Phys-272 Lecture 17 • Motional Electromotive Force ( emf ) • Induced Electric Fields • Displacement Currents • Maxwell’s Equations From Faraday's Law to Displacement Current AC generator Magnetic Levitation Train Review of Sources of B fields A long wire carries a current I. Clicker The B field at distance r from the wire I 2 Question I a) B= µo I/ 2 π r b) B= µo Ir / 2 π Verify using c) B= µ I/ 2 π r o Ampere’s Law Question II Direction is (a)radial, perp to wire (b) tangential (c) along the wire (d) anti-parallel to the wire Motional Electromotive Force d r r dΦ ε = − ∫ B ⋅ dA = − B dt dt In Faraday’s Law, we can induce EMF in the loop when the magnetic flux , ΦB, changes as a function of time. There are two Cases when ΦB is changing, 1) Change the magnetic field (non-constant over time) 2) Change or move the loop in a constant magnetic field The slide wire generator is an example of #2 and the induction of EMF via moving parts of the loop is called, motional EMF . Slide Wire Generator; revisited again Suppose we move a conducting bar in a constant B field, then a force F=q v ×B moves + charge up and – charge down. The charge distribution produces an electric field and EMF, , between a & b. This continues until equilibrium is reached. r r r b r r r F vq × B r r E = = = v × B ε = ∫ E ⋅ ld = vBl q q a In effect the bar’s motional EMF is an equivalent to a battery EMF Slide Wire Generator; revisited again If the rod is on the U shaped conductor, the charges don’t build up at the ends but move through the U shaped portion. They produce an electric field in the circuit. The wire acts as a source of EMF – just like a battery. Called motional electromotive force . b r r ε = ∫ E ⋅ ld = vBl a Direct Current Homopolar Generator invented by Faraday Rotate a metal disk in a constant perpendicular magnetic field. The charges in the disk when moving receive a radial force. The causes current to flow from center to point b. R 1 ε = ∫ ωBr dr = ωBR 2 0 2 Faraday’s Law (continued) What causes current to flow in wire? Answer: an E field in the wire. A changing magnetic flux not only causes an EMF around a loop but an induced electric field. Can write Faraday’s Law: r r d r r dΦ ε = ∫ E ⋅ ld = − ∫ B ⋅ dA = − B dt dt Remember for a long straight wire of length l, V = El. Note: For electric fields from static charges, the EMF from a closed path is always zero. Not true here. There are two possible sources for electric fields ! Induced Electric Fields Suppose we have electromagnetic that has an increasing magnetic field Using Faraday’s Law we predict, N r r d r r dΦ ∫ E ⋅ ld = − ∫ B ⋅ dA = − B B dt dt If we take a circular path inside and E centered on the magnet center axis, the electric field will be tangent to the circle. (E field lines are circles.) NOTE such an E field can never be made by static charges S E field lines will look like an onion slice N.B. there are no wire loops, E fields can appear w/o loops If we place a loop there, a current would flow in the loop Induced Electric Fields; example If we have a solenoid coil with changing current there will be circular electric fields created outside the solenoid. It looks very much like the mag. field around a current carrying wire, but it is an E field and there are no wires or loops. E r Note the E fields are r d r r dΦ ∫ E ⋅ ld = − ∫ B ⋅ dA = − B predicted by Faraday eqn. dt dt Eddy Currents Changing magnetic fields in metal induce eddy currents. Example: Energy loss in transformers. To reduce use laminations. But eddy currents often useful. Maxwell’s Equations (integral form) Name Equation Description Gauss’ Law for r r Q Charge and Electricity ∫ E ⋅ dA = electric fields ε0 Gauss’ Law for r r Magnetic fields Magnetism ∫ B ⋅ dA = 0 r r Electrical effects dΦB Faraday’s Law ∫ E ⋅ ld = − from changing B dt field r Ampere’s Law B ⋅ dl = µ i Magnetic effects ∫ 0 from current Needs to be modified. + ? There is a serious asymmetry . Remarks on Gauss Law’s with different closed surfaces r r Qenclosed Gauss Law’s works for ∫ E ⋅ dA = ANY CLOSED SURFACE ε r r 0 ∫ B ⋅ dA = 0 cylinder square Surfaces for integration of E flux sphere bagel Remarks on Faraday’s Lawr withr different attached surfaces r r d∫ B ⋅ dA Faraday’s Law works for ∫ E ⋅ ld = − any closed Loop and ANY dt attached surface area Line integral Surface area defines the integration for B flux Closed loop cylinder disk Fish bowl This is proved in Vector Calculus with Stokes’ Theorem Generalized Ampere’s Law and displacement current r Ampere’s original law, ∫ B ⋅ dl = µ 0 I enclose , is incomplete. Consider the parallel plate capacitor and suppose a current ic is flowing charging up the plate. If Ampere’s law is applied for the given path in either the plane surface or the bulging surface we we should get the same results, but the bulging surface has ic=0, so something is missing. Generalized Ampere’s Law and displacement current Maxwell solved dilemma by adding an addition term called displacement current, iD = ε dΦE/dt , in analogy to Faraday’s Law. r dΦ E ∫ B ⋅ dl = µ0 ()ic + iD = µ0 ic + ε 0 dt Current is once more continuous: iD between the plates = iC in the wire. q = CV εA = ()Ed d = εEA = εΦE dq dΦ = i = ε E dt c dt Summary of Faraday’s Law r r dΦ ∫ E ⋅ ld = − B dt B If we form any closed loop, the line integral of the electric field equals the time rate change of E magnetic flux through the surface enclosed by the loop. If there is a changing magnetic field, then there will be electric fields induced in closed paths. The electric fields direction will tend to reduce the changing B field. Note; it does not matter if there is a wire loop or an imaginary closed path, an E field will be induced. Potential has no meaning in this non-conservative E field. Charge is flowing onto this parallel plate capacitor at a rate dQ/dt=2 A II I III What is the displacement current in regions I and III ? A) 2 A B) 1 A C) 0 D) -2A Charge is flowing onto this parallel plate capacitor at a rate dQ/dt=2 A II I III What is the displacement current in region II ? A) -2/3A B) 1 A C) 2 A D) O A Summary of Ampere’s Generalized Law r dΦ ∫ B ⋅ dl = µ i + ε E 0 c 0 dt Current ic If we form any closed loop, the line B integral of the B field is nonzero if there is (constant or changing) current through the loop. If there is a changing electric field E through the loop, then there will be magnetic fields induced about a closed loop path. B Maxwell’s Equations James Clerk Maxwell (1831-1879) • generalized Ampere’s Law • made equations symmetric: – a changing magnetic field produces an electric field – a changing electric field produces a magnetic field • Showed that Maxwell’s equations predicted electromagnetic waves and c =1/ √ε0µ0 • Unified electricity and magnetism and light. All of electricity and magnetism can be summarized by Maxwell’s Equations. More important applications of Faraday’s Law Mutual Inductance If we have a constant current i 1 in coil 1, a constant magnetic field is created and this produces a constant magnetic flux in coil 2. Since the ΦB2 is constant, there NO induced current in coil 2. If current i 1 is time varying, then the ΦB2 flux is varying and this induces an emf ε2 in coil 2, the emf is dΦ ε = −N B2 2 2 dt We introduce a ratio, called mutual inductance , of flux in coil 2 divided by the current in coil 1. N2ΦB2 M21 = i1 Mutual Inductance N2ΦB2 mutual inductance , M 21 = , can now be used in Faraday’s eqn . i1 M21 i1 = N2 ΦB2 di dΦ di M 1 = N B2 = −ε ; ε = −M 1 21 dt 2 dt 2 2 21 dt We can also the varying current i 2 which creates a changing flux ΦB1 in coil 1 and induces an emf ε1. This is given by a similar eqn. di ε = −M 2 1 12 dt It can be shown ( we do not prove here ) that, M12 = M 21 = M The units of mutual inductance is T ⋅ m2/A = Weber/A = Henry (after Joseph Henry, who nearly discovered Faraday’s Law) Mutual Inductance The induced emf, di 2 ε1 = −M has the following features; dt • The induced emf opposes the magnetic flux change (Lenz’s Law) • The induced emf increases if the current changes very fast • The induced emf depends on M, which depends only the geometry of the two coils and not the current. • For a few simple cases, we can calculate M, but usually it is just measured. Problem 30.1 Two coils have mutual inductance of 3.25 ×10 −4 H. The current in the first coil increases at a uniform rate of 830 A/s .