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Back to Circuits for a bit …. 11-11-06 Induction - Fall 2006 1 What the heck are we doing? Today Monday More of the same, Wednesday 7:30 AM we had our problem review session Continue on with Induction & Inductors Watch those piling up WebAssigns! EXAMINATION #3 After Holiday 11-11-06 Complete the remaining six chapters in the syllabus. Induction - Fall 2006 2 The loop is pushed into a region where the magnetic field is into the page. The motion creates an induced current in the loop which in turn produces a magnetic field at A and B. 25% 25% 25% 25% The field at A & B are the same. 2. The field at A is stronger than the one at B 3. The field at B is stronger than the one at A. 4. None of these. 11-11-06 Induction - Fall 2006 e. th es s. .. on e of is N at B ld Th e fie fie e Th Th e fie ld ld at A at A & is B ... s. .. 1. 3 Today we will consider a Coil 11-11-06 Induction - Fall 2006 4 We will base our discussion on Faraday’s Law d emf V E ds dt Lentz 11-11-06 Induction - Fall 2006 5 Consider the following: There are N turns in the solenoid. There is a current flowing in the direction shown. The power supply is set to 20 volts and the resistor is 10 ohms. The coil wire has negligible resistance. 11-11-06 Induction - Fall 2006 6 The steady state current (20 volts, 10 ohms) in the circuit is 11-11-06 Induction - Fall 2006 in fo . es m or e pe r am d ee N th a or e M Le ss th an 2 2 am pe am pe re re s 4. 2 3. 25% 25% 25% 25% n 2. 2 amperes Less than 2 amperes More than 2 amperes Need more info. s 1. 7 The DIRECTION of the magnetic field in the coil is A B Induction - Fall 2006 N 11-11-06 ot en o ug h Fr om B in fo ... to B Fr om 3. to 2. From A to B From B to A Not enough information is given. A 1. A 33% 33% 33% 8 Back to the coil diagram … B Recall from the last discussion that the magnetic field in the coil is given by: n = #turns per unit length Coil is infinitely long 11-11-06 B 0 ni Sort of Induction - Fall 2006 9 Back to the coil diagram … B 0 nI B The Flux through a single turn of the coil is BA or 0niA. Now, let’s increase the applied voltage linearly at a rate of V/t. The current will change at a rate I/t. 11-11-06 Induction - Fall 2006 10 Back to the coil diagram … B 0 nI B For the single coil (as t0) FARADAY Says: d d dI emf ( 0 nIA) 0 nA dt dt dt 11-11-06 Induction - Fall 2006 11 Looking into the coil from the end with the red arrow, the emf around the coil induced current will be B 50 50 % % 11-11-06 Induction - Fall 2006 co a In In a cl oc . .. 2. In a clockwise direction In a counterclockwise direction u. .. 1. 12 So … the induced emf B Will create a current that will oppose the change in the current. The induced emf will therefore oppose the applied voltage (also an emf) from the power supply. 11-11-06 Induction - Fall 2006 13 So for the single coil emf d single dt d dI ( 0 nIA) 0 nA dt dt becomes dI dt for a coil of N turns, " back emf" is dI dI emf - 0 nNA L dt dt I N 0 nIA N sin gle L 0 nNA N 0 nA I I I emf1coil 0 nA 11-11-06 Induction - Fall 2006 14 Definition of Inductance L N B L i UNIT of Inductance = 1 henry = 1 T- m2/A B is the flux near the center of one of the coils making the inductor 11-11-06 Induction - Fall 2006 15 An inductor in the form of a solenoid contains 420 turns, is 16.0 cm in length, and has a cross-sectional area of 3.00 cm2. What uniform rate of decrease of current through the inductor induces an emf of 175 μV? 11-11-06 Induction - Fall 2006 16 N B nl 0 nIA 2 L 0 n lA I I 11-11-06 Induction - Fall 2006 17 Look at the following circuit: Switch is open NO current flows in the circuit. All is at peace! 11-11-06 Induction - Fall 2006 18 At the INSTANT that the switch is closed, the current through the resistor is: Induction - Fall 2006 ’t te ll ro Ze 11-11-06 an Can’t tell C 3. 33% R 2. Zero E/R 1. 33% E/ 33% 19 Three years after the switch is closed, the current through the resistor is: i.. w e D on ’t c ar e .. w E/ R 3. ro 2. E/R Zero Don’t care .. we will be out by then! Ze 1. 33% 33% 33% 11-11-06 Induction - Fall 2006 20 Graph? IR E/R Probably looks something Like this. time 11-11-06 Induction - Fall 2006 21 Close the circuit… After the circuit has been closed for a long time, the current settles down. Since the current is constant, the flux through the coil is constant and there is no Emf. Current is simply E/R (Ohm’s Law) 11-11-06 Induction - Fall 2006 22 Close the circuit… When switch is first closed, current begins to flow rapidly. The flux through the inductor changes rapidly. An emf is created in the coil that opposes the increase in current. The net potential difference across the resistor is the battery emf opposed by the emf of the coil. 11-11-06 Induction - Fall 2006 23 Looking at the math dI emf L dt 11-11-06 Ebattery V (notation) dI V IR L 0 dt Induction - Fall 2006 24 Just as we did with the capacitor, we can solve this equation and we get: E t / I (1 e ) R L time constant R 11-11-06 Induction - Fall 2006 25 The growth 11-11-06 Induction - Fall 2006 26 Death of the current: dI IR L 0 dt solution E t / I e R 11-11-06 Induction - Fall 2006 27 Graph 11-11-06 Induction - Fall 2006 28 Consider the Solenoid Again… l B ds i 0 enclosed Bl 0 nli or n turns per unit length 11-11-06 Induction - Fall 2006 B 0 ni 29 Inductance & Geometry N B nlBA nl 0 niA L i i i or L 0 n 2 Al or inductance 2 L/l n A unit length Depends only on geometry just like C and is independent of current. 11-11-06 Induction - Fall 2006 30 Max Current Rate of increase = max emf VR=iR ~current 11-11-06 Induction - Fall 2006 31 E (1 e Rt / L ) R L (time constant) R i 11-11-06 Induction - Fall 2006 32 IMPORTANT QUESTION Switch closes. No emf Current flows for a while It flows through R Energy is conserved (i2R) WHERE DOES THE ENERGY COME FROM?? 11-11-06 Induction - Fall 2006 33 For an answer Return to the Big C E=e0A/d +dq +q 11-11-06 -q We move a charge dq from the (-) plate to the (+) one. The (-) plate becomes more (-) The (+) plate becomes more (+). dW=Fd=dq x E x d Induction - Fall 2006 34 The calc q dW (dq) Ed (dq) d (dq) d e0 e0 A d d q2 W qdq e0 A e0 A 2 or 1 2 Ad 1 2 1 2 W (A) e 0 2 Ad e 0 E Ad 2e 0 A 2 e0 2 e0 2 d 2 energy 1 2 u e0 E unit volum e 2 11-11-06 Induction - Fall 2006 The energy is in the FIELD !!! 35 What about POWER?? di E L iR dt i : di 2 iE Li i R dt power to circuit power dissipated by resistor Must be dWL/dt 11-11-06 Induction - Fall 2006 36 So dWL di Li dt dt 1 2 WL L idi Li 2 1 2 WC CV 2 11-11-06 Induction - Fall 2006 Energy stored in the Capacitor 37 WHERE is the energy?? l B ds i 0 enclosed 0l Bl 0 nil B 0 ni or B 0 Ni l BA 11-11-06 Induction - Fall 2006 0 Ni l A 38 Remember the Inductor?? N L i N Number of turns in inductor i current. Φ Magnetic flux throu gh one turn. ????????????? 11-11-06 Induction - Fall 2006 39 So … N i N i L 1 1 N 1 W Li 2 i 2 N i 2 2 i 2 0 NiA l 1 0 NiA 1 2 2 2 A W Ni 0 N i 2 l 2 0 l L 11-11-06 Induction - Fall 2006 40 1 A W N i 2 0 l 2 0 2 2 From before : B 0 Ni l 1 A 1 2 W Bl B V (volume) 2 0 l 2 0 2 2 or W 1 2 u B V 2 0 11-11-06 ENERGY IN THE FIELD TOO! Induction - Fall 2006 41 IMPORTANT CONCLUSION A region of space that contains either a magnetic or an electric field contains electromagnetic energy. The energy density of either is proportional to the square of the field strength. 11-11-06 Induction - Fall 2006 42 11-11-06 Induction - Fall 2006 43