Chapter 6 Magnetic Fields in Matter 6.1.2. Torques and Forces on Magnetic Dipoles 6.1 Magnetization 6.1.1 Diamagnets, Paramagnets, Ferromagnets A magnetic dipole experiences a torque in a magnetic field, just as an electric dipole does in an electric field. 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 Electrons 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 (paramagnetism). 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 atom lock together in pairs with The magnitude of the force 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 Electron Speeds Up or Slows Down 1 e2 v2 Electrons not only spin; they also revolve around the nucleus. = m without the magnetic field. 4πε R2 e R Let’s assume the orbit is a circle of radius R. The current 0 looks like steady (really?) The centripetal force comes from two sources: the electric force and the magnetic force. e e ev Current I = = = T 2πR 2πR 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 Diamagnetism 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 atoms, 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′′′++ 1 ˆ ∂∂∂xyz′′′()()()xx−+−+−′′′222 yy zz ∇′ = r The vector potential of a single dipole m is 2 xyˆˆˆ′′′′′′()()()xx−+ yy −+ z zz − rˆ ==2223/22 r r (()(xx−+−+−′′′ yy )( zz )) r µ0 m×rˆ A(r) = µ0 1 2 The vector potential is A(r) = ∫ M(r′)×(∇′ )dτ ′ 4π r 4π r In the magnetized object, each volume element carries a Using the product rule ∇×( fA) = ∇f × A + f (∇× A) dipole moment Mdτ’, so the total vector potential is and integrating by part, we have µ0 1 M(r′) µ M(r′)× ˆ A(r) = ∫∫[∇′×M(r′)]dτ ′ − ∇′×[ ]dτ ′ A(r) = 0 r dτ ′ 4π r r ∫ 2 how? Prob. 1.60 4π r µ 1 µ 1 = 0 ∫ [∇′×M(r′)]dτ ′ + 0 ∫ [M(r′)×nˆ′]da′ 11 4π r 4π r 12 Vector potential and Bound Currents µ 1 µ 1 A(r) = 0 ∫ [∇′×M(r′)]dτ ′ + 0 ∫ [M(r′)×nˆ′]da′ 4π 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) = 0 ∫ b dτ ′ + 0 ∫ b da′ vS 4π v r 4π S r ((∇⋅vc × ))ddττ = c ⋅ ( ∇× v ) ∫∫ Let Evc=×, vv The electrical analogy ()vc×⋅dd a =−⋅ c v × a vv∫∫ volume charge density ρ = −∇⋅P SS b Since cvva is a constant vector, so (∇× )ddτ =− × surface charge density σ =⋅Pnˆ ∫∫v 13 b 14 vS µ Rω(cossinsin)− ψθ′′ φxˆ Example 5.11 A spherical shell, of radius R, carrying a Ar()= 0 Rdd2 sinθθφ′′′ 4π ∫ 22 uniform surface charge σ, is set spinning at angular velocity rR+−2cos rRθ ′ µ Rωψθφ(cos sin′′ cos− sin ψθ cos ′ )yˆ ω. Find the vector potential it produce at point r. + 0 R 2 sinθ ′′′ddθφ ∫ 22 4π rR+−2cos rRθ ′ µ Rωψθ(sin sin′′ sin φ )zˆ + 0 Rdd2 sinθθφ′′′ ∫ 22 4π rR+−2cos rRθ ′ −R3σωsin ψµ yˆ cosθ′ Ar()= 0 sinθ′′′ddθφ ∫ 22 4π rR+−2cos rRθ′ Sol : First, let the observer is in the z axis and ω is tilted at an angle ψ 3 π −R σωsin ψµ0yˆ −cosθ′ ′ = (2π )d cosθ′ µ0 Kr() 4π ∫ 22 Vector potential is Ar( ) = ∫ da′ 0 rR+−2cos rRθ′ 4π 3 1 r −µσωψR sin yˆ u The surface current density Kr(′′ ) = σ v = 0 du ∫ 22 2 −1 rR+−2 rRu 15 16 − µ R3σω 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. µ0 Rσ (ω×r) inside A(r) = 2 Sol : Choosing the z axis along the direction of M, µ 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 Magnetostatics 6.3.1 Ampere’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. 1 DEPD=+ε , ∇⋅=ρ (∇×B) = J = J f + Jb = J f + ∇×M 0 f µ0 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.