#set document(title: "12.6 Solenoids and Toroids", 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")) == 12.6#h(0.6em)Solenoids and Toroids Two of the most common and useful electromagnetic devices are called solenoids and toroids. In one form or another, they are part of numerous instruments, both large and small. In this section, we examine the magnetic field typical of these devices. === Solenoids A long wire wound in the form of a helical coil is known as a #strong[solenoid]. Solenoids are commonly used in experimental research requiring magnetic fields. A solenoid is generally easy to wind, and near its center, its magnetic field is quite uniform and directly proportional to the current in the wire. shows a solenoid consisting of #emph[N] turns of wire tightly wound over a length #emph[L]. A current #emph[I] is flowing along the wire of the solenoid. The number of turns per unit length is #emph[N]/#emph[L]; therefore, the number of turns in an infinitesimal length #emph[dy] are (#emph[N]/#emph[L])#emph[dy] turns. This produces a current #math.equation(block: true, alt: "d I equals the fraction N I over L d y .")[$d I = frac(N I, L) d y .$] We first calculate the magnetic field at the point #emph[P] of . This point is on the central axis of the solenoid. We are basically cutting the solenoid into thin slices that are #emph[dy] thick and treating each as a current loop. Thus, #emph[dI] is the current through each slice. The magnetic field #math.equation(block: false, alt: "d B →")[$d arrow(B)$] due to the current #emph[dI] in #emph[dy] can be found with the help of and : #math.equation(block: true, alt: "d B → equals the fraction μ sub 0 R squared d I over 2 open parenthesis y squared plus R squared close parenthesis to the power 3 / 2 j ^ equals open parenthesis the fraction μ sub 0 I R squared N over 2 L j ^ close parenthesis the fraction d y over open parenthesis y squared plus R squared close parenthesis to the power 3 / 2")[$d arrow(B) = frac(μ_(0) R^(2) d I, 2 attach(( y^(2) + R^(2) ), t: 3 "/" 2)) hat(j) = ( frac(μ_(0) I R^(2) N, 2 L) hat(j) ) frac(d y, attach(( y^(2) + R^(2) ), t: 3 "/" 2))$] where we used to replace #emph[dI]. The resultant field at #emph[P] is found by integrating #math.equation(block: false, alt: "d B →")[$d arrow(B)$] along the entire length of the solenoid. It’s easiest to evaluate this integral by changing the independent variable from #emph[y] to #math.equation(block: false, alt: "θ .")[$θ .$] From inspection of , we have: #math.equation(block: true, alt: "sin θ equals the fraction y over the square root of y squared plus R squared .")[$"sin" #h(0.1em) θ = frac(y, sqrt(y^(2) + R^(2))) .$] #figure(figph[Figure A is a drawing of a solenoid that is a long wire wound in the shape of a helix. Figure B shows that the magnetic field at the point P on the axis of the solenoid is the net field due to all of the current loops.], alt: "Figure A is a drawing of a solenoid that is a long wire wound in the shape of a helix. Figure B shows that the magnetic field at the point P on the axis of the solenoid is the net field due to all of the current loops.", caption: [(a) A solenoid is a long wire wound in the shape of a helix. (b) The magnetic field at the point #emph[P] on the axis of the solenoid is the net field due to all of the current loops.]) Taking the differential of both sides of this equation, we obtain #math.equation(block: true, alt: "cos θ d θ, equals [ minus the fraction y squared over open parenthesis y squared plus R squared close parenthesis to the power 3 / 2 plus the fraction 1 over the square root of y squared plus R squared ] d y; equals the fraction R squared d y over open parenthesis y squared plus R squared close parenthesis to the power 3 / 2 .")[$"cos" #h(0.1em) θ #h(0.2em) d θ & = [ − #h(0.05em) frac(y^(2), attach(( y^(2) + R^(2) ), t: 3 "/" 2)) + frac(1, sqrt(y^(2) + R^(2))) ] d y \ & = frac(R^(2) d y, attach(( y^(2) + R^(2) ), t: 3 "/" 2)) .$] When this is substituted into the equation for #math.equation(block: false, alt: "d B → ,")[$d arrow(B) ,$] we have #math.equation(block: true, alt: "B → equals the fraction μ sub 0 I N over 2 L j ^ ∫ θ sub 1 θ sub 2 cos θ d θ equals the fraction μ sub 0 I N over 2 L open parenthesis sin θ sub 2 minus sin θ sub 1 close parenthesis j ^ ,")[$arrow(B) = frac(μ_(0) I N, 2 L) hat(j) display(limits(∫)^(θ_(2))_(θ_(1)) "cos" #h(0.1em) θ #h(0.2em) d θ =) frac(μ_(0) I N, 2 L) ( "sin" #h(0.1em) θ_(2) − "sin" #h(0.1em) θ_(1) ) hat(j) ,$] which is the magnetic field along the central axis of a finite solenoid. Of special interest is the infinitely long solenoid, for which #math.equation(block: false, alt: "L → ∞ .")[$L → ∞ .$] From a practical point of view, the infinite solenoid is one whose length is much larger than its radius #math.equation(block: false, alt: "open parenthesis L ≫ R close parenthesis .")[$( L ≫ R ) .$] In this case, #math.equation(block: false, alt: "θ sub 1 equals the fraction − π over 2")[$θ_(1) = frac("−" π, 2)$] and #math.equation(block: false, alt: "θ sub 2 equals the fraction π over 2 .")[$θ_(2) = frac(π, 2) .$] Then from , the magnetic field along the central axis of an infinite solenoid is #math.equation(block: true, alt: "B → equals the fraction μ sub 0 I N over 2 L j ^ [ sin open parenthesis π / 2 close parenthesis minus sin open parenthesis − π / 2 close parenthesis ] equals the fraction μ sub 0 I N over L j ^")[$arrow(B) = frac(μ_(0) I N, 2 L) hat(j) [ "sin" ( π "/" 2 ) − "sin" ( "−" π "/" 2 ) ] = frac(μ_(0) I N, L) hat(j)$] or #math.equation(block: true, alt: "B → equals μ sub 0 n I j ^ ,")[$arrow(B) = μ_(0) n I hat(j) ,$] where #emph[n] is the number of turns per unit length. You can find the direction of #math.equation(block: false, alt: "B →")[$arrow(B)$] with a right-hand rule: Curl your fingers in the direction of the current, and your thumb points along the magnetic field in the interior of the solenoid. We now use these properties, along with Ampère’s law, to calculate the magnitude of the magnetic field at any location inside the infinite solenoid. Consider the closed path of . Along segment 1, #math.equation(block: false, alt: "B →")[$arrow(B)$] is uniform and parallel to the path as shown. Along segments 2 and 4, #math.equation(block: false, alt: "B →")[$arrow(B)$] is perpendicular to part of the path and vanishes over the rest of it. Therefore, segments 2 and 4 do not contribute to the line integral in Ampère’s law. Along segment 3, #math.equation(block: false, alt: "B → equals 0")[$arrow(B) = 0$] because the magnetic field is zero outside the solenoid. If you consider an Ampère’s law loop outside of the solenoid, the current flows in opposite directions on different segments of the loop. Therefore, there is no enclosed current and no magnetic field according to Ampère’s law. Thus, there is no contribution to the line integral from segment 3. As a result, we find #math.equation(block: true, alt: "∮ B → times d l → equals ∫ sub 1 B → times d l → equals B l .")[$display(∮ arrow(B)) · d arrow(l) = display(∫_(1) arrow(B) · d arrow(l)) = B l .$] #figure(figph[Figure shows the closed rectangular path and the infinite solenoid. Segment 1 is inside the solenoid and is parallel to the path. Segments 2 and 4 are perpendicular to the path. Segment 3 is outside the solenoid.], alt: "Figure shows the closed rectangular path and the infinite solenoid. Segment 1 is inside the solenoid and is parallel to the path. Segments 2 and 4 are perpendicular to the path. Segment 3 is outside the solenoid.", caption: [The path of integration used in Ampère’s law to evaluate the magnetic field of an infinite solenoid.]) The solenoid has #emph[n] turns per unit length, so the current that passes through the surface enclosed by the path is #math.equation(block: false, alt: "n l I")[$n l I$]. Therefore, from Ampère’s law, #math.equation(block: true, alt: "B l equals μ sub 0 n l I")[$B l = μ_(0) n l I$] and #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "B equals μ sub 0 n I")[$B = μ_(0) n I$] ] within the solenoid. This agrees with what we found earlier for #emph[B] on the central axis of the solenoid. Here, however, the location of segment 1 is arbitrary, so we have found that this equation gives the magnetic field everywhere inside the infinite solenoid. When a patient undergoes a #strong[magnetic resonance imaging] (MRI) scan, the person lies down on a table that is moved into the center of a large solenoid that can generate very large magnetic fields. The solenoid is capable of these high fields from high currents flowing through superconducting wires. The large magnetic field is used to change the spin of protons in the patient’s body. The time it takes for the spins to align or relax (return to original orientation) is a signature of different tissues that can be analyzed to see if the structures of the tissues is normal . #figure(figph[Photo shows an MRI system. It consists of the cylindrical solenoid that is used to generate a large magnetic field.], alt: "Photo shows an MRI system. It consists of the cylindrical solenoid that is used to generate a large magnetic field.", caption: [In an MRI machine, a large magnetic field is generated by the cylindrical solenoid surrounding the patient.]) #examplebox("Example 1")[Magnetic Field Inside a Solenoid][ A solenoid has 300 turns wound around a cylinder of diameter 1.20 cm and length 14.0 cm. If the current through the coils is 0.410 A, what is the magnitude of the magnetic field inside and near the middle of the solenoid? Strategy We are given the number of turns and the length of the solenoid so we can find the number of turns per unit length. Therefore, the magnetic field inside and near the middle of the solenoid is given by . Outside the solenoid, the magnetic field is zero. Solution The number of turns per unit length is #math.equation(block: true, alt: "n equals the fraction 300 turns over 0.140 m equals 2.14 times 10 cubed turns/m .")[$n = frac(300 #h(0.2em) "turns", 0.140 #h(0.2em) "m") = 2.14 #h(0.2em) × #h(0.2em) 10^(3) "turns/m" .$] The magnetic field produced inside the solenoid is #math.equation(block: true, alt: "B, equals, μ sub 0 n I equals open parenthesis 4 π times 10 to the power −7 T times m/A close parenthesis open parenthesis 2.14 times 10 cubed turns/m close parenthesis open parenthesis 0.410 A close parenthesis; B, equals, 1.10 times 10 to the power −3 T .")[$B & = & μ_(0) n I = ( 4 π #h(0.2em) × #h(0.2em) 10^(−7) "T" ⋅ "m/A" ) ( 2.14 #h(0.2em) × #h(0.2em) 10^(3) "turns/m" ) ( 0.410 #h(0.2em) "A" ) \ B & = & 1.10 #h(0.2em) × #h(0.2em) 10^(−3) "T" .$] Significance This solution is valid only if the length of the solenoid is reasonably large compared with its diameter. This example is a case where this is valid. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ What is the ratio of the magnetic field produced from using a finite formula over the infinite approximation for an angle #math.equation(block: false, alt: "θ")[$θ$] of (a) #math.equation(block: false, alt: "85 ° ?")[$85 "°" "?"$] (b) #math.equation(block: false, alt: "89 ° ?")[$89 "°" "?"$] The solenoid has 1000 turns in 50 cm with a current of 1.0 A flowing through the coils #solutionbox[ a. 1.00382; b. 1.00015 ] ] === Toroids A toroid is a donut-shaped coil closely wound with one continuous wire, as illustrated in part (a) of . If the toroid has #emph[N] windings and the current in the wire is #emph[I], what is the magnetic field both inside and outside the toroid? #figure(figph[Figure A shows a toroid that is a coil wound into a donut-shaped object. Figure B shows a loosely wound toroid that does not have cylindrical symmetry. Figure C shows a tightly wound toroid with the symmetry that is very close to the cylindrical. Figure D shows several paths. Paths D1 and D3 are external to the toroid. Path D2 lies within the toroid.], alt: "Figure A shows a toroid that is a coil wound into a donut-shaped object. Figure B shows a loosely wound toroid that does not have cylindrical symmetry. Figure C shows a tightly wound toroid with the symmetry that is very close to the cylindrical. Figure D shows several paths. Paths D1 and D3 are external to the toroid. Path D2 lies within the toroid.", caption: [(a) A toroid is a coil wound into a donut-shaped object. (b) A loosely wound toroid does not have cylindrical symmetry. (c) In a tightly wound toroid, cylindrical symmetry is a very good approximation. (d) Several paths of integration for Ampère’s law.]) We begin by assuming cylindrical symmetry around the axis #emph[OO]’. Actually, this assumption is not precisely correct, for as part (b) of shows, the view of the toroidal coil varies from point to point (for example, #math.equation(block: false, alt: "P sub 1 , P sub 2 ,")[$P_(1) , #h(0.2em) P_(2) ,$] and #math.equation(block: false, alt: "P sub 3")[$P_(3)$]) on a circular path centered around #emph[OO]’. However, if the toroid is tightly wound, all points on the circle become essentially equivalent \[part (c) of \], and cylindrical symmetry is an accurate approximation. With this symmetry, the magnetic field must be tangent to and constant in magnitude along any circular path centered on #emph[OO]’. This allows us to write for each of the paths #math.equation(block: false, alt: "D sub 1 , D sub 2 ,")[$D_(1) , D_(2) ,$] and #math.equation(block: false, alt: "D sub 3")[$D_(3)$] shown in part (d) of , #math.equation(block: true, alt: "∮ B → times d l → equals B open parenthesis 2 π r close parenthesis .")[$display(∮ arrow(B)) · d arrow(l) = B ( 2 π r ) .$] Ampère’s law relates this integral to the net current passing through any surface bounded by the path of integration. For a path that is external to the toroid, either no current passes through the enclosing surface (path #math.equation(block: false, alt: "D sub 1")[$D_(1)$]), or the current passing through the surface in one direction is exactly balanced by the current passing through it in the opposite direction #math.equation(block: false, alt: "open parenthesis path D sub 3 close parenthesis .")[$( "path" #h(0.2em) D_(3) ) .$] In either case, there is no net current passing through the surface, so #math.equation(block: true, alt: "∮ B open parenthesis 2 π r close parenthesis equals 0")[$display(∮ B ( 2 π r ) = 0)$] and #math.equation(block: true, alt: "B equals 0 (outside the toroid) .")[$B = 0 #h(1em) "(outside the toroid)" .$] The turns of a toroid form a helix, rather than circular loops. As a result, there is a small field external to the coil; however, the derivation above holds if the coils were circular. For a circular path within the toroid (path #math.equation(block: false, alt: "D sub 2")[$D_(2)$]), the current in the wire cuts the surface #emph[N] times, resulting in a net current #emph[NI] through the surface. We now find with Ampère’s law, #math.equation(block: true, alt: "B open parenthesis 2 π r close parenthesis equals μ sub 0 N I")[$B ( 2 π r ) = μ_(0) N I$] and #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "B equals the fraction μ sub 0 N I over 2 π r (within the toroid) .")[$B = frac(μ_(0) N I, 2 π r) #h(1em) "(within the toroid)" .$] ] The magnetic field is directed in the counterclockwise direction for the windings shown. When the current in the coils is reversed, the direction of the magnetic field also reverses. The magnetic field inside a toroid is not uniform, as it varies inversely with the distance #emph[r] from the axis #emph[OO]’. However, if the central radius #emph[R] (the radius midway between the inner and outer radii of the toroid) is much larger than the cross-sectional diameter of the coils #emph[r], the variation is fairly small, and the magnitude of the magnetic field may be calculated by where #math.equation(block: false, alt: "r equals R .")[$r = R .$] === Summary - The magnetic field strength inside a solenoid is #math.equation(block: true, alt: "B equals μ sub 0 n I (inside a solenoid)")[$B = μ_(0) n I #h(1em) "(inside a solenoid)"$] where #emph[n] is the number of loops per unit length of the solenoid. The field inside is very uniform in magnitude and direction. - The magnetic field strength inside a toroid is #math.equation(block: true, alt: "B equals the fraction μ sub o N I over 2 π r (within the toroid)")[$B = frac(μ_(o) N I, 2 π r) #h(1em) "(within the toroid)"$] where #emph[N] is the number of windings. The field inside a toroid is not uniform and varies with the distance as 1/#emph[r]. === Conceptual Questions Is the magnetic field inside a toroid completely uniform? Almost uniform? Explain why #math.equation(block: false, alt: "B → equals 0")[$arrow(B) = 0$] inside a long, hollow copper pipe that is carrying an electric current parallel to the axis. Is #math.equation(block: false, alt: "B → equals 0")[$arrow(B) = 0$] outside the pipe? #solutionbox[ If there is no current inside the loop, there is no magnetic field (see Ampère’s law). Outside the pipe, there may be an enclosed current through the copper pipe, so the magnetic field may not be zero outside the pipe. ] === Problems A solenoid is wound with 2000 turns per meter. When the current is 5.2 A, what is the magnetic field within the solenoid? #solutionbox[ #math.equation(block: true, alt: "B equals 1.3 times 10 to the power −2 T")[$B = 1.3 #h(0.2em) × #h(0.2em) 10^(−2) "T"$] ] A solenoid has 12 turns per centimeter. What current will produce a magnetic field of #math.equation(block: false, alt: "2.0 times 10 to the power −2 T")[$2.0 #h(0.2em) × #h(0.2em) 10^(−2) "T"$] within the solenoid? If a current is 2.0 A, how many turns per centimeter must be wound on a solenoid in order to produce a magnetic field of #math.equation(block: false, alt: "2.0 times 10 to the power −3 T")[$2.0 #h(0.2em) × #h(0.2em) 10^(−3) "T"$] within it? #solutionbox[ roughly eight turns per cm ] A solenoid is 40 cm long, has a diameter of 3.0 cm, and is wound with 500 turns. If the current through the windings is 4.0 A, what is the magnetic field at a point on the axis of the solenoid that is (a) at the center of the solenoid, (b) 10.0 cm from one end of the solenoid, and (c) 5.0 cm from one end of the solenoid? (d) Compare these answers with the infinite-solenoid case. #figure(figph[Figure A is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, opposite to the second winding. Figure B is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, between the first and the second winding. Figure C is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, five centimeters below the first winding.], alt: "Figure A is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, opposite to the second winding. Figure B is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, between the first and the second winding. Figure C is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, five centimeters below the first winding.", caption: none) Determine the magnetic field on the central axis at the opening of a semi-infinite solenoid. (That is, take the opening to be at #math.equation(block: false, alt: "x equals 0")[$x = 0$] and the other end to be at #linebreak() #math.equation(block: false, alt: "x equals ∞ .")[$x = ∞ .$]) #solutionbox[ #math.equation(block: true, alt: "B equals the fraction 1 over 2 μ sub 0 n I")[$B = frac(1, 2) μ_(0) n I$] ] By how much is the approximation #math.equation(block: false, alt: "B equals μ sub 0 n I")[$B = μ_(0) n I$] in error at the center of a solenoid that is 15.0 cm long, has a diameter of 4.0 cm, is wrapped with #emph[n] turns per meter, and carries a current #emph[I]? A solenoid with 25 turns per centimeter carries a current #emph[I]. An electron moves within the solenoid in a circle that has a radius of 2.0 cm and is perpendicular to the axis of the solenoid. If the speed of the electron is #math.equation(block: false, alt: "2.0 times 10 to the power 5 m/s ,")[$2.0 #h(0.2em) × #h(0.2em) 10^(5) "m/s" ,$] what is #emph[I]? #solutionbox[ 0.0181 A ] A toroid has 250 turns of wire and carries a current of 20 A. Its inner and outer radii are 8.0 and 9.0 cm. What are the values of its magnetic field at #math.equation(block: false, alt: "r equals 8.1 , 8.5 ,")[$r = 8.1 , 8.5 ,$] and #math.equation(block: false, alt: "8.9 cm?")[$8.9 #h(0.2em) "cm?"$] A toroid with a square cross section 3.0 cm #math.equation(block: false, alt: "times")[$#h(0.2em) × #h(0.2em)$] 3.0 cm has an inner radius of 25.0 cm. It is wound with 500 turns of wire, and it carries a current of 2.0 A. What is the strength of the magnetic field at the center of the square cross section? #solutionbox[ 0.0008 T ]