Motional Electromotive Force

Transcription

1 Motional Electromotive Force The charges inside the moving conductive rod feel the Lorentz force The charges drift toward the point a of the rod The accumulating excess charges at point a create an electric field Eventually a force equilibrium is reached q E = q v B This leads to a potential difference between points a and b V ab = E L = v B L

2 Motional Electromotive Force When the conductive rod is connected via a stationary U-shaped loop, no charge build-up occurs and the charges move in the U-shaped loop forming a current The moving rod acts as a motional EMF ε = E L = v B L If the total resistance is R I R = v B L Generalization leads to the Motional EMF ε = ( v B ) dl

3 Induced Electric Fields A long solenoid produces the field B = µ 0 n I For the magnetic flux we obtain Φ B = B A = µ 0 n I A For the EMF for a changing current ε = dφ B = µ 0 n A di Leading to a current in the loop of I = ε R

4 Induced Electric Fields A force is moving the charges around the loop, but it cannot be magnetic as the loop is not inside a magnetic field The force must be due to an induced electric field From chapter 23 we know for a conservative force ε = E dl And E dl = 0

5 Induced Electric Fields Here the total work done to a charge moving around the loop by the electric field is q ε So, the electric field cannot be conservative and we call this a nonelectrostatic field We obtain for a stationary path E dl = dφ B

6 Eddy Currents A rotating metal disk is partially exposed to a magnetic field A current forms in the exposed section with backflow in not exposed sections of the disk The current direction is downward because the force acting on the charges is F = q ( v B ) This results in a force created by the moving charges opposing the direction of motion of the disk F = I ( L B )

7 Displacement Current and Maxwell s Equations We just found that a varying magnetic field gives rise to an induced electric field The symmetry of Nature also leads to varying electric fields giving rise to magnetic fields Study Ampere s Law B dl = µ 0 I encl Ampere s Law is correct for the Plane surface For the bulging surface, no current is going through that surface, so B dl = 0

8 Displacement Current and Maxwell s Equations However, while charges are building up on the capacitor surface, the electric field and electric flux increase q = C v = ε A d ( E d ) = ε E A = ε Φ E dq i C = = ε dφ E = i D This is a new form of fictitious current, called displacement current, invented by James Clark Maxwell

9 Displacement Current and Maxwell s Equations This leads to the generalized Ampere s Law B dl = µ 0 ( i C + i D ) encl This can be tested with this capacitor arrangement and following the circular path with a and b The total current through that circle is i D encl = i C r 2 π r 2 = i C π R 2 R 2 B dl = 2 π r B = µ 0 i r 2 µ C B = 0 r i 2 π R 2 C R 2

10 Maxwell s Equations E da = Q encl ε 0 Gauss s Law for electric field B da = 0 Gauss s Law for magnetic field B dl = µ 0 ( i C + ε 0 dφ E ) encl Ampere s Law E dl = dφ B Faraday s Law

11 Superconductivity and the Meissner Effect Superconductive materials not only have zero resistance, but also are diamagnetic The Meissner effect is the expulsion of external magnetic field lines from the superconductor

12 Goals for Chapter 30 To introduce and illustrate mutual inductance To consider self-inductance To calculate magnetic-field energy To describe and study R-L circuits To describe and study L-C circuits To describe and study L-R-C circuits

13 Mutual Inductance A current in coil 1 produces a magnetic field The magnetic field lines produce a magnetic flux through coil 2 If the current, and hence the magnetic field, change in coil 1, the magnetic flux through coil 2 will also change Changing flux through coil 2 leads to an EMF induced in coil 2 ε 2 = N 2 dφ B2 The flux through coil 2 and the current in coil 1 are proportional N 2 Φ B2 = M 21 i 1 Φ B2 = Flux through one single turn of coil 2 M 21 = Mutual inductance

14 Mutual Inductance Using dφ N B2 2 = M 21 di 1 We obtain ε 2 = M 21 di 1 For the mutual inductance we can write N M 21 = 2 Φ B2 = i 1 N 1 Φ B1 i 2 The mutual inductance is a constant depending on the geometry of the two coils If a magnetic material is also present the mutual inductance depends on K m