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Electromagnetism: Magnetostatics and Electrodynamics

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Electromagnetism: Magnetostatics and Electrodynamics Electromagnetism: Magnetostatics and Electrodynamics Simon Ross Centre for Particle Theory, Department of Mathematical Sciences Durham University, South Road, Durham DH1 3LE February 26, 2015 Contents 1 Magnetostatics: Lorentz force law 2 2 Biot-Savart law 4 3 Divergence and curl of B~ 6 3.1 Integral form and applications of Ampere's law . 7 4 Vector potential 8 4.1 Multipole expansion for the vector potential . 9 5 Time-dependent fields 11 5.1 Faraday's law . 13 5.2 Maxwell's modification . 14 5.3 Magnetic charge . 15 6 Energy 16 6.1 Mutual inductance . 16 6.2 Energy in the magnetic field . 17 6.3 Energy conservation: Poynting's theorem . 18 7 Wave solutions, light 20 7.1 Waves ......................................... 20 7.2 Electromagnetic waves . 21 8 Sources: Potentials and gauge transformations 23 8.1 Gauge transformations . 24 8.2 Retarded potential solutions . 26 8.3 Dipole fields . 29 9 Electrodynamics and special relativity 30 9.1 Lorentz transformations . 31 9.2 Four-vectors . 32 9.3 Electrodynamics in four-vector notation . 34 1 1 Magnetostatics: Lorentz force law Study of magnetic fields has a much longer history than electric fields. The story becomes interesting to us with the realization in the 19th century that currents, that is, moving charges, create and feel a force from magnetic fields. We will first consider situations where the magnetic fields are time-independent: mag- netostatics. Magnetic fields are produced and felt by currents, that is moving charges, so in magnetostatics the charges cannot all be stationary, but we can consider steady currents, situations where ρ(~r) and ~j(~r) are independent of time. In general, we had a continuity equation, expressing conservation of charge, which said @ρ + r · ~j = 0: @t for steady currents, this simplifies to r ·~j = 0. This divergencelessness expresses the conserva- tion of current; it must circulate, if charge density is time-independent there are no sources or sinks of current. [Picture] This means we can't start with point charges; a point charge in motion has a time-dependent charge and current density. The elementary situation will be a line current flowing in a wire. [Picture]. Take the net charge density on the wire to vanish, the current density is ~j = Iδ2(r). Note the line need not be straight. Orsted (1819) found that a wire produces a magnetic field, which can be measured by compasses placed near the wire: 2I 1 B~ = ~e : (1.1) c r φ But the basic observation (analogue of Coulomb's law) is that there is a force between currents. Ampere (1825): If I have two parallel wires carrying steady currents, there is a force between them. [Picture]. Force per unit length ~e F~ / I I r (1.2) 1 2 r Similar in functional form to the force between long linear charge distributions, but depending on the currents, not charge density. Units: usually one has a constant µ0 here. In the units where 4π0 = 1 that I used last 1 4π term, this turns out to be related to the speed of light: 0µ0 = c2 , so in my units µ0 = c2 . I will write this in terms of c from the outset. Relative to Griffiths, I have also changed the definition of B~ by scaling it by a factor of c so that E~ and B~ have the same units (C=L2). Note I has units of C=T . We can then think of the force on the other wire as due to a velocity-dependent force on the charges moving in the wire, in the presence of the magnetic field B~ created by the first wire. The force is orthogonal to the velocity and B~ , so it goes as ~v ^ B~ . Experimentally, the force on a single point charge moving in a magnetic field B~ is ~v F~ = q ^ B:~ c This is the Lorentz force law. We could imagine (ahistorically) that we introduce the mag- netic field to explain a force on moving charges, just as we introduced the electric field to 2 explain a force on static charges. In the presence of both kinds of fields, the total force on a point charge is 1 F~ = q(E~ + ~v ^ B~ ): (1.3) c Remarks: • We derived this for static fields, but the above expression is true in general, for point particles moving in any electric and magnetic fields. • For charge and current distributions, the force on a volume element is 1 dF~ = ρE + ~j ^ B;~ c where ~j = ~vρ is the current density. • Because the force is orthogonal to the velocity, the magnetic field does no work. ~ • Exercise (for problems class) motion in a constant field B0. • There is a factor of 1=c here relative to the expression in Griffiths. This is because I define the magnetic field differently. Returning to the wires, we can work out the force on a current-carrying wire from the above expression. Suppose the wire carries no net charge density and a current I. Then for a small length element, 1 1 dF~ = ~j ^ B~ = Id~l ^ B;~ c c as the direction of the current density is in the direction of the wire. [Picture] So 1 Z F~ = I d~l ^ B:~ (1.4) c Example: For the situation we had before with two parallel long straight wires, the mag- netic field was 2I 1 B~ = 1 ~e ; c r φ which is constant along the second wire, so 2I I 1 Z F~ = − 1 2 ~e d~l; c2 r r reproducing the force per unit length I had before. ~ For a closed loop of wire in a constant field B0, I I F~ = d~l ^ B~ : c 0 This vanishes because H d~l = 0. 3 2 Biot-Savart law Experimentally, we see that currents produce magnetic fields. We want to write down an integral expression for the field in general, analogous to the one we had for the electric field. In this case, however, we can't build it up by superposition of point charges. Consider the expression for a straight line current: 2I 1 B~ (~r) = ~e : c r φ This is orthogonal to the direction of the current and to the separation vector ~r. Motivates us to propose that for a current element Id~l at ~r0, I d~l ^ (~r − ~r0) dB~ (~r) = : c j~r − ~r0j3 Note that unlike in the electric case, I can't consider this in isolation. The conservation law r · ~j = 0 means there can't be isolated current elements. For a single element, we take B~ to fall off as the inverse square of the distance, as in the electric case. This will reproduce the inverse distance for a long straight wire once we integrate over the source. Integrating, we have the Biot-Savart law I Z d~l ^ (~r − ~r0) B~ (~r) = : (2.1) c j~r − ~r0j3 For a continuous current density, 1 Z ~j(~r0) ^ (~r − ~r0) B~ (~r) = dV 0: (2.2) c j~r − ~r0j3 Compare to Z ρ(~r0)(~r − ~r0) E~ (~r) = dV 0: j~r − ~r0j3 Similar distance dependence, local source dependence, but different vector character. Example: Long straight wire, carrying current I. Work in cylindrical polar coordinates, 0 0 ~ so ~r = z ~ez, ~r = z~ez + r~er. dl = dz~ez. So ~ 0 dl ^ (~r − ~r ) = rdz~eφ; and Z 1 0 ~ I dz B = r 0 2 2 3=2 ~eφ c −∞ ((z − z ) + r ) Set z0 = z + rx; dz0 = rdx Z 1 ~ I 1 dx B = 2 3=2 ~eφ c r −∞ (1 + x ) So see there's no z dependence, 1=r falloff as claimed before. Now Z 1 dx x = [p ]1 = 2; 2 3=2 2 −∞ −∞ (1 + x ) 1 + x 4 so 2I 1 B~ = ~e : c r φ Example: Circular current loop of radius R, carrying current I: calculate B~ at a point on the axis of the loop a distance z from its center. Work in cylindrical polar coordinates again: ~ 0 ~ by symmetry, B = B~ez. ~r = z~ez, ~r = R~er, dl = Rdφ~eφ, so ~ 0 2 [dl ^ (~r − ~r )] · ~ez = R dφ, and Z 2π 2 I 2 dφ I 2πR B = R 2 2 3=2 = 2 2 3=2 : c 0 (R + z ) c (R + z ) Can't work out field off axis in closed form, but will discuss analogue of the electric dipole approximation later. Example: Solenoid. Consider a long cylinder with current I flowing in a wire with N turns per unit length, that is current density NI per unit length, radius R. Field on axis: add up contributions from many loops as above. At one end, NI Z L dz 2πNI z 2πNI L B = 2πR2 = [p ]L = p : 2 2 3=2 2 2 0 2 2 c 0 (R + z ) c R + z c R + L In the middle? Use superposition, it's like two such cylinders. 2πNI L=2 4πNI L B = 2 = p : c pR2 + L2=4 c 4R2 + L2 4πNI Inside a very long solenoid, B = c , which we will recover in a different way shortly. Example: Force between two closed circuits The magnetic field due to a closed loop carrying current I2 is I ~ ~ I dl2 ^ (~r1 − ~r2) B(~r1) = 3 ; c 2 j~r1 − ~r2j so total force on loop 1 due to loop 2 is I I I ~ ~ ~ I1 ~ ~ I1I2 dl1 ^ (dl2 ^ (~r1 − ~r2)) F1 = dl1 ^ B = 3 : c c 1 2 j~r1 − ~r2j ~ ~ This is not obviously antisymmetric. Would have expected F2 = −F1. Use A~ ^ (B~ ^ C~ ) = B~ (A~ · C~ ) − C~ (A~ · B~ ) to write as "I I ~ I I ~ ~ # ~ I1I2 ~ dl1 · (~r1 − ~r2) (~r1 − ~r2)dl1 · dl2 F1 = dl2 3 − 3 : c j~r1 − ~r2j 1 2 j~r1 − ~r2j The second term has the expected symmetry.
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