Download Magnetostatics and Magnetic Dipoles: Lecture Notes for PHY481 - Prof. Phillip Duxbury and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 25 Chapter 7,8 of PS, Chapter 5 of Griffiths A. Basic equations of magnetostatics The analogy with electrostatics is very close, with the major difference that the sources of magnetic fields in magnetostatics are current rings and the intrinsic magnetic moment of elementary particles. + → N , − → S enables us to draw magnetic field lines in analogy with electrostatics. Basic equations, ∮ ~B · d~a = 0; ~∇ · ~B = 0 (no name) (1) which implies that ~B = ~∇∧ ~A. In magnetostatics we choose the Coulomb gauge, ~nabla· ~A = 0. ∮ ~B · d~l = µ0i; ~∇∧ ~B = µ0~j (Ampere′s law) (2) Boundary conditions on ~B, Bextn −Bintn = 0 (from (1)); Bextt −Bintt = µ0K (from (2)) (3) Superposition laws d ~B = µ0 4π id~l ∧ r̂ r2 or ~B(~r) = µ0 4π ∫ ~j(~r′) ∧ (~r − ~r′)d3r′ |~r − ~r′|3 (4) and ~A = µ0 4π ∫ ~j(~r′)d3r′ |~r − ~r′| or ∇2 ~A = −µ0~j (5) This last equation looks like 3 Poisson’s equations. In parts of space where there are no current sources, we have 3 Laplace’s equations. We can then solve these equations with various boundary conditions. To do this we need the boundary conditions on the vector potential, that are, ~Aext = ~Aint; ∂ ~A ∂n |ext − ∂ ~A ∂n |ext = −µ0 ~K (6) Another useful relation, ∮ ~A · d~l = ∫ (~∇ · ~A) · d~a = ∫ ~B · d~a = φB (7) In some cases this makes it easy to find the vector potential from the magnetic field, e.g. a solenoid, where B = µ0ni, so that φB = Ba, and A(r) = µ0nia/(2πr). The direction is the 1 same as the direction of the current (from supperposition law above). B. Magnetic dipoles There are two sources of magnetic dipoles: currents and intrinsics magnet moments of elementary particles. In either case, we have a magnetic dipole moment ~m that gives the magnitude and direction of the magnetic dipole. This magnetic dipole moment is directly analagous to the electric dipole moment ~p, and we have, ~B = µ0 4π 3(~m · r̂)r̂ − ~m r3 ; ~τ = ~m · ~B; U = −~m · ~B (8) Two key new things here are: (i) that the magnetic moment of a current ring is given by, ~m = i~a, (9) where ~a is the area of the current loop and the direction is normal to the current loop. (ii) In addition we have a new relation for the vector potential of a magnetic dipole, ~A = µ0 4π ~m ∧ r̂ r2 . (10) It is a useful excersize to show that ~∇ ∧ ~A gives the dipole field formula (this was an assigned problem (2.8)). C. Lorentz force law and force on current carrying wire A charged particle of charge q with velocity ~v moving in a magnetic field ~B and electric field ~B experiences a force given by, ~F = q( ~E + ~v ∧ ~B) (Lorentz force law) (11) A wire segment carrying current i and with vector length d~l in a magnetic field ~B experiences a force, d~F = id~l ∧ ~B (12) D. Additional equations related to current The current and current density are related through, i = ∫ ~j · d ~A (13) 2
