in the radial distance from the axis of the toroid? #solutionbox[ #math.equation(block: true, alt: "the fraction d B over B equals minus the fraction d r over r")[$frac(d B, B) = − #h(0.05em) frac(d r, r)$] ] Show that the expression for the magnetic field of a toroid reduces to that for the field of an infinite solenoid in the limit that the central radius goes to infinity. A toroid with an inner radius of 20 cm and an outer radius of 22 cm is tightly wound with one layer of wire that has a diameter of 0.25 mm. (a) How many turns are there on the toroid? (b) If the current through the toroid windings is 2.0 A, what is the strength of the magnetic field at the center of the toroid? #solutionbox[ a. 5026 turns; b. 0.00957 T ] A wire element has #math.equation(block: false, alt: "d l → , I d l → equals J A d l equals J d v ,")[$d arrow(l) , I d arrow(l) = J A d l = J d v ,$] where #emph[A] and #emph[dv] are the cross-sectional area and volume of the element, respectively. Use this, the Biot-Savart law, and #math.equation(block: false, alt: "J equals n e v")[$J = n e v$] to show that the magnetic field of a moving point charge q is given by: #linebreak() #math.equation(block: false, alt: "B → equals the fraction μ sub 0 over 4 π the fraction q v times r ^ over r squared")[$arrow(B) = frac(μ_(0), 4 π) #h(0.2em) frac(q v #h(0.2em) × #h(0.2em) hat(r), r^(2))$] A reasonably uniform magnetic field over a limited region of space can be produced with the Helmholtz coil, which consists of two parallel coils centered on the same axis. The coils are connected so that they carry the same current #emph[I]. Each coil has #emph[N] turns and radius #emph[R], which is also the distance between the coils. (a) Find the magnetic field at any point on the #emph[z]-axis shown in the accompanying figure. (b) Show that #emph[dB]/#emph[dz] and #math.equation(block: false, alt: "the fraction d squared B over d z squared")[$frac(d^(2) B, d z^(2))$] are both zero at #emph[z] = 0. (These vanishing derivatives demonstrate that the magnetic field varies only slightly near #emph[z] = 0.) #figure(figph[This picture shows two parallel coils centered on the same axis that carry the same current I. Each coil has radius R, which is also the distance between the coils.], alt: "This picture shows two parallel coils centered on the same axis that carry the same current I. Each coil has radius R, which is also the distance between the coils.", caption: none) #solutionbox[ #math.equation(block: true, alt: "B sub 1 open parenthesis x close parenthesis equals the fraction μ sub 0 I R squared over 2 open parenthesis R squared plus z squared close parenthesis to the power 3 / 2")[$B_(1) ( x ) = frac(μ_(0) I R^(2), 2 attach(( R^(2) + z^(2) ), t: 3 "/" 2))$] ] A charge of #math.equation(block: false, alt: "4.0 μC")[$4.0 #h(0.2em) "μC"$] is distributed uniformly around a thin ring of insulating material. The ring has a radius of 0.20 m and rotates at #math.equation(block: false, alt: "2.0 times 10 to the power 4 rev/min")[$2.0 #h(0.2em) × #h(0.2em) 10^(4) "rev/min"$] around the axis that passes through its center and is perpendicular to the plane of the ring. What is the magnetic field at the center of the ring? A thin, nonconducting disk of radius #emph[R] is free to rotate around the axis that passes through its center and is perpendicular to the face of the disk. The disk is charged uniformly with a total charge #emph[q]. If the disk rotates at a constant angular velocity #math.equation(block: false, alt: "ω ,")[$ω ,$] what is the magnetic field at its center? #solutionbox[ #math.equation(block: true, alt: "B equals the fraction μ sub 0 σ ω over 2 R")[$B = frac(μ_(0) σ ω, 2) R$] ] Consider the disk in the previous problem. Calculate the magnetic field at a point on its central axis that is a distance #emph[y] above the disk. Consider the axial magnetic field #math.equation(block: false, alt: "B sub y equals μ sub 0 I R squared / 2 open parenthesis y squared plus R squared close parenthesis to the power 3 / 2")[$B_(y) = μ_(0) I R^(2) "/" 2 ( y^(2) + R^(2) )^(3 "/" 2)$] of the circular current loop shown below. (a) Evaluate #math.equation(block: false, alt: "∫ − a a B sub y d y .")[$display(∫_("−" a)^(a) B_(y) d y .)$] Also show that #math.equation(block: false, alt: "lim a → ∞ ∫ − a a B sub y d y equals μ sub 0 I .")[$limits("lim")_(a → ∞) display(∫_("−" a)^(a) B_(y)) d y = μ_(0) I .$] (b) Can you deduce this limit without evaluating the integral? (#emph[Hint:] See the accompanying figure.) #figure(figph[This picture shows the circular current loop I with the magnetic field B perpendicular to the plane of the loop.], alt: "This picture shows the circular current loop I with the magnetic field B perpendicular to the plane of the loop.", caption: none) #solutionbox[ derivation ] The current density in the long, cylindrical wire shown in the accompanying figure varies with distance #emph[r] from the center of the wire according to #math.equation(block: false, alt: "J equals c r ,")[$J = c r ,$] where #emph[c] is a constant. (a) What is the current through the wire? (b) What is the magnetic field produced by this current for #math.equation(block: false, alt: "r less than or equal to R ?")[$r ≤ R ?$] For #math.equation(block: false, alt: "r greater than or equal to R ?")[$r ≥ R ?$] #figure(figph[This figure shows a long, straight, cylindrical wire with a radius R that has current I flowing through it.], alt: "This figure shows a long, straight, cylindrical wire with a radius R that has current I flowing through it.", caption: none) A long, straight, cylindrical conductor contains a cylindrical cavity whose axis is displaced by #emph[a] from the axis of the conductor, as shown in the accompanying figure. The current density in the conductor is given by #math.equation(block: false, alt: "J → equals J sub 0 k ^ ,")[$arrow(J) = J_(0) hat(k) ,$] where #math.equation(block: false, alt: "J sub 0")[$J_(0)$] is a constant and #math.equation(block: false, alt: "k ^")[$hat(k)$] is along the axis of the conductor. Calculate the magnetic field at an arbitrary point P in the cavity by superimposing the field of a solid cylindrical conductor with radius #math.equation(block: false, alt: "R sub 1")[$R_(1)$] and current density #math.equation(block: false, alt: "J →")[$arrow(J)$] onto the field of a solid cylindrical conductor with radius #math.equation(block: false, alt: "R sub 2")[$R_(2)$] and current density #math.equation(block: false, alt: "− J → .")[$"−" arrow(J) .$] Then use the fact that the appropriate azimuthal unit vectors can be expressed as #math.equation(block: false, alt: "θ ^ sub 1 equals k ^ times r ^ sub 1")[$hat(θ)_(1) = hat(k) #h(0.2em) × #h(0.2em) hat(r)_(1)$] and #math.equation(block: false, alt: "θ ^ sub 2 equals k ^ times r ^ sub 2")[$hat(θ)_(2) = hat(k) #h(0.2em) × #h(0.2em) hat(r)_(2)$] to show that everywhere inside the cavity the magnetic field is given by the constant #math.equation(block: false, alt: "B → equals the fraction 1 over 2 μ sub 0 J sub 0 k times a ,")[$arrow(B) = frac(1, 2) μ_(0) J_(0) k #h(0.2em) × #h(0.2em) a ,$] where #math.equation(block: false, alt: "a equals r sub 1 minus r sub 2")[$a = r_(1) − r_(2)$] and #math.equation(block: false, alt: "r sub 1 equals r sub 1 r ^ sub 1")[$r_(1) = r_(1) hat(r)_(1)$] is the position of #emph[P] relative to the center of the conductor and #math.equation(block: false, alt: "r sub 2 equals r sub 2 r ^ sub 2")[$r_(2) = r_(2) hat(r)_(2)$] is the position of #emph[P] relative to the center of the cavity. #figure(figph[This figure shows a large circle with a radius R1 that has a circular hole of radius R2 in it at a distance a from the center. Point P is located in a hole at the distance r2 from the center of a hole and at a distance r1 from the center of a large circle.], alt: "This figure shows a large circle with a radius R1 that has a circular hole of radius R2 in it at a distance a from the center. Point P is located in a hole at the distance r2 from the center of a hole and at a distance r1 from the center of a large circle.", caption: none) #solutionbox[ derivation ] Between the two ends of a horseshoe magnet the field is uniform as shown in the diagram. As you move out to outside edges, the field bends. Show by Ampère’s law that the field must bend and thereby the field weakens due to these bends. #figure(figph[This figure shows a horse shoe magnet with the magnetic lines going from the North end to the South end.], alt: "This figure shows a horse shoe magnet with the magnetic lines going from the North end to the South end.", caption: none) Show that the magnetic field of a thin wire and that of a current loop are zero if you are infinitely far away. #solutionbox[ As the radial distance goes to infinity, the magnetic fields of each of these formulae go to zero. ] An Ampère loop is chosen as shown by dashed lines for a parallel constant magnetic field as shown by solid arrows. Calculate #math.equation(block: false, alt: "B → times d l →")[$arrow(B) · d arrow(l)$] for each side of the loop then find the entire #math.equation(block: false, alt: "∮ B → times d l → .")[$display(∮ arrow(B)) · d arrow(l) .$] Can you think of an Ampère loop that would make the problem easier? Do those results match these? #figure(figph[This figure shows an Ampere loop that is located in the constant magnetic field. One of the sides of the loop forms an angle theta with the magnetic line.], alt: "This figure shows an Ampere loop that is located in the constant magnetic field. One of the sides of the loop forms an angle theta with the magnetic line.", caption: none) A very long, thick cylindrical wire of radius #emph[R] carries a current density #emph[J] that varies across its cross-section. The magnitude of the current density at a point a distance #emph[r] from the center of the wire is given by #math.equation(block: false, alt: "J equals J sub 0 the fraction r over R ,")[$J = J_(0) frac(r, R) ,$] where #math.equation(block: false, alt: "J sub 0")[$J_(0)$] is a constant. Find the magnetic field (a) at a point outside the wire and (b) at a point inside the wire. Write your answer in terms of the net current #emph[I] through the wire. #solutionbox[ a. #math.equation(block: false, alt: "B equals the fraction μ sub 0 I over 2 π r")[$B = frac(μ_(0) I, 2 π r)$]; b. #math.equation(block: false, alt: "B equals the fraction μ sub 0 J sub 0 r squared over 3 R")[$B = frac(μ_(0) J_(0) r^(2), 3 R)$] ] A very long, cylindrical wire of radius #emph[a] has a circular hole of radius #emph[b] in it at a distance #emph[d] from the center. The wire carries a uniform current of magnitude #emph[I] through it. The direction of the current in the figure is out of the paper. Find the magnetic field (a) at a point at the edge of the hole closest to the center of the thick wire, (b) at an arbitrary point inside the hole, and (c) at an arbitrary point outside the wire. (#emph[Hint:] Think of the hole as a sum of two wires carrying current in the opposite directions.) #figure(figph[This figure shows a circle with a radius a that has a circular hole of radius b in it at a distance d from the center.], alt: "This figure shows a circle with a radius a that has a circular hole of radius b in it at a distance d from the center.", caption: none) Magnetic field inside a torus. Consider a torus of rectangular cross-section with inner radius #emph[a] and outer radius #emph[b]. #emph[N] turns of an insulated thin wire are wound evenly on the torus tightly all around the torus and connected to a battery producing a steady current #emph[I] in the wire. Assume that the current on the top and bottom surfaces in the figure is radial, and the current on the inner and outer radii surfaces is vertical. Find the magnetic field inside the torus as a function of radial distance #emph[r] from the axis. #figure(figph[Figure shows a torus or donut-like shape with an inner radius of 'a' and outer radius of 'b'. Insulated wire of 'N' turns are wound evenly around the torus and connected to a battery (not shown).], alt: "Figure shows a torus or donut-like shape with an inner radius of 'a' and outer radius of 'b'. Insulated wire of 'N' turns are wound evenly around the torus and connected to a battery (not shown).", caption: none) #solutionbox[ #math.equation(block: true, alt: "B open parenthesis r close parenthesis equals μ sub 0 N I / 2 π r")[$B ( r ) = μ_(0) N I "/" 2 π r$] #linebreak() #figure(figph[This figure shows a torus with the inner radius a and an outer radius b. A thin wire is wound evenly on the torus.], alt: "This figure shows a torus with the inner radius a and an outer radius b. A thin wire is wound evenly on the torus.", caption: none) ] Two long coaxial copper tubes, each of length #emph[L], are connected to a battery of voltage #emph[V]. The inner tube has inner radius #emph[a] and outer radius #emph[b], and the outer tube has inner radius #emph[c] and outer radius #emph[d]. The tubes are then disconnected from the battery and rotated in the same direction at angular speed of #math.equation(block: false, alt: "ω")[$ω$] radians per second about their common axis. Find the magnetic field (a) at a point inside the space enclosed by the inner tube #math.equation(block: false, alt: "r less than a ,")[$r < a ,$] and (b) at a point between the tubes #math.equation(block: false, alt: "b less than r less than c ,")[$b < r < c ,$] and (c) at a point outside the tubes #math.equation(block: false, alt: "r greater than d .")[$r > d .$] (#emph[Hint:] Think of copper tubes as a capacitor and find the charge density based on the voltage applied, #math.equation(block: false, alt: "Q equals V C ,")[$Q = V C ,$] #math.equation(block: false, alt: "C equals the fraction 2 π ε sub 0 L over ln open parenthesis c / b close parenthesis .)")[$C = frac(2 π ε_(0) L, "ln" ( c #h(0.1em) "/" #h(0.1em) b )) ".)"$] === Challenge Problems The accompanying figure shows a flat, infinitely long sheet of width #emph[a] that carries a current #emph[I] uniformly distributed across it. Find the magnetic field at the point P, which is in the plane of the sheet and at a distance #emph[x] from one edge. Test your result for the limit #math.equation(block: false, alt: "a → 0 .")[$a → 0 .$] #figure(figph[This picture shows a flat, infinitely long sheet of width a that carries a current I uniformly distributed across it. Point P is in the plane of the sheet and at a distance x from one edge.], alt: "This picture shows a flat, infinitely long sheet of width a that carries a current I uniformly distributed across it. Point P is in the plane of the sheet and at a distance x from one edge.", caption: none) #solutionbox[ #math.equation(block: true, alt: "B equals the fraction μ sub 0 I over 2 π a ln the fraction x plus a over x .")[$B = frac(μ_(0) I, 2 π a) "ln" frac(x + a, x) .$] ] A hypothetical current flowing in the #emph[z]-direction creates the field #math.equation(block: false, alt: "B → equals C [ open parenthesis x / y squared close parenthesis i ^ plus open parenthesis 1 / y close parenthesis j ^ ]")[$arrow(B) = C [ ( x "/" y^(2) ) hat(i) + ( 1 "/" y ) hat(j) ]$] in the rectangular region of the #emph[xy]-plane shown in the accompanying figure. Use Ampère’s law to find the current through the rectangle. #figure(figph[This figure shows the rectangular region of the xy-plane; z axis is perpendicular to the plane. Points a1 and a2 are located at the x axis. Points b1 and b2 are located at the y axis. There is an equal distance between all points.], alt: "This figure shows the rectangular region of the xy-plane; z axis is perpendicular to the plane. Points a1 and a2 are located at the x axis. Points b1 and b2 are located at the y axis. There is an equal distance between all points.", caption: none) A nonconducting hard rubber circular disk of radius #emph[R] is painted with a uniform surface charge density #math.equation(block: false, alt: "σ .")[$σ .$] It is rotated about its axis with angular speed #math.equation(block: false, alt: "ω .")[$ω .$] (a) Find the magnetic field produced at a point on the axis a distance #emph[h] meters from the center of the disk. (b) Find the numerical value of magnitude of the magnetic field when #math.equation(block: false, alt: "σ equals 1 C/m squared ,")[$σ = 1 "C/m"^(2) ,$] #math.equation(block: false, alt: "R equals 20 cm , h equals 2 cm ,")[$R = "20 cm" , #h(0.2em) h = "2 cm" ,$] and #math.equation(block: false, alt: "ω equals 400 rad/sec ,")[$ω = 400 #h(0.2em) "rad/sec" ,$] and compare it with the magnitude of magnetic field of Earth, which is about 1/2 Gauss. #solutionbox[ a. #math.equation(block: false, alt: "B equals the fraction μ sub 0 σ ω over 2 [ the fraction 2 h squared plus R squared over the square root of R squared plus h squared −2 h ]")[$B = frac(μ_(0) σ ω, 2) [ frac(2 h^(2) + R^(2), sqrt(R^(2) + h^(2))) "−2" h ]$]; b. #math.equation(block: false, alt: "B equals 4.09 times 10 to the power −5 T ,")[$B = 4.09 #h(0.2em) × #h(0.2em) 10^(−5) "T" ,$] 82% of Earth’s magnetic field ]