sign in sign up
<<
Home , Magnetization

Chapter 6 Magnetic Fields in Matter 6.1.2. Torques and on Magnetic Dipoles 6.1 Magnetization 6.1.1 Diamagnets, Paramagnets, Ferromagnets A experiences a torque in a , just as an electric dipole does in an . All the magnetic phenomena are due to electric charges in motion: Any current loop could be built up from infinitesimal rectangles, with all the “internal” side canceling. There is orbiting around nuclei magnetic dipoles Electrons spinning about their axes no actual loss of generality in using the shape. When a magnetic field is applied, a net alignment of these magnetic dipoles occurs, and the medium becomes magnetically polarized, or magnetized. The magnetic polarization M, unlike electrical polarization P, might be parallel to B (paramagnets) or opposite to B (diamagnets). Let’s calculate the torque on a rectangular current loop in A few substances (ferromagnets) retain their magnetization a uniform magnetic field. even after the external field has been removed. 1 2

Torques and Forces on Magnetic Dipoles Torques and Dipole Moment

Center the loop at the origin, and tilt it an angle θ from the N = IabBsinθxˆ = mBsinθxˆ = m×B z axis towards the y axis. Let B point in the z direction. where m = Iab is the magnetic dipole moment of the loop.

This equation is identical in form to the electrical analogy. N = p×E The torque is again in such a direction as to line the dipole up parallel to the field (). Sloping sides: the forces cancel. Horizontal sides: the forces cancel but they generate a torque. N = L×F = aF sinθxˆ QM: The Pauli exclusion principle dictates that the electrons within a given lock together in pairs with The magnitude of the on each of these segments is: opposite spins, and this effectively neutralizes the torque F = IbB on the combination. 3 4 Forces in Nonuniform Magnetic Field Forces on an Infinitesimal Current Loop and Model In a uniform field, the net force on a current loop is zero: F = ∇(m ⋅B) (= −∇U, where U = −(m⋅B)) =0 F = I (dl×B) = I (dl)×B = 0 ∫ ∫ Identical to the electrical formula F = ∇(p⋅E) In a nonuniform field this is no longer the case, because the magnetic field B could not come outside the integral. Does the magnetic dipole consist of a pair of opposite magnetic monopoles just like an electric dipole? Fringing field effect : F = 2πIRBcosθ

5 6

6.1.3. Effect of a Magnetic Field on Atomic Orbits Speeds Up or Slows Down 1 e2 v2 Electrons not only ; they also revolve around the nucleus. = m without the magnetic field. 4πε R2 e R Let’s assume theπ orbit is a circle of radius R. The current 0 looks like steady (really?π ) The centripetal force comes from two sources: the electric force and the magnetic force. e e ev Current I = = = T 2 R 2 R The negative charge 1 e2 v′2 v of the electron ′ 2 + ev B = me evR 4πε0 R′ R′ Orbital dipole moment m = − zˆ (m = IπR2 ) 2 Assume R′ ≅ R m (v′2 − v2 ) m (v′ + v)(v′ − v) ev′B = e = e R R eRB ∴∆v = (v′ − v) ≈ When B is turn on, the electron speeds up. 7 2me 8 The Dipole Moment and The 6.1.4 Magnetization

A change in the orbital speed means a change in the In the present of a magnetic field, matter becomes magnetized. dipole moment. Upon microscopic examination, it contains many tiny dipoles, 1 e2 R2 ∆m = − e∆vR = − B with a net alignment along some direction. 2 2m e Two mechanisms account for this magnetic polarization: The change in m is opposite to the direction of B. 1.Paramagnetism: the dipoles associated with the spins of In the presence of a magnetic field, each atom picks up a unpaired electrons experience a torque tending to line them up little “extra” dipole moment, and the increments are all parallel to the field. antiparallel to the field. This is the mechanism responsible 2. Diamagnetism: the orbital speed of the electrons is altered for diamagnetism. in such a way as to change the orbital dipole moment in the This is a universal phenomenon, affecting all , but it is direction opposite to the field. typically much weaker than paramagnetism. We describe the state of magnetic polarization by the vector quantity: 9 M ≡ magnetic dipole moment per unit volume. 10

6.2 The Field of a Magnetized Object Vector potential and Bound Currents 6.2.1 Bound Currents Can the equation be expressed in a more illuminating form, Suppose we have a piece of magnetized as in the electrical case? Yes! material (i.e. M is given). What field does By exploiting the identity, ∂∂∂ 1 this object produce? ()xyzˆˆˆ′′′++ 222 1 rˆ ∂∂∂xyz′′′()()()xx−+−+−′′′ yy zz ∇′ = xyˆˆˆ′′′′′′()()()xx−+ yy −+ z zz − ˆ The vector potential of a single dipole m is 2 ==r r r 2223/22 (()(xx−+−+−′′′π yy )( zz )) r µ0 m×rˆ A(r) = µ0 1 4π 2 The vector potential is A(r) = ∫ M(r′)×(∇′ )dτ ′ r µ 4 r Using the product rule ∇×( fA) = ∇f × A + f (∇× A) In the magnetized object, each volume element carries a π dipole moment Mdτ’, so the total vector potential is and integrating by part, we have µ τ  1 M(r′)  π A(r) = π0  [∇′×M(r′)]d ′ − ∇′×[ ]d ′ µ0 M(r′)×rˆ ∫∫ A(r) = dτ ′ 4  r r τ  ∫ 2 how? Prob. 1.60 4 r  1 τ  µ 1 = 0 ∫ [∇′×M(r′)]d ′ + 0 ∫ [M(r′)×nˆ′]da′ 11 4  r  4π r 12 Vector potential and Bound Currents π µ 1 µ 1 A(r) = 0 ∫ [∇′×M(r′)]dτ ′ + 0 ∫ [M(r′)×nˆ′]da′ 4 r 4π r

Jb = ∇′×M(r′) K b = M(r′)×nˆ′ volume current surface current

With these definitions, bound currents µ J µ K Gauss's law ∫∫ (∇⋅EEa )ddτ =v ⋅ A(r) = 0π∫ b dτ ′ + 0 ∫ b da′ vS 4 v r 4π S r  ((∇⋅vc × ))dd = c ⋅ ( ∇× v ) ∫∫ Let Evc=× ,  vvττ The electrical analogy ()vc×⋅dd a =−⋅ c v × a vv∫∫ volume ρ = −∇⋅P  SS b Since cvva is a constant vector, so (∇× )dd =− × surface charge density σ =⋅Pnˆ ∫∫v 13 b 14 vS

τ π µ ωψθ φ µ π Rω(cossinsin)− ψθ′′ φxˆ Example 5.11 A spherical shell, of radius R, carrying a Ar()= 0 π Rdd2 sin ′′′ 4 µ∫ 22 ′ uniform surface charge σ, is set spinning at angular velocity rRωψθ+−2cos rR φ ψθθ R (cos sin′′ cos− sin cos ′ )yˆ ω. Find the vector potential it produce at point r. + 0 θ R 2 sinθ ′′′ddθφ ∫ 22 θ θθφ 4 rR+−2cos rR ′ θθφ R π(sin sin′′ sin )zˆ + 0 Rdd2 sin ′′′ ∫ σω ψµ22 4 rR+−2cos rR ′ µ π −R3µσωψσω ψµsin yˆ cos ′ Ar()= 0 θ sinθ′′′ddθφ π ∫ 22 4 rR+−2cos rRθ′ Sol : First, let the observer is in the z axis and ω is tilted at an angle ψ 3 π σ −R sin 0yˆ −cos ′ ′ = (2π )θ θ d cosθ′ 0 Kr() 4 ∫ 22 Vector potential is Ar ( ) = ∫ da′ 0 rR+−2cos rR ′ 4 3 1 r − R sin yˆ u The surface Kr (′′ ) = v = 0 du ∫ 22 2 −1 rR+−2 rRu

15 16 µ σω − R3 sinψyˆ (R2 + r 2 + Rr) | R − r | −(R2 + r 2 − Rr)(R + r) A(r) = 0 (− ) Example 6.1 Find the magnetic field of a 2 3R2r 2 uniformly magnetized sphere of radius R. µ σ  0 R  (ω×r) inside A(r) = µ 2 Sol : Choosing the z axis along the direction of M,  R4σ  0 3 (ω×r) outside  J′b = ∇×M = 0  2r we have  K′b = M ×nˆ′ = M sinθφˆ Reverting to the “natural” coordinate, we have The surface current density is analogous to that of a spinning spherical shell with uniform surface current density.

K′b = M ×nˆ′ = M sinθφˆ ⇔ K′ = σv′ = σRω sinθφˆ σRω → M 2 B = µ M (inside) 3 0

Surprisingly, the field inside the spherical shell is uniform. 17 Can you find a more direct method? 18

6.2.2 Physical Interpretation of Bound Current Physical Interpretation of Bound Current

Bound surface current Kb: Bound current density Jb: What is the current in terms of M? What if the magnetization is not uniform? In terms of the magnetization M, its dipole The adjacent current loops do not completely cancel out. moment is m=Mat=Ia. So, M=I/t=Kb

∂M z Consider a thin slab of uniformly magnetized material, with Case (a) I x = [M z (y + dy) − M z (y)]dz = dydz the dipoles represented by tiny current loops. ∂y ∂M All the “internal” currents cancel. However, at the edge there y Case (b) I x = [M y (z + dz) − M y (z)]dy = dydz is no adjacent loop to do the canceling. ∂z ∂M ∂M ∴(J ) = z − y ⇒ J = ∇×M b x ∂y ∂z b

19 20 6.3 The Auxiliary Field H The Role of H in 6.3.1 ’s Law in Magnetized Materials What is the difference between bound current and free H plays a role in magnetostatics analogous to D in the current? electrostotics. J = Jb + J f Ampere’s law can be written: D allows us to write Gauss’s law in terms of free change alone. µ 1 DEPD=+ε , ∇⋅ =ρ (∇×B) = J = J f + Jb = J f + ∇×M 0 f 0 H permits us to express Ampere’s law in terms of free current 1 alone. 1 ⇒ ∇×( B − M) = J f HBMHJ=−∇×=, f µ0 µ0 H What we can control directly. In terms of H, then the Ampere’s law reads

∇× H = J f (differential form) Why can’t we turn the bound currents on or off independently? H ⋅dl = I f (integral form) ∫ 21 22

Example 6.2 A long copper rod of radius R carries a uniformly distributed (free) current I. Find H inside and Sol : outside the rod. Use the πAmpere's law in the integral form and properly choose a suitableπ Amperian loop. π s2 s ≤ R : H (2 s) = I = I fenc 2 π R sI so Hπ = φˆ 2 R2 I s > R : H (2 s) = I, so H = φˆ 2πs

How to determine the magnetic field B? How to choose a suitable Amperian loop? Symmetry.

23 24 H and B, D and E 6.3.2 A Deceptive Parallel

∇× H = J f In free space Which equation is more useful? ∇×BJ = µ0 ∇ ⋅D = ρ f ∇⋅B =0

We can easily control the free current I, but not the In matter ∇× H = J free charge. So H can be determined accordingly. f B On the other hand, the potential difference V can be ∇⋅H = ∇⋅( − M) = −∇⋅M ≠0 read from the voltmeter, which can be used to µ0 determine E. At what condition the divergence of H is equal to zero? The name of H: Some author call H, not B, the “magnetic field”, but it is not a good choice. Let’s just call it “H”. M // B i.e. M // B // H for uniform material only.

25 26

6.3.3 Boundary Conditions Homework #11 The magnetostatic boundary conditions can be rewritten in terms of H and the free surface current Kf.

// // ∇× H = J f ⇒ Habove − Hbelow = K f ×nˆ ⊥ ⊥ ⊥ ⊥ Problems: 6.4, 6.10, 6.13, 6.15 ∇⋅H = −∇⋅M ⇒ H above − Hbelow = −(M above − M below )

The corresponding boundary condition in terms of B and total surface current K.

// // ∇×BJ =µ0 ⇒ BBabove−=× below 0 () Knˆ ⊥⊥ ∇⋅B =0 ⇒ BBabove−= below 0 How to express the boundary conditionsµ at metal or interface? 27 28 6.4 Linear and Nonlinear Media Material Susceptibility 6.4.1 and Permeability The magnetization of paramagnetic and diamagnetic materials is sustained by the field, i.e. when B is removed, M disappears.

MH= χm ,

where the proportionality constant χm is called the magnetic susceptibility.

χm Why not use MB= ? Because MH= χm ∝ I f µ0 µ Materials that obey M = χ mH are called linear media. Β = µ(H + M) =µ (1+ )H = H, 0 µ 0 χ m where = 0 (1+ χ m ) is call the permeability of the material. µ 29 30

Divergence of the Magnetization Example 6.3 An infinite solenoid (n turns per unit length, current I) is filled with linear material of Does the linear media avoid the defect that the divergence susceptibility χ : Find the magnetic field inside the m of M is zero? No! solenoid. Even though M, H, and B are parallel, the divergence of M Sol: The problem exhibits solenoidal symmetry. Thus, we is not zero at the boundary. Consider the following example. can employ the Ampere’s law. H ⋅dl = I (integral form) ∫ f ∫ M ⋅da ≠ 0 HA = nAI ∴ H = nIzˆ Gaussian pillbox ⇒ ∇⋅M ≠ 0 B = µ0 (1+ χm )nIzˆ The enhancement of the magnetic field strength depends on and J = ∇×M = ∇× χ H = χ J the susceptibility of the material. b m m f

Is there a material that the field is significantly enhanced? 31 32 6.4.2 Ferromagnetic Domains

Ferromagnets---which are not linear---require no external fields to sustain the magnetization unlike paramagnets and diamagnets. In a ferromagnet, each dipole “like” to point in the same direction as its neighbors. All the spins point the same way.

Domain boundaries: Domains parallel to the field grow, and the others shrink. Why isn’t every wrench and nail a powerful ? If the field is strong enough, one domain takes over entirely, Domains. and the iron is said to be “saturated”. 33 34

Hysteresis Loop and Phase Transitions

Temperature effect: The dipoles within a given domain line up parallel to one another. However, the random thermal motions complete with this ordering.

Curie temperature: As the temperature increases, the alignment is gradually destroyed. At certain temperature the iron completely turns into paramagnet. This temperature is called the curie temperature.

Hysteresis: The path we have traced out. phase transitions In the experiment, we adjust the current I, i.e. control H. In practice M is huge compared to H. 35 36 Homework #12

Problems: 6.17, 6.21, 6.23, 6.26

37 38

Supplementary Material (optional): Properties of materials (I) Introduction to ferrite materials Nonreciprocal electrical property: the transmission coefficient through the device is not the same for different The ferrites are crystals direction of propagation. 鐵磁性 having small electric conductivity compared to ferromagnetic materials, Unequal propagation constant: The left and right 反鐵磁性 circularly polarized waves have different propagation constant along the direction of external magnetic field B0. Thus they are useful in high-frequency situations because of the absence Anisotropic magnetic properties: The permeability of the of significant eddy ferrite is not a single scalar quantity, but instead is a tensor, 鐵氧體磁性 current losses. which can be represented as a matrix.

39 40 Classical picture of the magnetization process Properties of ferrite materials (II) --- By treating the spinning electron as a gyroscopic top. Ferrites are ceramiclike materials with specific resistivities that may be as much as 1014 greater than that of metals and If an electron is located in a uniform static magnetic field with dielectric constants around 10 to 15 or greater. B , a torque is given by 0 m e T = m×B0 = − p×B0 = B0 × P p mo Ferrites are made by sintering a mixture of metal oxides dP e and have the general chemical composition MO·Fe2O3, = T = B0 × P = 0 × P dt mo where M is a divalent metal such as Mn, Mg, Fe, Zn, Ni, Cd, ω etc. eB0 where ω0 = is called the Larmor fequency; mo = Relative permeabilities of several thousand are common. P = is angular ; and The magnetic properties of ferrites arise mainly from the 2 e= magnetic dipole moment associated with the electron m = is magnetic dipole moment. spin. 41 2mo 42

Quantum mechanics’ viewpoint sz=±1/2 magnetization

In the absence of any damping forces, the actual precession angle will be determined by the initial position As the strength of the bias field H is increased, more magnetic of the magnetic dipole, and the dipole will precess about 0 dipole moments will align with H B at this angle indefinitely (free precession). 0 0 until all are aligned, and M reaches an upper limit. In reality, however, the existence of damping forces will cause the magnetic dipole to spiral in from its initial angle The material is then said to be magnetically saturated, and until m is aligned with B0. Ms is denoted as the saturation magnetization. Ms typically ranges from 4πMs=300 to 5000 Gauss. This explains why sz equals ±1/2 in the Quantum Mechanics. Below saturation, ferrite materials can be very lossy at But where does the damping force come from? microwave frequencies, and rf interaction is reduced.

43 The ferrites are usually operated in the saturated state. 44 Curie temperature Properties of some ferrite materials

The saturation magnetization of a material is a strong function of temperature, decreasing as temperature increases. This effect can be understood by noting that the vibrational energy of an atom increases with temperature, making it

more difficult to align all the magnetic dipoles. Why use 4πMs? B = 4πM+H = µH (Gaussian unit)

At a high enough temperature a zero net magnetization The unit of B is gauss; the unit of H is Oersted. They have same dimension. results. This temperature is called the Curie temperature, Tc. What does ∆H and Mr mean? 45 Ferrite linewidth and remanent magnetization 46

Anisotropic magnetic properties (I) ω Anisotropicω magnetic properties (II) ω If H is the applied ac field, the total magnetic field is H = H zˆ + H, t 0 Omittingω higher order terms, the equations ωcan be reduced to 2 dMx ω dH ω = − M + H , d Mx 2 y ω where H << H . The field produced a total magnetization is 0 y m y +ω M = + H , 0 dt dt2 0 x m dt 0 m x ω the ferrite given by M t = M s zˆ + M . dM 2 y ω d M dH = −( 0M x + mHµx ), y 2 x ω M is the dc saturation magnetization and M is the dt γ 2 + 0 M y = − m + 0ωmH y. s dt dt additional ac magnetization (in the xy plane) caused by dM ω z = 0, where = H and = µ γM applied field. dt 0 0 0 m 0 s ω jωt The componentµ γ equations of motion: If Mω and H ∝ e , the above equtions can be reduced to dM theω phasor equationsω ω : x = − M (H + Hµ ) + (M + M )H ω χ 0 y 0 z γ 0 s z y χ χ 0 dt µ γ 2 2 ωω  xx xy  ( 0 − )Mx = 0 ωωmHx + j mH y , χ  dMy M = []H = yx χyy 0 H, = − M (H + H ) − (M + M )H ( 2 − 2 )M = − j H +ω ω H .   0 x 0 z µ γ 0 s z x 0 y m x ω 0 m y  0 0 0 dt µ γ ω   dM χ ω ω χ jωω z µ where = χ = 0 m and = −χ = m = − 0 MxH y + 0γM y Hx 47 xx yy 2 2 xy yx ω2 2 48 dt 0 − 0 −ω µ Anisotropic magnetic properties (III) Forced precession of spinning electron (I) µ To relate B and H, we have If a small ac magnetic field is superimposed on the static µ  j 0  µ µ   field H0, the magnetic dipole moment will undergo a forced Bµ= (M + H) = []H ⇒ []= ( U + [][]µ ) = − j 0 0 χ 0χ κ  precession. µ κ  0 0 0  κ χ µ Of particular interest is the case where the ac magnetic  µ µ ω = (1χ+ ) = (1+ ) = (1+ω 0 m ) field is circularly polarized in the plane perpendicular to H0.  0 xx 0 yy 0 2 2 µ  µ ω 0 − χ ω  µ ωω A right-hand circularly polarized wave can be expressed in  = − j ) = j = ω m )  0 xy 0 yx 0 2 2 phasor form as  0 −ω H + = H + (xˆ − jyˆ) A material having a permeability tensor of this form is called and in time-domain form as gyrotropic. ω + + jωt + How to apply this concept to a circularly polarized wave? H = Re{H e }= H (xˆ cos t + yˆ sinωt)

49 50

Forced precession of spinning electron (II) Real and imaginary permeability & propagation and attenuation constant.

propagation constant ω β ω ± ω = εµ ω ± ω ω ω RHCPω LHCP ω ω + m +  µ M x = H , µ ω 0 −  + m + + m  ⇒ M = H ⇒ = 0 (1+ ω ) RHCP − j − −ω M + = m H +. 0 0 y  0 −  µ− ωm = µ0 (1+ ω ) LHCP 51 52 0 +ω Faraday rotation --- a nonreciprocal effect Microwave gyrator Consider linearly polarized electric field at z=0, represented Gyrator with a as the sum of a RHCP and a twist section. LHCP wave: E E E = xˆE = 0 (xˆ − jyˆ) + 0 (xˆ + jyˆ) (z=0) 0 2 2 β These two polarized waves propagate with different propagation constants. β E β E E = 0 (xˆ − jyˆ)e− j β+A + 0 (xˆ + jyˆ)e− j −A Gyrator without (z=A) 2 2 β a twist section. β  + − − + − β−  − j( + +β− )A/2 θ = E0 xˆcos( )A − yˆ sin( )A e  2 2  β E − β = tan−1 y = −( + − )A. This effect is called Faraday rotation. E 2 x 53 54

Faraday-rotation isolator Resonance isolator

55 56 Four-port circulator Three-port circulator

57 58