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CHAPTER 5 11/16/2016  A brief history of magnetostatics MAGNETOSTATICS

 The Law Magnetostatics Chapter 5

 The concept of current Lee Chow  The Biot-Savart Law Department of University of Central Florida  The divergence and of B Orlando, FL 32816  The magnetic vector potential

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A brief history of magnetostatics In 1820, Hans Oersted discovered the

11/16/2016 relationship between and 11/16/2016 , namely a current The Chinese compass was invented carrying wire will deflect the needle of a around 4th century BC. Chapter Magnetostatics Chapter 5 compass. Magnetostatics Chapter 5 In the same year, Jean-Baptiste Biot and Felix Savart found that the magnetic In 1600, William Gilbert field varies inversely with the distance published “De Magnete”, one of from the wire and curl around the the first book on current. The equation describing the and . Gilbert was magnetic field generated by a current regarded by some as the father bears their names. of electricity and magnetism. ′ 3 4

Ampere learned about Oersted’s result in Then in 1831, proposed 1820 and started to develop a theory to that change in the can also understand the relationship of electricity 11/16/2016 produce ----Faraday’s law of 11/16/2016 and magnetism. Ampere showed that two induction. The “field line” concept also parallel wires carrying current will attract came from Faraday. Chapter Magnetostatics Chapter 5 Magnetostatics Chapter 5 or repel each other depending on whether the currents are in the same or opposite direction. ·

In 1826, Ampere discovered the Ampere’s law. James Maxwell in 1864 published his Maxwell’s equations which combined Gauss’s law, Faraday’s law, and the · modified Ampere’s law. Maxwell coined the phrase “the ” for his theory. 5 6

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Let’s put things into perspective. Maxwell Equations

11/16/2016 Electrons were discovered by J. J. 11/16/2016 Thomson in 1897. He found that the • ∮ · · cathode rays in a cathode ray tube Magnetostatics Chapter 5 travel further in air than ions. He Magnetostatics Chapter 5 estimated that the mass of cathode rays • ∮ · · to be at least 1000 times lighter than hydrogen. • ∮ · In 1887, Michelson and Morley performed the experiment bears their • ∮ · name. They show that there is no “aether” and pave the way for the development of . 7 8

The Lorentz force law Example 5.2 Cycloid motion A uniform magnetic field is in the x-direction while

When a charge q moving in a magnetic field , it 11/16/2016 11/16/2016 experiences a force called the Lorentz force another uniform electric field is in the z-direction as shown. Find the trajectory of a charge particle starting at

Chapter Magnetostatics Chapter 5 rest at the origin. Magnetostatics Chapter 5 · Initially at rest at the origin, the charge will experience a force in the z- Since the Lorentz force is always perpendicular to the velocity, therefore no work done by Lorentz force. direction , so initial , , velocity will be in the z- direction. The Lorentz If both E and B are present force will move the charge in the y-direction. 9 10

The solution to the previous equation can be ̂ = ̂

11/16/2016 11/16/2016 y(t) = Substitute back into (2), we arrive at

Magnetostatics Chapter 5 Magnetostatics Chapter 5 Let , (1) sin Apply the initial conditions, at t = 0, , . Take derivative of (1) and then substitute into (2), then and integrate over time once, we end up and

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Current &

11/16/2016 Current in a wire (1D) is defined as 11/16/2016 the amount of charges passing through a point per unit time. Let Magnetostatics Chapter 5 Magnetostatics Chapter 5 Ampere = Coulomb per second If line is and velocity is V, R This is the equation for a circle of radius Since the current is almost always confined in a R with the center at (Rωt, R) conducting wire, we can use the direction of the wire to (Rωt, R) indicate the direction of current 13 14

A 3-D current model (Ideal gas model) The concept of Lorentz force on a “charge” can be extended to “current”, since “current” can be viewed as “charge

11/16/2016 density” times velocity. 11/16/2016

Chapter Magnetostatics Chapter 5 Magnetostatics Chapter 5

Since direction of I is the same as dl, and I is usually a scalar constant,

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Example 5.3 When the loop starts to rise, the charge no A rectangular loop of wire, supporting a mass m, hanging longer moving horizontally, it acquires a vertically with one end in a uniform magnetic field as shown. vertical component u, so also tilts left What current would be needed to balance the gravitational force? 11/16/2016 11/16/2016 as shown. is still perpendicular to , no work done on charge q. The vertical component, qwB which can be written as Magnetostatics Chapter 5 Magnetostatics Chapter 5 The horizontal force, , so the work done by What happens if we increase the current? battery to overcome is

The magnetic force now will be larger λ than gravitational force mg, and the mass will rise.

This is similar to sliding a block Does that means the magnetic force up a frictionless ramp, pushing actual do work? 17 18 horizontally

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When charge flows on the surface of a conductor (two dimensional), From definition of current density we describe it by the “surface current density” K as · 11/16/2016 11/16/2016 They are From divergence theorem, we have different Chapter Magnetostatics Chapter 5 Chapter Chapter Magnetostatics 5 ⋅·

When the flow of charge is 3D, we describe it by the volume From conservation of charge, the amount of charge leaving a volume current density, · · = · · ·· Combine with previous equation, we have For magneto- · 19 · 20

The magnetic field produced by a steady current, The Biot-Savart Law 11/16/2016 ′ 11/16/2016 Steady current is the source that produce magnetic field in magnetostatics. It plays exactly the same role as stationary charge in . The Biot-Savart law Chapter Magnetostatics Chapter 5 where is the permeability of the free space. Magnetostatics Chapter 5 plays the exactly same role in magnetostatics as the Coulomb’s law which links the source to the field Electrostatics Magnetostatics . ⁄ (Speed of light) Source Stationary charges Steady current Equation Coulomb’s Law Biot-Savart Law The unit of magnetic field Field* Diverge from a point Curl around a line (MKS unit, or SI unit) · Field*---Electric field produced by one point charge, or magnetic field produced by a current on a straight wire. 21 (cgs unit) 22 ·

Example 5.5 Find the magnetic field a distance s from a long straight wire · · carrying a steady current I. 11/16/2016 11/16/2016 The Magnetic field is given by · Chapter Magnetostatics Chapter 5 Magnetostatics Chapter 5 ′ For infinite straight line, ′ ·α · Now if we have two long parallel wires carrying current · , · I and I the force between them is 1 2,

and 23 24

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= The divergence and Curl of

Force per unit length is given by 11/16/2016 For a straight wire carrying a current I, the magnetic field 11/16/2016 at a distance R is given by Magnetostatics Chapter 5 Magnetostatics Chapter 5 Example 5.6 Find the magnetic field above the center of a The line of this magnetic field is circular loop. · ′ This is Ampere’s law. This is true for any shape of closed loop around the current. 25 26

Formal derivation of and · We can show that in cylindrical coordinates We start out with the Biot-Savart Law 11/16/2016 11/16/2016 , ′ Substitute into the equation below Magnetostatics Chapter 5 Magnetostatics Chapter 5 · · ··· · · Apply product rule (4) on page 21 of Griffiths to eq. (1) above From Stoke’s theorem and definition of current density 0 0 · · · · · · Ampere’s Law in Page 57 differential form 27 · 28

First, we can switch from to and change a sign as below Next we’ll take the curl of the Biot-Savart law, ′

11/16/2016 11/16/2016 ′ · ·′

The integrant can be re-arranged Magnetostatics Chapter 5 Next, we will look at the x-component of the above equation Magnetostatics Chapter 5 and use integration by parts () 0 0 0 0 · · · · ′ · · ′ ·

· · The second term on the right hand side goes to zero can be seen on next page. 29 (J is zero at large boundary) 30

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Applications of Ampere’s Law Example 5.8 Find the magnetic field of an infinite uniform surface current flowing over the xy plane. The Ampere’s law can be written in integral form 11/16/2016 11/16/2016

The way we solve this problem is

· Magnetostatics Chapter 5 very similar to the infinite Magnetostatics Chapter 5 uniform surface charge problem of Example 2.4 on page 73. Ampere’s Law in magnetostatics plays exactly the same role as the Gauss’s Law in electrostatics. But Ampere’s Apply Ampere’s Law, we have law is a line integral while the Gauss’s law is a surface integral. ·

/ Electrostatics Coulomb’s Law Gauss’s Law Magnetostatics Biot-Savart Law Ampere’s Law 31 32

Example 5.9 Find the magnetic field of a very long solenoid Magnetostatics and Electrostatics consisting of n closely wound turns per unit length.

(a) First we argue that there is no magnetic field in the 11/16/2016 11/16/2016 radial direction. · (b) Second we argue that there is no magnetic field in the

azimuthal direction (. Magnetostatics Chapter 5 Magnetostatics Chapter 5 (c) Third, we argue that outside the solenoid the magnetic Magnetostatics Electrostatics field is equal to zero. · · (d) Last we argue that the magnetic field inside the

solenoid is uniform and in the z-direction. (e) Use Ampere’s Law, we found the field inside to be · · 33 34

Magnetic Vector Potential Re-arrange the previous equation, we can see that

From the fact that ·, we can set up a vector 11/16/2016 11/16/2016 ′ potential, such that Chapter Magnetostatics Chapter 5 We let the expression in the parenthesis equal to the vector Magnetostatics Chapter 5 potential of the field. For comparison, , To find the expression for , we start from Biot-Savart law, V(r) is also shown below. ′ ′ ′ = ′ If we add a gradient term to , and let , the 0 magnetic field will remain the same as shown below. 0 ′ 35 ′ 36

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The situation here is similar to the electrostatics case, This means that if is not divergence-free, then we add a where add a constant to the potential does not change the to it until it is divergence-free. To find out the exact electric field at all. functional form of , we can either solve the Poisson 11/16/2016 11/16/2016 In magnetostatics, adding a gradient term to the vector equation at the end of last page, or we can explore the potential ′, is called gauge transformation. similarity between electrostatic and eq. on page 33.

Chapter Magnetostatics Chapter 5 · Magnetostatics Chapter 5 What this means is that Ampere’s Law becomes It can be used to choose a particular vector potential. For 0 example, we can choose a vector potential that is divergence-free. · · Coulomb Gauge · This will lead to · 37 We reduce the magnetostatics problem into solving 38 Laplace equation, just like in the electrostatics case.

Example 5.11 A spherical shell of radius R, carrying a uniform surface charge σ, is set spinning at angular velocity ′ ′

ω. Find the vector potential at point r. 11/16/2016 11/16/2016 Rotate the axis such that the direction is the same as the ′ ′ ′ . The surface current is given by , and is the Magnetostatics Chapter 5 Magnetostatics Chapter 5 velocity of the shell, = ′ ′ The is pointing at an angle ψ from the direction. ′

The integration is over the surface of the sphere

′ ′ 39 40

And ′is independent of ′, So the integral becomes

so all integrations involving sin′ or cos′ will be zero, 11/16/2016 11/16/2016 and the only term left is ′

Chapter Magnetostatics Chapter 5 Magnetostatics Chapter 5 (for convenience we drop the prime notation)

·

For R > s (inside the shell), equation (1) becomes

Let , and 41 42

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For R < s, (outside of the shell), equation (1) becomes Since ·, we can re-arrange the equations 11/16/2016 11/16/2016 (inside) Chapter Magnetostatics Chapter 5 Magnetostatics Chapter 5 (outside) So the vector potential becomes Now we convert back to the original coordinates with , and , and , and (inside) (outside) 43 44

The magnetic field inside and outside of the shell can be In Coulomb gauge, we have In magnetostatics, we have derived from the vector potential

11/16/2016 11/16/2016 · · Magnetostatics Chapter 5 Magnetostatics Chapter 5 Since the equations on the left, are almost identical to the equation on the right side, we can use the functional form of Biot-Savart law to express the vector potential as follow. Similarly the magnetic field outside can be found

′ ′ (This is true in Coulomb gauge only) Biot-Savart Law Compare to eq. 3.103, we can see that it has the form as an field 45 46 due to an .

Summary: Magnetostatic Boundary conditions

In summary, we show the relationship between source (J),

11/16/2016 Start with ·, we look at the normal 11/16/2016 Field (B) and the vector potential (A) in magnetostatics component of the magnetic field across a below. It is quite similar to the case of electrostatics. surface current Chapter Magnetostatics Chapter 5 Magnetostatics Chapter 5

·

Now start with , and do the line integral show on the figure to the right:

· ∥ ∥ 47 48 Missing link?

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Like the scalar potential in electrostatics, the vector potential is continuous across any boundary, because · (Coulomb gauge) guarantees that the normal components of are continuous, while means that The next 3 pages are from Chapter 2, which discuss the the tangential components of are also continuous, 11/16/2016 11/16/2016 boundary conditions across a surface charge distribution σ · · ⟹ Chapter Magnetostatics Chapter 5 Magnetostatics Chapter 5

(right at the boundary) but We can combine the two equations on page 48 Apply Gauss’s Law

This can be compared with the equation in electrostatics · 49 50

Tangential component of the E field. · · Now we turn our attention to the 11/16/2016 11/16/2016 tangential component of E field near a surface charge distribution. Magnetostatics Chapter 5 Magnetostatics Chapter 5 · From symmetry argument, the magnitude of the field ∥ ∥ ∥ ∥ above and below should be the same, but the directions = ⋅ ⋅ are different. So the parallel component of the E field is continuous across the surface charges. In general, we can write the boundary condition as follow:

The normal component of the electric field has a = (right at the boundary) 51 52 “discontinuity” across the charged boundary.

Multipole expansion of the vector potential Now let’s look at the first few terms of the previous equation. From page 33, we can see that the vector potential can be 11/16/2016 11/16/2016 expressed as For n = 0, · · ′ ′ Magnetostatics Chapter 5 Magnetostatics Chapter 5 For n = 1 From eq. 3.94, we can expand , . · · ′ For n = 2 Substitute into the expression for we end up with · · · ′ 53 This is multipole expansion for the vector potential . 54

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For n = 0, it is corresponds to the monopole term and Now let “d” represent “differential” wrt the prime coordinates

·′ ′ ·′ ′ ·′ ′ 11/16/2016 11/16/2016 Since the closed loop integral of any vector equals to zero,

This implied that magnetic monopole term equals to zero, Magnetostatics Chapter 5 Magnetostatics Chapter 5 or magnetic monopole moment does not exit. ·′ ′ For n = 1, this is the magnetic dipole term and ·′ ′ ·′ ′ ′·· and ′ · ′ ′ ·′ ′ ·′ ′

Note that prime system indicates the source system. Since 55 From vector triple product rule (Eq. 1.17, page 8) 56 is referring to current direction, so ′, but l ≠ r’

The magnetic dipole moment is defined as follow ′ ′ ·′ ′

11/16/2016 11/16/2016 Since ′ , we will use the results from above and ′ ′

Magnetostatics Chapter 5 Magnetostatics Chapter 5 ·′ For a planar loop ′ Substitute back into eq. on page 55, we have r’ dl ∮ ′ From above equation, we can see that the magnetic dipole · moment is independent of the choice of origin. This is not 57 58 As a comparison surprising because there is no magnetic monopole.

The Vector potential and magnetic field due to a magnetic dipole located at the origin can be expressed in spherical coordinates 11/16/2016 11/16/2016 Chapter Magnetostatics Chapter 5 Magnetostatics Chapter 5

Ideal Dipole Physical Dipole

59 Electric Dipole 60

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