Capstones in Physics: Electromagnetism 3. Magnetostatics 3.1. Lorentz Force Law 3.2. Biot-Savart law and Ampere’s law 3.3. Magnetic vector potential 3.4. Magnetic dipole 3.5. Magnetization ©2005 Oregon State University Philip J. Siemens (revised by Yun-Shik Lee) Magnetostatics page 3.2 3.1 LORENTZ FORCE LAW Lorentz force on a charge Q is F(r) = Q [E(r) + v ×B(r) ] Work done by the fields when particle moves by dr = v dt is dW = dr •F(r) = Q dt [E(r)•v +v •(v ×B(r) )] =Q dt E(r)•v (vanishes, since v ⊥v ×B ) † IDEALIZED CURRENTS THE IDEALIZATION OF THE WIRE = line charge density λ J dA dL = I dL I = current dL ρ dA dL = λ dL ρ = charge density v = average velocity dA J = current density = ρ v dFmagnetic = I dL × B(R) , Fmagnetic = I ∫ dL × B(R) THE IDEALIZATION OF THE SHEET K = surface current density J dA dd = K d A σ = surface charge density ρ dA dd = σ d A ρ = charge density d A v = average velocity d d J = current density = ρ v dFmagnetic = dA K × B(R) , Fmagnetic = ∫ dA K × B(R) ©1999 Oregon State University Capstones in Physics Philip J. Siemens Magnetostatics page 3.3 3.2. BIOT-SAVART LAW AND AMPERE’S LAW 3.2.A. Biot-Savart law for B due to current distribution J steady state case: assume charge density ρ and current J independent of time; then E and B are also time independent May also be useful approximation when slowly varying! In this case the Maxwell equations for E and B decouple: E depends only on ρ, B depends only on J dvol = d3 R = dX dY dZ dQ = ρ dvol J = ρ V R J(R) ⇓ r–R V dQ = J dvol r µo for time-independent sources, B(r) = d3R J(R)×(r−R)/|r−R|3 4π ∫ proof: check ∇ • B = 0, ∇ × B = µoJ important special case: thin wire, cross section dA, total current I = J dA d3R J(R) = dL dA J = I dL where dL = path element ©1999 Oregon State University Capstones in Physics Philip J. Siemens Magnetostatics page 3.4 BIOT-SAVART EXAMPLE µoI B(r) = dL×(r−R)/|r−R|3, z 4π ∫ dL = dZ = zˆ dZ, I r R =Z = Z zˆ x r–R = r –Z zˆ , R r - R zˆ × (r –Z zˆ ) = zˆ × r, |r–R|2=r⊥2+Z2 ∞ 1 µ0I µ0I 2 µoI B(r) = ( zˆ ×r) dZ 3/ 2 = ( zˆ ×r) = φˆ 4π ∫ 2 2 4π 2 2π|r | −∞ ()r⊥ + Z r⊥ ⊥ check: ∫ dL •B = µo I on a circular path centered on the wire 3.2.B. AMPERE’S LAW r r The equation for the curl of B : ∇ × B = µ0 J r r r r ∇ × B ⋅ dar = B ⋅ dl = µ J ⋅ dar = µ I Stokes theorem : ∫ ( ) ∫ 0 ∫ 0 A much simpler way to find B When the source distribution has lots of symmetry, we can often use Ampere's law directly to find the magnetic field Example: straight-line currents we can argue from symmetry that the field B(r) • is independent of φ or z , so it depends only on r⊥ + ρ , B(r) = B(ρ) • is in the azimuthal direction, along φˆ B(ρ) = φˆ B(ρ) To find the remaining unknown function B(ρ) , ©1999 Oregon State University Capstones in Physics Philip J. Siemens Magnetostatics page 3.5 we apply Ampere's law to a cleverly chosen path chosen so that it is • either along the field direction, but with constant field, or • perpendicular to the field A circular path of radius ρ centered on the wire has dL = φˆ ρ dφ, µoI dL •B =⌠2πdφ ρ φˆ • φˆ B(ρ) = 2πρ B(ρ) = µ I ⇒ B(ρ) = ∫ ⌡0 o 2πρ Example Toroidal coil square cross section c N turns, current I a following Lorraine, Corson & Lorraine b Apply Ampere's cicuital law in integral form to paths: ∫ dL •B = µo ∫ dA •J = µoIenclosed paths a, b and c: dL is azimuthal, so ∫ dL •B = 2πρ Bφ where ρ is the radius of the path µo dL •B = µ dA •J = µ I , so B = I ∫ o ∫ o enclosed φ 2πρ enclosed path a: can find a surface bounded by this path with no current penetrating it, so Bφ =0. path c: likewise Bφ =0; note the surface can be chosen at will, only its boundary is fixed. For example, choose a plane surface, then equal currents flow up and down through the surface µo path b: current penetrating surface is NI , so B = ΝI φ 2πρ ©1999 Oregon State University Capstones in Physics Philip J. Siemens Magnetostatics page 3.6 3.3. MAGNETIC VECTOR POTENTIAL Because ∇ • B = 0, there exists a vector function A such that B = ∇ × A The ampere’s law r r r 2 r r ∇ × B = ∇ × (∇ × A)= ∇(∇ ⋅ A)− ∇ A = µ0 J You can add to the magnetic potential any function whose curl vanishes with no effect on B. ⇒ ∇ • A = 0 2 r r The Ampere’s law becomes ∇ A = −µ0 J : Poisson’s equation. r r r r µ0 J (R) 3 r A()r = r d R Therefore, 4π ∫ rr − R For line and surface currents r r r r r r µ0 I µ0 I dl r r µ0 K()R A()r = r dl = r ; A()r = r da 4π ∫ rr − R 4π ∫ rr − R 4π ∫ rr − R Mutipole expansion of the vector potential r r µ I dl µ I ∞ 1 r A()rr = 0 = 0 Rn P (cosα)dl ∫ r r ∑ n+1 ∫ n 4π r − R 4π n=0 r µ0 I 1 r 1 r = dl + 2 Rcosα dl +L 4π r ∫∫r r µ I r µ I r r µ I A ()rr = 0 Rcosα dl = 0 rˆ⋅ Rdl = − 0 rˆ× dar dip 4πr 2 ∫ 4πr 2 ∫ 4πr 2 ∫ r µ mr ×rˆ A ()rr = 0 mr = I dar = I ar ⇒ dip 4π r 2 where ∫ ©1999 Oregon State University Capstones in Physics Philip J. Siemens Magnetostatics page 3.7 3.4. MAGNETIC DIPOLE 3.4.A. MAGNETIC DIPOLES SIMPLE DIPOLE: a loop of current I, area a magnetic field of simple loop dipole moment is m =– I a Far field has same pattern as electric dipole 1 3 more generally, m = 2 ∫ d R R × J(R) , which is independent of origin. Dipole magnetic vector potential and field µ m×rˆ A ()rr = 0 dip 4π r 2 µ µ m B(r) = 0 []3(m ⋅ r))rˆ − m = 0 (2rˆcosθ + zˆsinθ ) 4π r3 4π r3 for r → ∞, m || zˆ ©1999 Oregon State University Capstones in Physics Philip J. Siemens Magnetostatics page 3.8 SIMPLE LOOP: a loop of current I, area dS = πa2 z µoI compute B(r)= dL×(r−R)/|r−R|3 4π ∫ coordinates: choose y = 0, r L in xy plane ⇒ Az = 0, θ dL = a dφ φˆ = a dφ(–xˆ sinφ+yˆ cosφ) symmetry ⇒ By = 0 ψ a y dipole moment is L m + I dS = I πa2 zˆ x φ dL numerator: dL × (r–L) = dL × r – dL × L ; substitute r = x xˆ + z zˆ → dL × (r–L) = a dφ {x (–zˆ cosφ) +z(–yˆ sinφ + xˆ cosφ)} + a2 dφ zˆ the yˆ component integrates to zero by symmetry, so we can use dL × (r–L) → a dφ { (z cosφ)xˆ + (a –x cosφ) zˆ } denominator: |r–L|2 = r2+a2 – 2a r cos ψ = r2+a2 – 2a x cos φ since r cos ψ = projection of r along L = x cos φ far away: look at r »a , let ε + a/r « 1, then 1 1 x 1 x = (1–2ε cosφ+ε2) –3/2≈ (1+3ε cosφ) |r–L|3 r3 r r3 r µoIa x ⇒ B(r) ≈ ⌠2πdφ {(z cosφ) xˆ + (a –x cosφ) zˆ } ( 1+ 3ε cosφ ) 4πr3 ⌡0 r x z now ⌠2πdφ = 2π, ⌠2πdφ cosφ = 0, ⌠2πdφ cos2φ = π, = sinθ, = cosθ ⌡0 ⌡0 ⌡0 r r µoIa xz x2 µoIa2 B(r)≈ 3ε xˆ +(2a –3ε )zˆ = 3sinθ cosθxˆ +(2–3sin2θ)zˆ 4r3 { r r } 4r3 { } µoIa2 but rˆ = sinθ xˆ + cosθ zˆ , so B(r) ≈ 3cosθ rˆ –zˆ , or 4r3 { } µo B(r) ≈ 3(m•rˆ ) rˆ –m 4πr3 { } ©1999 Oregon State University Capstones in Physics Philip J. Siemens Magnetostatics page 3.9 3.4.B. TORQUE ON A DIPOLE dipole moment m = I A Force on dL(R) is dL dF(R) = I dL(R)×B dA Torque on dL = R×dF(R) Total torque τ = I ∫ R×(dL(R)×B) vector identity 2: τ = I ∫ [ dL(r)(r • B) – B (dL(r) •r) ] 1 2 last term integrates out: r = 2 ∇(r ) consider part of τ proportional to Bx : dL(r) = dr ⇒ I Bx ∫ dL(r) x = I ∫ dr x = I Bx ∫(xˆdx + yˆdy + zˆdz)x ˆ ˆ ˆ = I Bx (x∫ xdx + y∫ xdy +z∫ xdz) = I Bx(xˆ ( 0 ) + yˆ Az – zˆ Ay) compare m × B = I A × B , part proportional to Bx is I ( yˆ Az – zˆ Ay) Bx √ ©1999 Oregon State University Capstones in Physics Philip J. Siemens Magnetostatics page 3.10 3.5. MAGNETIZATION INTRINSIC MAGNETIC MOMENTS Properties of elementary particles include: 1 • Electric charge (quantized in units 3 e ) • Mass = energy in rest frame/c2 (no known natural unit) • Angular momentum (quantized in units −h /2 = h /π) • Magnetic moment (no known natural unit) – always parallel to the intrinsic angular momentum – both are relativistic effects, arise in Dirac's theory NUCLEI AND ATOMS HAVE MAGNETIC MOMENTS They arise partly from the intrinsic moments of the electrons electron and quarks, nucleus and partly from their motion in little current loops.