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Static Magnetic Fields

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Presentation on theme: "Static Magnetic Fields"— Presentation transcript:

1 Static Magnetic Fields
Simple observations Biot-Savart Law & Examples Ampere’s Law & Examples Ampere’s Law in point form The Curl Stoke’s Theorem & Examples Maxwell’s Equations for Static Fields Magnetic Vector Potential

2 Experimental - Magnetic Forces Between Currents
Magnetic forces arise from charges in motion. Forces between current-carrying wires help determine what 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 Field Wire 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 attractive I1 up, I2 down, force repulsive I1 up, I2 into plane, no force Note B = H in free space, similar to D =εoE.

4 Biot-Savart Law Magnetic field contribution dH created by
“point source” current element dL. Note Inverse-square distance dependence Cross product yields vector pointing into page Similarity 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 is To determine total field at P from closed circuit path, sum contributions from current elements over 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 +𝑹 𝑹=𝜌 𝒂 𝝆 − 𝑧 ′ 𝒂 𝒛 Unit 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 products 𝑎 𝑧 × 𝑎 𝜌 = 𝑎 𝜑 𝑎 𝑧 × 𝑎 𝑧 =0

8 Example 1 – H around Long Wire III
Current into page. Magnetic field streamlines concentric circles decrease with inverse distance from the z axis End view of wire Ampere’s law near long wire 𝐻= 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 +𝑹 𝑹= 𝑧 𝑜 𝒂 𝒛 −𝑎 𝒂 𝝆 Current 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 density K [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 path is 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 c Practical use requires knowledge of symmetry of path

16 Example 1 - Ampere’s Law Applied to Long Wire
Symmetry suggests H will be circular, constant-valued at constant radius, and centered on current (z) axis. Choosing path a, and integrating H around circle of 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 opposite currents, I. Line assumed to be infinitely long, and circular symmetry suggests H will be entirely  - directed, and vary only with radius . Four Regions Field within inner conductor Field between conductors (same as long wire) Field within outer conductor Field 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 is Combining

19 Example 2 - Field between Conductors
a < < b As 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 becomes So H is:

21 Example 2 - Field Outside Both Conductors
Outside the transmission line no current is enclosed by the integration path, so The 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 be same as field magnitude below sheet 2. Also from previous page: so field is constant outside current plane Combining 1 and 2 Half 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 𝒁 Where N/d is number of turns per/length, and (N/d)IΔz is the total current through the path. 𝐻 𝑧 ∆𝑧= 𝑁𝐼 𝑑 ∆𝑧 The 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 shear Combining 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 by by 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 field that 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 of path integral, thumb points in the direction of normal (the curl).

40 Example – Curl in Rectangular Coordinates

41 Curl in Other Coordinate Systems
Cylindrical coordinates Spherical coordinates

42 2 of 4 Maxwell’s Equations
Gauss’s Law 𝛻∙𝐃= ρ v Ampere’s Law (static fields)

43 Visualization of Curl Consider placing a small “paddle wheel” in a flowing stream of water, as shown below. The wheel axis 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 3 components of curl. Note the curl is directed normal to both the field and the direction of its variation.

44 Stoke’s Theorem - Add Individual Curls
Surface S is partitioned into sub-regions, each of small area ΔS Line 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 is around outer periphery of surface S. No cancellation here: Result is Stoke’s Theorem This is path integral of H over outer perimeter as interior paths cancel This is integral of curl of H over surface S

46 Summary - Two Theorems Stoke’s Theorem – Chapter 7
Surface integral = Volume integral(Divergence) 𝑆 𝑫∙𝒅𝑺 = 𝑣𝑜𝑙 𝜵∙𝑫 𝑑𝑣 Stoke’s Theorem – Chapter 7 Divergence Theorem – Chapter 3 Line 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

47 Example 1 – Stoke’s Calculation

48 Example 1 – Stoke’s Calculation II
𝐻=6𝑟 𝑠𝑖𝑛𝜑 𝒂 𝒓 +18𝑟 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜑 𝒂 𝝋 2nd and 5th curl term zero, as no HΘ 6th term zero, as Hr does not involve Θ Only 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=0 If 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 theorem 𝑣𝑜𝑙 (𝛻∙𝛻×𝐴) 𝑑𝑣= 𝑆 (𝛻×𝐴) 𝑑𝑆 for closed surface S By Stoke’s Theorem 𝑆 𝛻×𝐴 𝑑𝑆= 𝑙𝑜𝑜𝑝 𝐴∙𝑑𝐿 =0 for loop bounding closed S If 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=𝐽 Take divergence of both sides 𝛻∙𝛻×𝐻=𝛻∙𝐽=0 Thus current must follow eqn. of continuity 𝛻∙𝐽=0

54 Magnetic Potential? Magnetic Scalar Potential? 𝑯=−𝛻 𝑉 𝑚 Taking Curl
𝜵×𝑯=𝑱=𝛻× −𝛻 𝑉 𝑚 ≡0 Current must be zero (Not much use) Magnetic Vector Potential 𝑯= 1 𝜇 𝑜 𝜵×𝑨 𝜵×𝑯=𝑱=𝛻×𝛻×𝐴≠0 No 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 zero This 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 Fields Conservative property of static electric fields (needs changing B field) Ampere’s Circuital Law (needs displacement current) Gauss’ Law for Magnetic Fields In 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 Fields Conservative 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 represented by a differential length of line charge. On the right is a differential current element. The setups for obtaining potential are identical between the two cases. Line Charge Line Current Scalar Electrostatic Potential Vector 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 must close 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 in other coordinate systems) The vector field A is sometimes described as “fuzzy image”of its generating current.

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