#### Transcript Document

Lectures 16-17 (Ch. 29) Electromagnetic Induction 1. The law of EM induction 2. The Lenz’s rule 3. Motional emf: slide wire generator, Faraday’s disc dynamo, Ac and dc current generators Electromagnetic Induction, 1830-1832 Change of magnetic flux through the loop of wire induces current (i.e. emf) in the loop. Michael Faraday (1791-1867) Joseph Henry (1797-1878) Law of EM Induction (Faraday’s law) B B dA B cosdA B dA d B A A A dt [ ] 1T 1m 2 1Wb ( weber) B Wilhelm Weber (1804 – 1891) Flux can be changed by change of B or A or angle between B and dA. In order to find the direction of the induced current it is convenient to write the faraday’s law in the form l0 in the loop d B where l0 is the unite vector of a circulation which direction is connected with dA by a RHR: dt It is convenient to choose dA in such direction that B dA 0 dA l0 Examples. Find direction of the induced current. d B l0 dt l0 l0 l0 0 Induced current is in the direction opposite to l 0 l0 0 Induced current is in the direction of l 0 Lenz’s law Magnetic field produced by induced current opposes change of magnetic flux Heinrich Lenz (1804 –1865) Examples Example Motional emf Slide-wire generator d B dA l0 B dt dt dA lvdt l0 Blv Origin of this emf is in separation of charges in a rod caused by its motion in B. Fm qv B a Fm dl q b ( v B ) dl a b vBl Motional emf exists in the conductor moving in B. It does not require the existence of the closed circuit. The secondary magnetic force Fm ' I Bl m ’ External force is required to keep constant velocity of the rod Fext Fm ' I Bl 2 2 ( vBl ) Pel I 2 R R R dW dx (vBl) 2 Pmech Fext IBlv Blv dt dt R R Example. Find motional emf in the rod. I V d L Example. Find induced current in the loop with resistance R. V I Example A single rectangular loop of wire with the dimensions inside a region of B=0.5 T and part is outside the field. The total resistance of the loop is 0.2Ω . The loop is pulled from the field with a constant velocity of 5m/s. 1)What is the magnitude and direction of the induced current? 2) In which part of the loop an induced emf is developed? 3) Find the force required to pull the loop at a constant velocity. 4) Explain why such force is required. 0.1m 0.75m B 0.5m x v Example Find emf in each side of the loop and the net emf when the loop is the region: a) all inside the region of B b) partly outside of this region c) all outside of this region B=1T x 40cm v=2cm/s 30cm Faraday’s disk dynamo R BR2 v B dr B rdr 2 0 0 R v r F’m + Fm v AC –current generator (alternator) BA cos , t BA cost d BA sin t dt BAsin t I R R B 2 A2 sin 2 t B IABsin t R Induced current results in torque which slows down a rotation. External torque is required to maintain the rotation with a constant frequency. DC-current generator Applications 2007 Nobel Prize in Physics For the discovery of the giant magnetic resistance Big R B Peter Grünberg Albert Fert Tiny magnetic field triggers large change in electrical resistance. Small R B I Resistance strongly depends on the direction of the spin in the first ferromagnetic layer. When it is the same as in the next ferromagnetic layer R is small, when it’s opposite to it R is big. Better read-out heads for pocket-size devices: miniaturization of PC, ipods, etc. Eddy currents Meisner’s effect Eddy currents responsible for levitation and Meisner effect in superconductors S N v Bind B0 Eddy currents limit efficiency of transformers Induced nonelectrostatic electric field dI 0 dt d B dt Origin of emf? No motion, moreover no B outside solenoid, i.e. in the region of a wire loop. Then it should be E which results in induced current. dI 0 dt Fel dl d B q dt Fel qE , Fel dl 0 Nonconcervative force d B E dl dt E dl 0 Nonelectrostatic field B(t) should induce E by independently on the presence of the loop of the wire! Let’s find E(r). d B E dl dt 1.r R dB E 2r r dt B ni, K m 0 2 R E n di 2 dt 2) r R E r dB E 2r R dt n di R 2 E 2 dt r 2 r R Displacement current dB/dt produces E. Let’s show that dE/dt produces B! Consider the process of charging the capacitor. Calculate B in front of the plate of capacitor at r>R. 1 B dl 0 I encl 2 line Using the plane surface 1 we get B2r 0ic Using the bulging surface 2 we get B2r 0 We come to contradiction! What is wrong ? q CV , C 0 A ,V Ed , q 0 AE d id dq dE d E dE ic 0 A 0 id , jd 0 dt dt dt A dt B dl 0 (ic id ) 1.Now we get the same answer for both surfaces 1 and 2! 2. B≠0 between the plates! General form of Amper’s law d E B dl (i dt ) μ= Kmμ0, ε=Kε0, In free space K=1, Km =1 Let’s find B between the plates. 1.r R r2 B 2r 0 jd r 0ic 2 R id ic jd 2 A R 0ic r B 2 R 2 2.r R B 2r 0id 2 B r R 0ic B 2r Maxwell’s equations Two Gauss’s laws + Faraday’s law +Amper’s law qencl E dA B dA 0 d B E d l dt d E B dl (iencl dt ) James Clerk Maxwell (1831 –1879) Maxwell introduced displacement current, wrote these four equations together, predicted the electromagnetic waves propagating in vacuum with velocity of light and shown that light itself is e.m. wave. 1865 Maxwell’s theory 1887 Hertz’s experiment 1890 Marconi radio (wireless communication)