Lecture 22. Inductance. Magnetic Field Energy. Outline: Self-induction and self-inductance. Inductance of a solenoid. The energy of a magnetic field. Alternative definition of inductance. Mutual Inductance. 1 - CH2 exam average ~54 %. - To the right are VERY tentative correlations between percentage scores and letter grades in Physics 227. The instructors reserve the right to adjust these boundaries (both up and down) before final grades are assigned at the end of the semester. A ≥ 85 80 ≤ B+ < 85 70 ≤ B < 80 65 ≤ C+ < 70 53 ≤ C < 65 40 ≤ D < 53 40 < F 2 Maxwell’s Equations ο± Gauss’s Law: charges are sources of electric field (and a non-zero net electric field flux through a closed surface); field lines begin and end on charges. πππππ π0 οΏ½ πΈ β ππ΄β = οΏ½ π΅ β ππ΄β = 0 οΏ½ πΈ β ππβ = − οΏ½ π΅ β ππβ = π0 οΏ½ π π π π ο± No magnetic monopoles; magnetic field lines form closed loops. π π π π ο± Faraday’s Law of electromagnetic induction; a time-dependent ΦB generates E. ππππ π οΏ½ π΅ β ππ΄β ππ π π’π’π’ πΦπ΅ β°=− ππ ο± Generalized Ampere’s Law; B is produced by both currents and time-dependent ΦE. ππππ π π’π’π’ π½β + π0 ο± The force on a (moving) charge. πΉβ = π πΈ + π£β × π΅ ππΈ β ππ΄β ππ 3 Induced E.M.F. and its consequences πΦπ΅ πΉπΉπΉπΉπΉπΉπΉ: β° = − ππ 1. 2. ππππππππ π© π 3. πΌ π‘ πΌ π‘ π© π 8 (Self) Inductance Variable battery πΌ π‘ π© π The back e.m.f. : π =− Inductance: (or self-inductance) πΦπ΅ πΉπΉπΉπΉπΉπΉπΉ: β° = − ππ The flux depends on the current, so a wire loop with a changing current induces an “additional” e.m.f. in itself that opposes the changes in the current and, thus, in the magnetic flux (sometimes called the back e.m.f.). π π±π© π π = −π³ π π π π π± π³≡ π° πΏ – the coefficient of proportionality between Φ and πΌ, purely geometrical quantity. Units: T⋅m2/A = Henry (H) The self-inductance L is always positive. This definition applies when there is a single current path (see below for an inductance of a system of distributed currents π½ π ). Inductance as a circuit element: 9 Inductance of a Long Solenoid Long solenoid of radius r and length l with the total number of turns π: πΏ≡ Φπ΅ πΌ l Note that The magnetic flux generated by current πΌ: π π0 π 2 πΌ 2 Φπ΅ = π΅ππ π = π΅ = π0 πΌ = ππ = π0 π2 πΌ l ππ 2 l l 2 Φπ΅ π0 π 2 2 πΏ= = ππ = π0 π2 l ππ 2 πΌ l (n – the number of turns per unit length) - the inductance scales with the solenoid volume l ππ 2 ; π - the inductance scales as N2: B is proportional to l and the net flux ∝ N. Let’s plug some numbers: l =0.1m, N=100, r=0.01m 100 πΏ = π0 π2 l ππ 2 = 4π β 10−7 β 0.1 How significant is the induced e.m.f.? For ππ ~104 π΄/π (let’s say, βπΌ~ 1 mA in 0.1 µs) ππ 2 β 0.1 β π 0.01 2 ≈ 40ππ ππ 4 β 10−5 π» × 1 β 104 π΄ β° =πΏ = = 0.4π ππ π 10 Inductance vs. Resistance Inductor affects the current flow only if there is a change in current (di/dt ≠ 0). The sign of di/dt determines the sign of the induced e.m.f. The higher the frequency of current variations, the larger the induced e.m.f. 13 Energy of Magnetic Field To increase a current in a solenoid, we have to do some work: we work against the back e.m.f. Let’s ramp up the current from 0 to the final value I (π π‘ is the instantaneous current value): V π π‘ π π‘ π π‘ = πΏ ππ = π π‘ ππ The net work to ramp up the current from 0 to πΌ : ππ π‘ ππ π π‘ =π π‘ π π‘ =πΏ ππ π‘ π π‘ ππ P(power) - the rate at which energy is being delivered to the inductor from external sources. π πΌ 1 2 π = οΏ½ π π‘ ππ = οΏ½ πΏπΏ β ππ = πΏπΌ 2 π 0 π π πΌ = π³π° π This energy is stored in the magnetic field created by the current. πΏ = π0 π2 l ππ 2 = π0 π2 β π£π£π£π£π£π£ 1 π΅2 2 2 ππ΅ = π0 π π£π£π£π£π£π£ β πΌ = π΅ = π0 ππ = β π£π£π£π£π£π£ = π’π΅ β π£π£π£π£π£π£ 2π0 2 The energy density in a magnetic field: π©π ππ© = πππ (compare with π’πΈ = This is a general result (not solenoid-specific). π0 πΈ 2 2 ) 14 Quench of a Superconducting Solenoid MRI scanner: the magnetic field up to 3T within a volume ~ 1 m3. The magnetic field energy: π = π£π£π£π£π£π£ β π’π΅ = 1π The capacity of the LHe dewar: ~2,000 liters. 3 3π 2 ππ 2 β 4π β 10−7 π΄βπ ≈ 4ππ The heat of vaporization of liquid helium: 3 kJ/liter. Thus, ~ 1000 liters will be evaporated during the quench. Magnet quench: http://www.youtube.com/watch?v=tKj39eWFs10&feature=related 15 Another (More General) Definition of Inductance Alternative definition of inductance: 1 2 πΏπΌ = 2 οΏ½ πππ π π π π π π΅2 ππ 2π0 π³≡ ππΌπ© π°π Advantage: it works for a distributed current flow (when it’s unclear which loop we need to consider for the flux calculation). 16 Example: inductance per unit length for a coaxial cable πΌ π π πΌ Uniform current density in the central conductor of radius a: π 1 ππ΅ = οΏ½ π΅ π 2π0 0 2 ππ π΅ π = Straight Wires: Do They Have Inductance? Using π³ = ππΌπ© /π°π : π π0 πΌ 1 ππ΅ = οΏ½ 2ππ 2π0 π 2 π π0 πΌ 1 ππ = οΏ½ 2ππ 2π0 π Using π³ = π±/π°: π πΌ ~l π΅ ≈ 0 2ππ l l Φ ≈ lοΏ½ π 2 π0 πΌ ,π < π < π 2ππ π0 πΌπ , 2ππ2 π<π π0 πΌ 2 l β l β 2ππππ ≈ (π~l ) ≈ β l β ππ 4π π π0 πΌ ππ π° l ππ = β l β ππ 2ππ ππ π at home πΏ= 2ππ΅ πΌ2 l ππ π³π― ≈ l β ππ ππ π π0 l l −7 πΏπ» ≈ l β ππ ≈ 2 β 10 l π β ππ 2π π π l = 1ππ πΏ ≈ 10−9 π» 17 Inductance vs. Capacitance πΏ= Φπ΅ 2ππ΅ = 2 πΌ πΌ 1 2 1 Φπ΅ 2 ππ΅ = πΏπΌ = 2 2 πΏ π΅2 ππ΅ = β π£π£π£π£π£π£ = π’π΅ β π£π£π£π£π£π£ 2π0 πΆ= π 2ππΈ = 2 π π 1 2 1 π2 ππΈ = πΆπ = 2 2 πΆ π0 πΈ 2 ππΈ = β π£π£π£π£π£π£ = π’πΈ β π£π£π£π£π£π£ 2 18 Mutual Inductance Let’s consider two wire loops at rest. A time-dependent current in loop 1 produces a time-dependent magnetic field B1. The magnetic flux is linked to loop 1 as well as loop 2. Faraday’s law: the time dependent flux of B1 induces e.m.f. in both loops. loop 2 π©π The e.m.f. in loop 2 due to the time-dependent I1 in loop 1: β°2 = − πΌ1 loop 1 Primitive “transformer” πΦ1→2 ππ Φ1→2 = π1→2 πΌ1 The flux of B1 in loop 2 is proportional to the current I1 in loop 1. The coefficient of proportionality π1→2 (the so-called mutual inductance) is, similar to L, a purely geometrical quantity; its calculation requires, in general, complicated integration. Also, it’s possible to show that π1→2 = π2→1 = π so there is just one mutual inductance π. π can be either positive or negative, depending on the choices made for the senses of transversal about loops 1 and 2. Units of the (mutual) inductance – Henry (H). β°2 = −π ππΌ1 ππ π= Φ1→2 πΌ1 19 Calculation of Mutual Inductance l π1→2 = π2→1 = π Calculate M for a pair of coaxial solenoids (π1 = π2 = π, l 1 = l 2 = l, π1 and π2 - the densities of turns). π΅1 = π0 π1 πΌ1 the flux per one turn 2 π2 Φ π π π ππ 0 1 2 π= = π2 π0 π1 ππ 2 = l πΌ1 2 ππ 2 π π 0 πΏ = π0 π2 l ππ 2 = l Experiment with two coaxial coils. Φ = π΅1 ππ 2 = π0 π1 πΌ1 ππ 2 π= πΏ1 πΏ2 The induced e.m.f. in the secondary coil is much greater if we use a ferromagnetic core: for a given current in the first coil the magnetic flux (and, thus the mutual inductance) will be ~πΎπ times greater. ππΌ1 ππΌ1 β°2 = π∗ = πΎπ π ππ ππ no-core mutual inductance 20 Next time: Lecture 23. RL and LC circuits. §§ 30.4 - 6 21 Appendix I: Inductance of a Circular Wire Loop 2π Example: estimate the inductance of a circular wire loop with the radius of the loop, b, being much greater than the wire radius a. π The field dies off rapidly (~1/r) with distance from the wire, so it won’t be a big mistake to approximate the loop as a straight wire of length 2πb, and integrate the flux through a strip of the width b: b π0 πΌ π΅≈ 2ππ π Φ ≈ 2ππ οΏ½ 2πb π πΏ π = 1π, π = π0 πΌ π ππ = π0 ππΌ β ππ 2ππ π 10−3 π ≈ 4π β Note that we cannot assume that the wire radius is zero: π π0 πΌ π Φ~ οΏ½ ππ ~ππ 2ππ "0" - diverges! 10−7 πΏ= Φ π ≈ π0 π β ππ πΌ π ππ 1π β ππ103 ≈ 10ππ π΄βπ What to do: introduce some reasonably small non-zero radius. "0" 22