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Transcript

INDUCTION Spring ‘09 DIS HERE WEEK We begin the study of magnetic induction There will be a quiz on Friday There is a new WebAssign. Exam #3 is Wednesday4/8 It will include the material covered through April 6th. The end is in sight … Check the website for the Final Exam Schedule AMPERE’S LAW B d s i 0 enclosed B d s ( i i ) 0 1 2 USE THE RIGHT HAND RULE IN THESE CALCULATIONS LAST TIME: A SIMPLE EXAMPLE FIELD AROUND A LONG STRAIGHT WIRE B ds i 0 enclosed B 2r 0i 0i B 2r THE FIGURE BELOW SHOWS A CROSS SECTION OF AN INFINITE CONDUCTING SHEET CARRYING A CURRENT PER UNIT X-LENGTH OF L; THE CURRENT EMERGES PERPENDICULARLY OUT OF THE PAGE. (A) USE THE BIOT–SAVART LAW AND SYMMETRY TO SHOW THAT FOR ALL POINTS P ABOVE THE SHEET, AND ALL POINTS P´ BELOW IT, THE MAGNETIC FIELD B IS PARALLEL TO THE SHEET AND DIRECTED AS SHOWN. (B) USE AMPERE'S LAW TO FIND B AT ALL POINTS P AND P´. FIRST PART Vertical Components Cancel APPLY AMPERE TOL CIRCUIT B Infinite Extent B current per unit length Current inside the loop is therefore : i L THE “MATH” B Infinite Extent B B ds i 0 enclosed BL BL 0 L B 0 2 Infinite Sheet of Charge E 2 0 Infinite sheet of current B 0 2 COIL OR SOLENOID Valve Application Switches THE MAGNETIC FIELD A PHYSICAL SOLENOID/COIL MODEL INSIDE THE SOLENOID For an “INFINITE” (long) solenoid the previous problem and SUPERPOSITION suggests that the field OUTSIDE this solenoid is small! MORE ON LONG SOLENOID Field is ZERO far from coil! Field looks “UNIFORM” Field is ZERO far from coil THE REAL THING….. Finite Length Weak Field Stronger - Leakage ANOTHER WAY Far away Ampere : B ds i 0 enclosed 0h Bh 0 nih B 0 ni n=number of turns per unit length. nh=total number of turns. THE LENGTH OF A SOLENOID IS DOUBLED BUT THE NUMBER OF TURNS REMAINS THE SAME. A. B. C. D. E. The magnetic field is doubled. The magnetic field is cut to a quarter. The magnetic field does not change. The magnetic field is quadrupled. The magnetic field is cut in half. THE NUMBER OF TURNS IS DOUBLED BUT THE LENGTH REMAINS THE SAME. A. B. C. D. E. The magnetic field is quadrupled. The magnetic field is cut to a quarter. The magnetic field is doubled. The magnetic field does not change. The magnetic field is cut in half. B 0 ni APPLICATION Creation of Uniform Magnetic Field Region Minimal field outside except at the ends! TWO COILS “REAL” HELMHOLTZ COILS Used for experiments. Can be aligned to cancel out the Earth’s magnetic field for critical measurements. THE TOROID Slightly less dense than inner portion THE TOROID Ampere again. We need only worry about the INNER coil contained in the path of integratio n : B ds B 2r Ni (N total # turns) 0 so 0 Ni B 2r MAGNETIC FLUX NEW CONCEPT – WELL SORTA NEW MAGNETIC FLUX B dA M M B dA Like Gauss - Open Surface MAGNETIC FLUX IS A A. B. C. Vector Scalar Tensor MAGNETIC FLUX M B dA Like Gauss - Open Surface For a CLOSED Surface we might expect this to be equal to some constant times the enclosed poles … but there ain’t no such thing! B d A 0 CLOSED SURFACE A PUZZLEMENT .. closed path B ds 0ienclosed Let’s apply this to the gap of a capacitor. CONSIDER THE POOR LITTLE CAPACITOR… i i CHARGING OR DISCHARGING …. HOW CAN CURRENT FLOW THROUGH THE GAP In a FIELD description?? THROUGH WHICH SURFACE DO WE MEASURE THE CURRENT FOR AMPERE’S LAW? I=0 IN THE GAP… DISPLACEMENT CURRENT The ELECTRIC FLUX through S2 E EA q 0 d E 1 dq dt 0 dt Let dq I d (in gap) dt Displaceme nt Current dq d E Id 0 dt dt SO FAR .. We found that currents create magnetic fields. Stationary charges do not. Static magnetic fields do not create currents. What about CHANGING magnetic fields?? FROM THE DEMO .. FARADAY’S EXPERIMENTS ? ? INSERT MAGNET INTO COIL REMOVE COIL FROM FIELD REGION Summary THAT’S STRANGE ….. These two coils are perpendicular to each other REMEMBER THE DEFINITION OF TOTAL ELECTRIC FLUX THROUGH A CLOSED SURFACE: E surface d E Total Flux of the Electric Field LEAVING a surface is E E n outdA MAGNETIC FLUX - REMINDER Applies to an OPEN SURFACE only. “Quantity” of magnetism that goes through a surface. A Scalar B B dA surface CONSIDER A LOOP xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx Magnetic field passing through the loop is CHANGING. FLUX is changing. There must be an emf developed around the loop. A current develops (as we saw in demo) Work has to be done to move a charge completely around the loop. FARADAY’S LAW (MICHAEL FARADAY) Again, xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx for a current to flow around the circuit, there must be an emf. (An emf is a voltage) The voltage is found to increase as the rate of change of flux increases. FARADAY’S LAW (MICHAEL FARADAY) xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx Faraday' s Law d emf dt We will get to the minus sign in a short time. FARADAY’S LAW (THE MINUS SIGN) xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx Using the right hand rule, we would expect the direction of the current to be in the direction of the arrow shown. FARADAY’S LAW (MORE ON THE MINUS SIGN) xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx The minus sign means that the current goes the other way. This current will produce a magnetic field that would be coming OUT of the page. The Induced Current therefore creates a magnetic field that OPPOSES the attempt to INCREASE the magnetic field! This is referred to as Lenz’s Law. HOW MUCH WORK? xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx Work/Unit Charge d W / q V E ds dt A magnetic field and an electric field are intimately connected.) MAGNETIC FLUX B B dA This is an integral over an OPEN Surface. Magnetic Flux is a Scalar The UNIT of FLUX is the 1 weber = 1 T-m2 weber WE FINALLY STATED FARADAY’s LAW d emf V E ds dt FROM THE EQUATION Lentz d emf V E ds dt B B dA FLUX CAN CHANGE B B dA If B changes If the AREA of the loop changes Changes cause emf s and currents and consequently there are connections between E and B fields These are expressed in Maxwells Equations MAXWELL’S FOUR EQUATIONS Ampere’s Law Gauss Faraday No Monopoles closed surface B dA 0 B dA 0 ANOTHER VIEW OF THAT DAMNED MINUS SIGN AGAIN …..SUPPOSE THAT B BEGINS TO INCREASE ITS MAGNITUDE INTO THE PAGE xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx The Flux into the page begins to increase. An emf is induced around a loop A current will flow That current will create a new magnetic field. THAT new field will change the magnetic flux. THE STRANGE WORLD OF DR. LENTZ LENZ’S LAW Induced Magnetic Fields always FIGHT to stop what you are trying to do! i.e... Murphy’s Law for Magnets EXAMPLE OF NASTY LENZ The induced magnetic field opposes the field that does the inducing! DON’T HURT YOURSELF! The current i induced in the loop has the direction such that the current’s magnetic field Bi opposes the change in the magnetic field B inducing the current. Let’s do the Lentz Warp again ! AGAIN: LENZ’S LAW An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current. (The result of the negative sign!) … OR The toast will always fall buttered side down! AN EXAMPLE The field in the diagram creates a flux given by B=6t2+7t in milliWebers and t is in seconds. (a) What is the emf when t=2 seconds? (b) What is the direction of the current in the resistor R? THIS IS AN EASY ONE … B 6t 7t 2 d emf 12t 7 dt at t 2 seconds emf 24 7 31mV Direction? B is out of the screen and increasing. Current will produce a field INTO the paper (LENZ). Therefore current goes clockwise and R to left in the resistor. The diagram shows two parallel loops of wire having a common axis. The smaller loop (radius r) is above the larger loop (radius R) by a distance x >> R. Consequently, the magnetic field due to the current i in the larger loop is nearly constant throughout the smaller loop. Suppose that x is increasing at the constant rate of dx/dt = v. (a) Determine the magnetic flux through the area bounded by the smaller loop as a function of x. In the smaller loop, find (b) the induced emf and (c) the direction of the induced current. v DOING IT B is assumed to be constant through the center of the small loop and caused by the large one. q THE CALCULATION OF BZ dBz dB cos q cos q cos q R R 0 ids 4 R 2 x 2 1/ 2 x2 q 0 ids R dBz 4 R 2 x 2 R 2 x 2 ds Rd 2 Bz 0iR 2 2 R x 2 2 3/ 2 1/ 2 MORE WORK In the small loop: Bz A r 2 Bz r 2 0iR 2 2R x For x R (Far Away as prescribed ) 2 r 2 0iR 2 3 2x d 3r 2 0iR 2 V emf 4 2x dt 2 3/ 2 dx/dt=v WHICH WAY IS CURRENT IN SMALL LOOP EXPECTED TO FLOW?? B q WHAT HAPPENS HERE? Begin to move handle as shown. Flux through the loop decreases. Current is induced which opposed this decrease – current tries to re-establish the B field. MOVING THE BAR Flux BA BLx Dropping the minus sign... d dx emf BL BLv dt dt emf BLv i R R MOVING THE BAR TAKES WORK F BiL BL BLv R or v B 2 L2 v F R dW d POWER Fx Fv dt dt B 2 L2 v P v R B 2 L2 v 2 P R WHAT ABOUT A SOLID LOOP?? Energy is LOST BRAKING SYSTEM METAL Pull Back to Circuits for a bit …. DEFINITION Current in loop produces a magnetic field in the coil and consequently a magnetic flux. If we attempt to change the current, an emf will be induced in the loops which will tend to oppose the change in current. This this acts like a “resistor” for changes in current! REMEMBER FARADAY’S LAW d emf V E ds dt Lentz LOOK AT THE FOLLOWING CIRCUIT: Switch is open NO current flows in the circuit. All is at peace! CLOSE THE CIRCUIT… After the circuit has been close 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) 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. CLOSE THE CIRCUIT… d emf dt Ebattery V (notation) d V iR 0 dt MOVING RIGHT ALONG Ebattery V… (notation) d V iR 0 dt The flux is proportion al to the current as well as to the number of turns, N. For a solonoid, i Li N B d di L dt dt di V iR L 0 dt 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 CONSIDER A SOLENOID l B ds i 0 enclosed Bl 0 nli or n turns per unit length B 0 ni SO…. 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. INDUCTIVE CIRCUIT i Switch to “a”. Inductor seems like a short so current rises quickly. Field increases in L and reverse emf is generated. Eventually, i maxes out and back emf ceases. Steady State Current after this. THE BIG INDUCTION As we begin to increase the current in the coil The current in the first coil produces a magnetic field in the second coil Which tries to create a current which will reduce the field it is experiences And so resists the increase in current. BACK TO THE REAL WORLD… Switch to “a” i sum of voltage drops 0 : di E iR L 0 dt same form as the capacitor equation q dq E R 0 C dt SOLUTION E Rt / L i (1 e ) R time constant L R SWITCH POSITION “B” E0 di L iR 0 dt E t / i e R VR=iR ~current Max Current Rate of increase = max emf E (1 e Rt / L ) R L (time constant) R i 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?? FOR AN ANSWER RETURN TO THE BIG C We E=0A/d +dq +q -q 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 THE CALC q dW (dq) Ed (dq) d (dq) d 0 0 A d d q2 W qdq 0 A 0 A 2 or 1 2 Ad 1 2 1 2 W (A) 0 2 Ad 0 E Ad 2 0 A 2 0 2 0 2 d 2 energy 1 2 u 0 E unit volum e 2 The energy is in the FIELD !!! WHAT ABOUT POWER?? di E L iR dt i : di 2 iE Li i R dt power to circuit Must be dWL/dt power dissipated by resistor SO dWL di Li dt dt 1 2 WL L idi Li 2 1 2 WC CV 2 Energy stored in the Capacitor WHERE IS THE ENERGY?? l B ds i 0 enclosed 0l Bl 0 nil B 0 ni or B 0 Ni l BA 0 Ni l A REMEMBER THE INDUCTOR?? N L i N Number of turns in inductor i current. Φ Magnetic flux throu gh one turn. SO … N L i N i L 1 2 1 2 N 1 W Li i 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 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 ENERGY IN THE FIELD TOO! 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.