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Module I: Electromagnetic waves Lecture 4: Energy in electric and magnetic fields Amol Dighe TIFR, Mumbai Outline Energy in static electric field Energy in static magnetic field Energy of EM waves Coming up... Energy in static electric field Energy in static magnetic field Energy of EM waves Work done in vacuum I I Work done in increasing charge density by δρ(~x): Z Z ~ δW = φ(~x) δρ(~x)dV = 0 φ(~x)δ(∇ · E)dV (1) Integrate by parts: δW δW Z ~ ∇ · (φδ E)dV − 0 Z ~ · δ EdV ~ E = 0 = 0 Z ~ (∇φ) · (δ E)dV (2) Total work done W = 0 2 Z ~ 2 dV E (3) Work done in a dielectric I I Work done in increasing charge density by δρ(~x): Z Z ~ δW = φ(~x) δρ(~x)dV = φ(~x)δ(∇ · D)dV (4) Integrate by parts: δW = δW = Z ~ ∇ · (φδ D)dV − Z ~ · δ DdV ~ E Z ~ = E) ~ Only for a linear dielectric (D Z 1 ~ ~ W = E · DdV 2 ~ (∇φ) · (δ D)dV (5) (6) Matching with energy of collection of discrete charges I For N charges qi , we expect that the energy is 0 X qi φi U= 2 n (7) where φi is the potential at the i th charge qi due to the others. ~ i : electric field due to charge i Let E R 2 ~ dV , we get I Using U = (0 /2) E Z X 0 ~2 + ~i · E ~ j )dV U = (Σi E E (8) i 2 i6=j Z 0 X ~ = U0 + Ei · (−∇φi )dV (9) 2 i I The first term, U0 , does not depend on the distribution of charges, it is the “self-energy”. I The second term matches with what we expect. Coming up... Energy in static electric field Energy in static magnetic field Energy of EM waves Differences compared to the electric field I A steady current has to be maintained, for which energy should continue to be supplied by an external electric field, Eext . I When this current is increased by a small amount, a back-EMF Eind is induced, against which work needs to be done. (This is the analog of work done against a repulsive force when bringing a charge from infinity.) I This work done will be stored in the magnetic field produced. The rest of the energy supplied by the external EMF will be dissipated as heat in the conductor. I ~ ext + E ~ ind ) ⇒ ~J2 = σ(E ~ ext · ~J + E ~ ind · ~J) Total current ~J = σ(E I Rate of energy input: dUin ~ ext · ~JdV = =E dt I Z ~2 Z J ~ ind · ~JdV dV − E σ (10) The first term is rate of energy dissipation as heat, the second term is rate of storage of energy in the magnetic field Energy stored in static magnetic field I ~ =E ~ ind ) Rate of energy storage in magnetic field: (E Z Z dUm ~ · ~JdV = − E ~ · (∇ × H)dV ~ = − E dt Z Z ~ · (∇ × E)dV ~ ~ × H)dV ~ = − H + ∇ · (E Z = I ~ · (∂ B/∂t)dV ~ H Incremental energy stored: δUm = (11) (12) R ~ · δ BdV ~ H ~ is linear in B ~ Only if H Z Z 1 ~ 1 ~ ~ ~ Um = H · BdV = H · (∇ × A)dV 2 2 Z Z 1 ~ · (∇ × H)dV ~ ~ × A)dV ~ = A − ∇ · (H 2 Z 1 ~ ~ A · JdV = 2 (13) (14) ~ ×H ~ A note about E R H ~ ~ × H)dV ~ ~ × H) ~ · dS ∇ · (E = (E I We neglected the term I This is indeed valid for static electromagnetic fields I But for time-dependent fields, we shall see later that the leading ~ and H ~ go as 1/r , so over the surface of a sphere with terms in E large radius r , the integral actually will have a constant nonzero vale. I This will be the energy radiated away at infinity due to the ~ ≡E ~ ×H ~ will be defined as the Poynting changing currents. N vector, which gives the rate of loss of energy through radiation. Recap of topics covered in the lecture I Energy stored in static electric fields: in vacuum and in a dielectric I Energy stored in a static magnetic field I Note: some of the results are only valid when the material media have linear properties. I Energy stored and transported by an EM wave Coming up... Energy in static electric field Energy in static magnetic field Energy of EM waves Quadratic quantities and factors of 2 I In Electrodynamics, for convenience, we often use notation involving complex numbers (mainly exponentials), e.g. ~ =E ~ 0 ei(kx−ωt) , E ~ = −i B ~ 0 ei(kx−ωt) B (15) when we actually want to represent ~ E ~ B ~ 0 cos(kx − ωt) = Re(E ~ 0 ei(kx−ωt) ) = E ~ 0 sin(kx − ωt) = Re(i B ~ 0 ei(kx−ωt) ) = B (16) (17) I While performing calculations in complex notation and taking the real part of the final answer works as long as we are dealing with ~ or B, ~ one has to be careful while dealing quantities linear in E with quadratic (or higher order) quantities. I For example, in the complex notation above, ~ 2 i = h|E ~ ∗ · E|i ~ = |E ~ 0 |2 h|E| (18) while the actual answer should be (using real notation) ~ 0 |2 ~ 2 i = |E ~ 0 |2 hcos2 (kx − ωt)i = 1 |E h|E| 2 (19) Energy density stored in EM fields I We have already seen that the energy stored in electric field is ~ 2 (we showed this result for a static field). When Ue = (1/2)0 |E| the electric field represents a propagating wave, then taking into account the “factor of 2” for averaged quadratic quantities, we get hUe i = I (20) ~ 2 /µ0 (we The energy stored in magnetic field is Um = (1/2)|B| showed it for a static magnetic field). For a propagating wave, ~ = |/˛ω||E|. ~ Including the “factor of 2”, we get |B| hUm i = I 1 ~ 2 0 |E0 | 4 ~ 0 |2 1 |B 1 ||˛2 ~ 2 1 ~ 2 = |E0 | = 0 |E 0| 4 µ0 4 ω 2 µ0 4 (21) For a plane EM wave, energy stored in electric and magnetic field is equal. The total energy of an EM wave is hUi = hUe i + hUm i = 1 ~ 2 0 |E0 | 2 (22) Energy transported by the EM fields I The rate of energy transport is given by the Poynting vector, ~ =E ~ ×H ~ N I (23) The time-averaged value of this quantity is ~ = h|N|i 1 ~ ||˛ ~ 1 1 ~ ~ |E0 |.|H0 | = |E |E0 | = 0| 2 2 ωµ 2 r 0 ~ 2 |E0 | µ0 (24) Compared with the rate ofpenergy consumption in a conductor, ~ 0 |2 , the quantity 0 /µ0 is termed the conductance of (1/2)σ|E vacuum p I Similarly, /µ is the conductance of a medium through which an EM wave propagates I Recap of topics covered in this lecture I Energy stored in static electric and magnetic fields I Energy transported by the EM field