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**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

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**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.

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**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.

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**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]

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**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

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**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)

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**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

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**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

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**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):

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**Example 3 – H for right-angle segments**

e.g. motor winding?

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**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:

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**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.

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**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:

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**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:

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**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

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**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.

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**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)

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**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

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**Example 2 - Field between Conductors**

a < < b As with long straight wire: Result:

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**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:

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**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)

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**Example 2 - Field over entire Radius of Coax Line**

Combining previous results, and assigning dimensions as in the inset below:

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**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.

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**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.

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**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.

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**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.

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**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

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**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

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**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

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**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

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**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:

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**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.

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**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.

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**Line Integral from entire Closed Loop**

The total integral is now the sum: Combining previous results:

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**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

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**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.

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**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

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**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:

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**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).

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**Example – Curl in Rectangular Coordinates**

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**Curl in Other Coordinate Systems**

Cylindrical coordinates Spherical coordinates

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**2 of 4 Maxwell’s Equations**

Gauss’s Law 𝛻∙𝐃= ρ v Ampere’s Law (static fields)

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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.

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**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:

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**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

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**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

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**Example 1 – Stoke’s Calculation**

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**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

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**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

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**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

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**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!

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**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

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**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

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**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.

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**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

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**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

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**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.

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**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

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**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

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**Vector Potential Example**

For differential current element: Evaluated at point P: Taking curl in cylindrical coordinates: Same as Biot-Savart Law

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**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.

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**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

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**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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30.5 Magnetic flux 30. Fig 30-CO, p.927

30.5 Magnetic flux 30. Fig 30-CO, p.927

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