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
Energy Transformations in EM Waves and Circuits The energy of EM waves traveling in vacuum is stored in both E and B fields (in equal amounts). Circuit elements that store the E field energy - capacitors. Circuit elements that store the B field energy inductors. By combining capacitors and inductors, we can build the EM oscillators (e.g., generators of EM waves). 2
Maxwell s Equations 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. ssss E da = Q eeee ε 0 No magnetic monopoles; magnetic field lines form closed loops. ssss B da = 0 Faraday s Law of electromagnetic induction; a time-dependent Φ B generates E. looo E dl = d B da suuu E = dφ B Generalized Ampere s Law; B is produced by both currents and time-dependent Φ E. looo E B dl = μ 0 J + ε 0 suuu da The force on a (moving) charge. F = q E + v B 3
Induced E.M.F. and its consequences 1. FFFFFFF: E = dφ B 2. I t eeeeeeee B t 3. I t B t B t 8
(Self) Inductance Variable battery I t FFFFFFF: E = dφ B B t The flux depends on the current, so a wire loop with a changing current induces an aitional 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.). The back e.m.f. : E = dφ B = L L the coefficient of proportionality between Φ and I, purely geometrical quantity. Inductance: (or self-inductance) L Φ I Units: T m 2 /A = Henry (H) This definition applies when there is a single current path (see below for an inductance of a system of distributed currents J r ). 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 N: L Φ B I l The magnetic flux generated by current I: Φ B = Bπr 2 N = B = μ 0 N l I = μ 0N 2 I l πr 2 = μ 0 n 2 I l πr 2 L = Φ B I = μ 0N 2 l πr 2 = μ 0 n 2 l πr 2 (n the number of turns per unit length) Note that - the inductance scales with the solenoid volume l πr 2 ; - the inductance scales as N 2 : B is proportional to N l and the net flux N. Let s plug some numbers: l =0.1m, N=100, r=0.01m L = μ 0 n 2 l πr 2 = 4π 10 7 100 0.1 2 0.1 π 0.01 2 40μμ How significant is the induced e.m.f.? For ~104 A/s (let s say, I~ 10 ma in 1 µs) E = L = 4 10 5 H 1 10 4 A s = 0.4V 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. V i t V t Let s ramp up the current from 0 to the final value I (i t is the instantaneous current value): = L t P t = V t i t = L t i t i t The net work to ramp up the current from 0 to I : = P t f U = P t i P(power) - the rate at which energy is being delivered to the inductor from external sources. I = LL 0 = 1 2 LI2 U = 1 2 LI2 This energy is stored in the magnetic field created by the current. L = μ 0 n 2 l πr 2 = μ 0 n 2 vvvvvv U B = 1 2 μ 0n 2 vvvvvv I 2 = B = μ 0 nn = B2 2μ 0 vvvvvv = u B vvvvvv The energy density in a magnetic field: u B = B2 2μ 0 (compare with u E = This is a general result (not solenoid-specific). ε 0E2 2 ) 14
Quench of a Superconducting Solenoid MRI scanner: the magnetic field up to 3T within a volume ~ 1 m 3. The magnetic field energy: U = vvvvvv u B = 1m 3 3T 2 2 4π 10 7 WW A m 4MM The capacity of the LHe dewar: ~2,000 liters. 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: U B = 1 2 LI2 = B2 2μ 0 dτ all sssss L 2U B I 2 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 I I b a Uniform current density in the central conductor of radius a: b U B = 1 2μ 0 B r 0 2 dτ B r = μ 0 I, a < r < b 2ππ μ 0 Ir, r < a 2πa2 at home L = 2U B I 2 Straight Wires: Do They Have Inductance? Using L = 2U B /I 2 : U B = 1 2μ 0 a b μ 0I 2ππ 2 Using L = Φ/I: ~l B μ 0I 2ππ l dτ b = 1 2μ 0 a μ 0I 2ππ l Φ l μ 0I 2ππ a 2 l 2ππππ = μ 0I 2π l ll l a (b~l ) μ 0I 2 4π l ll l a L H μ 0 2π l ll l a 2 10 7 l m ll l a l = 1cc L 10 9 H L H μ 0 2π l ll l a 17
Inductance vs. Capacitance L = Φ B I = 2U B I 2 C = Q V = 2U E V 2 U B = 1 2 LI2 = 1 2 Φ B 2 L U B = B2 2μ 0 vvvvvv = u B vvvvvv Q 2 U E = 1 2 CV2 = 1 2 C U E = ε 0E 2 2 vvvvvv = u E vvvvvv 18
Mutual Inductance Let s consider two wire loops at rest. A time-dependent current in loop 1 produces a time-dependent magnetic field B 1. The magnetic flux is linked to loop 1 as well as loop 2. Faraday s law: the time dependent flux of B 1 induces e.m.f. in both loops. B 1 I 1 loop 2 loop 1 Primitive transformer The e.m.f. in loop 2 due to the time-dependent I 1 in loop 1: E 2 = dφ 1 2 Φ 1 2 = M 1 2 I 1 The flux of B 1 in loop 2 is proportional to the current I 1 in loop 1. The coefficient of proportionality M 1 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 M 1 2 = M 2 1 = M so there is just one mutual inductance M. M 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). E 2 = M I 1 M = Φ 1 2 I 1 19
Calculation of Mutual Inductance Calculate M for a pair of coaxial solenoids (r 1 = r 2 = r, l 1 = l 2 = l, n 1 and n 2 - the densities of turns). l B 1 = μ 0 n 1 I 1 Φ = B 1 πr 2 = μ 0 n 1 I 1 πr 2 M 1 2 = M 2 1 = M L = μ 0 n 2 l πr 2 = μ 0N 2 πr 2 Experiment with two coaxial coils. l the flux per one turn M = N 2Φ = N I 2 μ 0 n 1 πr 2 = μ 0N 1 N 2 πr 2 1 l M = L 1 L 2 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 ~K m times greater. E 2 = M di 1 = K m M di 1 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 2a b Example: estimate the inductance of a circular wire loop with the radius of the loop, b, being much greater than the wire radius a. 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 B μ 0I 2ππ 2πb Note that we cannot assume that the wire radius is zero: b Φ 2ππ μ 0I 2ππ = μ 0 bi ll b a a L b = 1m, a = 10 3 m 4π 10 7 L = Φ I μ 0b ll b a WW A m 1m ll103 10μμ b Φ~ μ 0I 2ππ "0" ~ll b "0" - diverges! What to do: introduce some reasonably small non-zero radius. 22