Presentation on theme: "Static Magnetic Fields"— Presentation transcript:
1 Static Magnetic Fields Simple observationsBiot-Savart Law & ExamplesAmpere’s Law & ExamplesAmpere’s Law in point formThe CurlStoke’s Theorem & ExamplesMaxwell’s Equations for Static FieldsMagnetic Vector Potential
2 Experimental - Magnetic Forces Between Currents Magnetic forces arise from charges in motion. Forces between current-carrying wires help determinewhat magnetic force field should look like:3 easily-observed situations:How do we describe field around wire 1 that can be used to determine force on wire 2?“Fields” (like E field) break the problem into two parts.
3 Note B = H in free space, similar to D =εoE. The Magnetic FieldWire 1 creates field H which circulates around 1 by Right-Hand Rule 1(Right thumb in direction of current, fingers curl in direction of H)Wire 2 interacts with field H to produce force by Right-Hand Rule 2(Hand in direction I2, then H, thumb points in direction of force)Examine 3 cases:I1 up, I2 up, force attractiveI1 up, I2 down, force repulsiveI1 up, I2 into plane, no forceNote B = H in free space, similar to D =εoE.
4 Biot-Savart Law Magnetic field contribution dH created by “point source” current element dL.NoteInverse-square distance dependenceCross product yields vector pointing into pageSimilarity with Coulomb’s Law >>Units H are [A/m]
5 Magnetic Field From Complete Current Loop At point P, the magnetic field from differential current element IdL isTo determine total field at P from closed circuitpath, sum contributions from current elementsover entire loop
6 Example 1 - H around Long Wire Evaluate magnetic field H on y axis (or xy plane) from infinite current filament along z axis.Vector from source to observation point:𝜌 𝒂 𝝆 = 𝑧 ′ 𝒂 𝒛 +𝑹𝑹=𝜌 𝒂 𝝆 − 𝑧 ′ 𝒂 𝒛Unit vector from source to observation point:Biot-Savart becomes:(into page by RHR)
7 Example 1 - H around Long Wire II Integrating over entire wire:Using cross products𝑎 𝑧 × 𝑎 𝜌 = 𝑎 𝜑𝑎 𝑧 × 𝑎 𝑧 =0
8 Example 1 – H around Long Wire III Current into page.Magnetic field streamlinesconcentric circlesdecrease with inverse distance from the z axisEnd view of wireAmpere’s law near long wire𝐻= 𝐼 2𝜋𝜌 𝒂 𝝆
9 Example 2 - H from Finite Current Segment Field is found in xy plane at Point 2. Biot-Savart integral is taken over finite wire length:Which simplifies to (Problem 7.8):
10 Example 3 – H for right-angle segments e.g. motor winding?
11 Example 4 - H from Current Loop Biot-Savart Law:Vector from source to observation point:𝑧 𝑜 𝒂 𝒛 =𝑎 𝒂 𝝆 +𝑹𝑹= 𝑧 𝑜 𝒂 𝒛 −𝑎 𝒂 𝝆Current length element:
12 Example 4 - H from Current Loop II Substituting R and ar in Biot-Savart Law:Carrying out cross products:Substitute for of angular-dependent radial unit vector:radial components not integrate to zero, only z-component remains.
13 Example 4 - H from Current Loop III Only z component remains, integral evaluates to:Numerator is product of current and loop area.We define magnetic moment as:
14 Two- and Three-Dimensional Currents For surface carrying uniform current densityK [A/m], current within width b is:So the differential current quantity is:And Biot-Savart law over 2D surface is:And Biot-Savart law over 3D surface (plus depth) is:
15 Ampere’s Circuital Law Ampere’s Circuital Law states that the line integral of H around any closed pathis equal to the current enclosed by that path.Line integral of H around closed paths a and b gives total current I,integral over path c only gives portion of current that lies within cPractical use requires knowledge of symmetry of path
16 Example 1 - Ampere’s Law Applied to Long Wire Symmetry suggests H will be circular, constant-valuedat constant radius, and centered on current (z) axis.Choosing path a, and integrating H around circleof radius gives enclosed current,I:Same as Biot-Savart Law.
17 Example 2 - Ampere’s Law for Coaxial Transmission Line Two concentric conductors carry equal and oppositecurrents, I.Line assumed to be infinitely long, and circular symmetrysuggests H will be entirely - directed,and vary only with radius .Four RegionsField within inner conductorField between conductors (same as long wire)Field within outer conductorField outside both conductors (zero, since net enclosed current zero)
18 Example 2 - Field Within Inner Conductor Current distributed uniformly inside conductors, the H assumed circular everywhere.Ampere’s Law inside inner conductor at radius :Current enclosed isCombining
19 Example 2 - Field between Conductors a < < bAs with long straight wire:Result:
20 Example 2 - Field Inside Outer Conductor Inside outer conductor, the enclosed current consists of the inner conductor current plus that portion of the outer conductor current at radii less than Ampere’s Circuital Law becomesSo H is:
21 Example 2 - Field Outside Both Conductors Outside the transmission line no current is enclosed by the integration path, soThe current is uniform with circular symmetry over the integration path, and thus must be 0:Applications:Coaxial line(Twisted pair)
22 Example 2 - Field over entire Radius of Coax Line Combining previous results, and assigning dimensions as in the inset below:
23 Example 3 - Ampere’s Law for Current Sheet Uniform plane current in y direction, H should be x-directed from RHR and symmetry.No Hy in direction of current (RHR)No Hz since overlapping filament components cancel (RHR)Applying Ampere’s Law to path 1 - 1’ – 2 - 2’.Thus magnetic field is discontinuous across current sheet by magnitude of the surface current density.
24 Example 3 - Ampere’s Law for Current Sheet II If loop 1 – 1’ – 3 – 3’ is outside current plane:1. By symmetry, field magnitude above sheet must besame as field magnitude below sheet2. Also from previous page:so field is constant outside current planeCombining 1 and 2Half the magnetic field / surface current discontinuity is on each side of the current sheet.
25 Example 3 - Ampere’s Law for Current Sheet III Magnetic field above current sheet is equal and opposite to field below sheet.Field in either region written as cross product:where aN is unit vector normal to current sheet,and points into region where field is evaluated.
26 Example 4 - Ampere’s Law for Solenoid Applying Ampere’s Law to rectangular path Δz long through side of solenoid:Where paths DA and BC are radially in and out, and CD is parallel at a great distance.N/d is number of turns per/length.
27 Example 4 - Ampere’s Law for Solenoid II Paths BC and DA are oppositely-directed and cancel, and path CD is evaluated at great distance where H is zero.𝐻 𝑧 = 𝑁𝐼 𝑑 𝒂 𝒁Where N/d is number of turns per/length, and (N/d)IΔz is the total current through the path.𝐻 𝑧 ∆𝑧= 𝑁𝐼 𝑑 ∆𝑧The field is thus
28 Example 5 - Ampere’s Law for Toroid A toroid is a doughnut-shaped set of windings around a core material. A cross-section with inner radius (ρo – a) and outer radius (ρo + a) is shown below.The windings are modeled as N individual current loops, each of which carries current I
29 Example 5 - Ampere’s Law for Toroid II Ampere’s Law is applied by taking a line integral around the circular path C at radius By symmetry H is assumed to be circular and a function of radius only:Ampere’s Law takes the form:Result:Performing line integrals in regions ρ < (ρo - a) and ρ > (ρo + a) enclose no net current, and lead to no magnetic field
30 Ampere’s Law in Point Form Consider magnetic field H at center of a small closed loop.We approximate field over closed path by extrapolating H to each of 4 sides.This will be the point form of Ampere’s Law
31 Line Integral H∙ΔL Along Front Segment Line integral along front segment 1-2:Extrapolating H to front segment:How the y component is changing as you move in the x direction 𝜕 𝐻 𝑦 𝜕𝑥 shearCombining 2 terms:
32 Line Integrals along Front and Back Segments The contribution from front side 1-2 is:The contribution from back side 3-4 is:Note signs used in extrapolating H to front and back,and in evaluating line integral direction.
33 Line Integrals along Side Segments The contribution from right side 2-3:The contribution from left side 4-1:Note signs used in extrapolating H to right and left,and in evaluating line integral direction.
34 Line Integral from entire Closed Loop The total integral is now the sum:Combining previous results:
35 Entire Line Integral related to Current Density Complete line integral now equated to total current passing through loop in z direction Jz ΔxΔy byby Ampere’s Law.Dividing by loop area gives:Expression becomes exact as Δx, Δy → 0
36 Line Integral in Other Loop Orientations Similar results can be obtained with the rectangular loop in the other two orthogonal orientations:Loop in yz plane:Loop in xz plane:Loop in xy plane:This gives all three components of current density field.
37 Ampere’s Law in Point Form Using the Definition of the Curl operator 𝜵×This is Ampere’s Circuital Law in point form. (for static fields)Adding all 3 components and loop orientations
38 Curl in Rectangular Coordinates Assembling the results of the rectangular loop integration exercise, we find the vector fieldthat comprises curl H:An easy way to calculate this is to evaluate the following determinant:which we see is equivalent to the cross product of the del operator with the field:
39 General - Curl of Vector Field In general, curl of vector field 𝛻×𝐻 is another field normal to original field.The curl component in the direction N, normal to the plane of the integration loop is:Direction of N uses right-hand rule: With right-hand fingers oriented in direction ofpath integral, thumb points in the direction of normal (the curl).
43 Visualization of CurlConsider placing a small “paddle wheel” in a flowing stream of water, as shown below. The wheelaxis points into the screen, and the water velocity decreases with increasing depth.The wheel will rotate clockwise, and give a curl component that points into the screen (right-hand rule).Positioning the wheel at all three orthogonal orientations yields measurements of all 3components of curl. Note the curl is directed normal to both the field and thedirection of its variation.
44 Stoke’s Theorem - Add Individual Curls Surface S is partitioned into sub-regions, each of small area ΔSLine integral around each ΔS is:Summing path integrals and curls:
45 Stoke’s Theorem – Cancel Internal Paths .Cancellation here:Add curl contributions from all ΔS elements,and note adjacent path integrals cancel!only contribution to overall path integral isaround outer periphery of surface S.No cancellation here:Result is Stoke’s TheoremThis is path integral of Hover outer perimeter asinterior paths cancelThis is integral of curlof H over surface S
46 Summary - Two Theorems Stoke’s Theorem – Chapter 7 Surface integral = Volume integral(Divergence)𝑆 𝑫∙𝒅𝑺 = 𝑣𝑜𝑙 𝜵∙𝑫 𝑑𝑣Stoke’s Theorem – Chapter 7Divergence Theorem – Chapter 3Line integral = Surface integral(Curl)A divergence is a 3d volume derivative going between opposite surfaces, a curl is a 2d shear derivative going around a circle
48 Example 1 – Stoke’s Calculation II 𝐻=6𝑟 𝑠𝑖𝑛𝜑 𝒂 𝒓 +18𝑟 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜑 𝒂 𝝋2nd and 5th curl term zero, as no HΘ6th term zero, as Hr does not involve ΘOnly 1st, 3rd, and 4th remain
49 Example 2 – Ampere’s Integral Form Begin with Ampere’s Law in point form (static fields):Integrate both sides over surface S:Left and right equal by Stokes’ Theorem.The center term is just net current through surface S.Equating to the middle
50 Example 3 - 3rd Maxwell’s Equation Already know for static electric field:Integrand must be zero:Thus conservative field has zero curl.(static fields)Note: when − 𝜕𝐵 𝜕𝑡 is added to right hand side, this becomes Faraday Induction!Using Stoke’s Theorem:𝐸∙𝑑𝐿=0= 𝑆 𝜵×𝑬
51 Example 4 – Vector Identity Prove 𝜵∙𝜵×𝑨=𝟎 *Show 𝛻∙𝛻×𝐴 is a scalar (divergence of anything is scalar)Show integral 𝑣𝑜𝑙 (𝛻∙𝛻×𝐴) 𝑑𝑣=0If integral is zero, and integrand is scalar, then integrand must be zero*Can also put in expressions for curl and divergence in 3 coordinate systems,and crank it through!
52 Example 4 – Vector Identity II By Divergence theorem𝑣𝑜𝑙 (𝛻∙𝛻×𝐴) 𝑑𝑣= 𝑆 (𝛻×𝐴) 𝑑𝑆for closed surface SBy Stoke’s Theorem𝑆 𝛻×𝐴 𝑑𝑆= 𝑙𝑜𝑜𝑝 𝐴∙𝑑𝐿 =0for loop bounding closed SIf surface S is closed then loop bounding surface is zero.If integral is zero, and integrand is scalar, then integrand must be zero𝜵∙𝜵×𝑨=0
53 Example 5 – Steady-state Current Ampere’s Law𝛻×𝐻=𝐽Take divergence of both sides𝛻∙𝛻×𝐻=𝛻∙𝐽=0Thus current must follow eqn. of continuity𝛻∙𝐽=0
54 Magnetic Potential? Magnetic Scalar Potential? 𝑯=−𝛻 𝑉 𝑚 Taking Curl 𝜵×𝑯=𝑱=𝛻× −𝛻 𝑉 𝑚 ≡0Current must be zero (Not much use)Magnetic Vector Potential𝑯= 1 𝜇 𝑜 𝜵×𝑨𝜵×𝑯=𝑱=𝛻×𝛻×𝐴≠0No such restriction.
55 Magnetic Vector Potential Then Divergence of B is identically zero:Which is the 4th Maxwell Equation – Gauss’s Law for Magnetism - no free magnetic poles:Define B and H in terms of magnetic vector potential A:And Ampere’s Law is:Which is NOT identically zero𝛻∙𝑩=𝛁∙𝛁×𝑨≡0
56 4th Maxwell Equation 𝑆 𝐵 ∙𝑑𝑆=0 May rewrite using Divergence Theorem:Thus the integrand is zeroThis result is known as Gauss’ Law for the magnetic field in point form.Since no free magnetic poles, integral of B over closed surface is zero:𝑆 𝐵 ∙𝑑𝑆=0
57 Maxwell’s Equations for Static Fields We have now completed the derivation of Maxwell’s equations in point form for no time variation:Gauss’ Law for Electric FieldsConservative property of static electric fields (needs changing B field)Ampere’s Circuital Law (needs displacement current)Gauss’ Law for Magnetic FieldsIn free space:2 additional terms are needed when the fields vary with time, which is another course.
58 Maxwell’s Equations for Static Fields II Maxwell’s Equations in integral form for static fields :Gauss’ Law for Electric FieldsConservative property of static electric fields(needs changing magnetic field)Ampere’s Circuital Law(needs displacement current)Gauss’ Law for Magnetic Fields
59 Expressions for Potential Consider a differential elements, shown here. On the left is a point charge representedby a differential length of line charge. On the right is a differential current element. The setupsfor obtaining potential are identical between the two cases.Line ChargeLine CurrentScalar Electrostatic PotentialVector Magnetic Potential
60 Vector Potential Example For differential current element:Evaluated at point P:Taking curl in cylindrical coordinates:Same as Biot-Savart Law
61 General Expressions for Vector Potential For large-scale charge or current distributions, we sum differential contributions by integrating over charge or current:The closed path integral indicates current mustclose on itself to form complete circuit.For surface or volume current distributions we have:Similar to scalar electric potential.
62 Magnetic Poisson’s Equation Start with:Vector identity defines the vector Laplacian:It can be shown that (Sec. 7.7):This gives:Counterpart to Poisson’s equation
63 Direction of A Magnetic Poisson’s equation In rectangular coordinates: Equation separates to give:Direction of A is same as current to which it is associated.(not simple inother coordinate systems)The vector field A is sometimes described as “fuzzy image”of its generating current.