EMF Notes 9; Electromagnetic Induction ELECTROMAGNETIC INDUCTION

Transcription

1 EMF Notes 9; Electromgnetic nduction EECTOMAGNETC NDUCTON (Y&F Chpters 3, 3; Ohnin Chpter 3) These notes cover: Motionl emf nd the electric genertor Electromgnetic nduction nd Frdy s w enz s w nduced electric field nductnce Mgnetic energy The Electric nd mgnetic fields re inter-relted. The electric nd mgnetic fields re not independent. n fct: A chnging mgnetic field induces n electric field The reverse is lso true: chnging electric field induces mgnetic field To see how this occurs, we first consider MOTONA emf. B (out of pge) Motionl emf Consider wire moving with velocity v in mgnetic field B. Assume for simplicity tht v nd B re perpendiculr The electrons in the wire experience force v F e ( v B) This produces seprtion of chrge with n excess of negtive chrge t one end nd positive chrge t the other. There is therefore n NDUCED emf etween the two ends v B B (out of pge) F -e(v B ) _

2 EMF Notes 9; Electromgnetic nduction nduced emf E work done to move unit positive chrge from to ginst the mgnetic force. F QvB vb Work done (Force)(Distnce) E vb MOTONA emf The electric genertor Assume tht the ends of the wire re connected up through some externl circuit (which we represent here y simple resistor). ε Current flows nd dissiptes ( ) vb electricl power P ε in. This current flowing in wire immersed in mgnetic field experiences force cting upon it of mgnitude F B s we hve seen nd directed in direction perpendiculr to oth nd B nd directed opposite to v ie the force opposes the motion. B (out of pge) Alterntively we my view this force s tht cting on individul electrons in motion. The current is due to electrons moving in the wire with some drift velocity v electron. The electrons therefore experience mgnetic force F e velectron B ( ) v v B (out) F -e(v electron B ) v electron This force OPPOSES the motion of the wire through the field.

3 EMF Notes 9; Electromgnetic nduction The overll force on the wire is; F mg ΣF for ll the electrons. ecll: Force on current-crrying wire in mgnetic field is; F mg B. So, the power delivered to the circuit comes from the effort needed to PUSH the wire through the mgnetic field ginst this force. The mgnetic field thus cts s meditor in the conversion of MECHANCA ENEGY into EECTCA ENEGY. This is the principle of the EECTC GENEATO. Exercise: Show tht the power needed to move the wire through the mgnetic field is equl to the power dissipted in the resistor. s s t W Fds B ds Bv Bv t s s t W P Bv E t Motionl emf nd mgnetic flux. We my tke n lterntive view; Agin using : E vb B (out of pge) n time, the conductor moves distnce v. The re swept out is therefore da v v The mgnetic flux pssing through this re is 3 v

4 EMF Notes 9; Electromgnetic nduction d BdA Bv Therefore d vb E d The induced emf is equl to the rte of sweeping out of mgnetic flux Note:. This pplies to ANY shpe of conductor in ANY mgnetic field.. The sme eqution pplies to sttionry conductor in chnging mgnetic field. Frdy's lw of induction. The induced emf round ny closed pth in mgnetic field is equl to minus the rte of chnge of the mgnetic flux intercepted y the pth: enz's lw. d E Why the negtive sign? This is represented y the negtive sign in the ove eqution for the induced emf. The induced emf is in such direction s to OPPOSE the chnge in mgnetic field tht produces it. f ψ is incresing then E cuses current to flow so s to generte mgnetic field which OPPOSES ψ. Current due to induced emf B induced (opposes B) Current due to induced emf B (incresing) B (decresing) f ψ is decresing then E cuses current to flow so s to generte mgnetic field which SUPPOTS ψ. B induced (supports B) 4

5 EMF Notes 9; Electromgnetic nduction nduced electric field. Consider gin wire moving in B (out of pge) mgnetic field. ook t the sitution from the point of view of someone who moves long with the wire: v there is no mgnetic force v E Therefore, this oserver interprets the force cting on the electrons in the wire s eing due to n NDUCED EECTC FED. E. d round the closed pth shown. The induced electric field is not conservtive. Frdy's w is usully written in this form: d E. d FAADAY'S AW MAXWE'S 3rd EQUATON n words: The line integrl of the electric field round closed pth is equl to minus the rte of chnge of mgnetic flux through the pth 5

6 EMF Notes 9; Electromgnetic nduction Mutul inductnce. Frdy's w chnging B field gives rise to n induced emf Consider two nery circuits: Chnging current in Circuit B Chnging mgnetic field through Circuit nduced emf in Circuit Current flows in Circuit This is the sis of phenomenon known s mutul inductnce. To understnd this in more quntittive fshion; et (t) crete mgnetic field B (t). et e the flux through circuit due to Clerly, B, so DEFNTON: The constnt of proportionlity etween nd is clled the MUTUA NDUCTANCE, M : ie. M M From Frdy s w, the induced emf is given y E d d M Note: M depends on the shpe, size, numers of turns nd reltive positions of the two circuits M M so we need only use M s the symol for mutul inductnce Ohnin uses for mutul inductnce n the S system, inductnce is mesured in Henrys (H) H nductnce tht produces n emf of Volt for rte of chnge of current of A s - H V s A - The usul unit for µ is H m - Circuit Circuit 6

7 Self inductnce. EMF Notes 9; Electromgnetic nduction Even single circuit produces mgnetic field tht psses through the circuit itself. So, if chnges chnges B chnges induced emf Definition: SEF NDUCTANCE Mgnetic flux through circuit (due to) Current in circuit itself nductnce nd enz s lw E d ecll: The negtive sign represents enz s w: the emf cuses current to flow so s to oppose the chnge in flux tht produces it. f ψ is NCEASNG DECEASNG then E will cuse current to flow so s to crete mgnetic field tht OPPOSES ψ. SUPPOTS 7

8 EMF Notes 9; Electromgnetic nduction (t) Circuit Circuit B (t) (t) B (t) (t) increses increses nduced emf in circuit drives current (t) (t) genertes B (t) which OPPOSES B (t) Circuit Circuit (t) B (t) (t) B (t) (t) decreses decreses nduced emf in circuit drives current (t) (t) genertes B (t) which SUPPOTS B (t) Procedure for finding inductnce.. Assume tht current flows in (one of the) circuit(s). Find the mgnetic field (e.g., use Ampere s w) 3. Find the mgnetic flux linked 4. Put M or equl to Exmples:. Mutul inductnce of two long coxil solenoids A long solenoid (enles us to ignore end effects) of rdius is plced inside second solenoid of rdius >. f there is stedy current through the second (outer) coil it will crete mgnetic field; B µ n N n Where (>> ) is the length of the solenoid The mgnetic flux through the centre of the outer coil is 8

9 EMF Notes 9; Electromgnetic nduction BA ( µ n )( π ) This flux links ll the turns N of the inner coil thus due to coil there is flux through coil NBA N ( µ n )( π ) et ( ) e the flux through circuit due to M NN µ π NB f we hd used current in coil to induce voltge in coil the flux ( µ n )( π ) NBA N where the re of coil is used insted of tht of coil. This is ecuse the mgnetic field of very long solenoid is restricted to its inside region the field outside eing pproximtely zero. ie. the cross section re of the smllest solenoid lwys ppers in the eqution for mutul inductnce.. Self inductnce of long solenoid The self inductnce,, is defined through the eqution; Where BA For solenoid the mgnetic field in the solenoid is Bxil µ n So the flux through one turn of rdius is; π µ n 9

10 EMF Notes 9; Electromgnetic nduction The totl flux enclosed is ( π µ n) N ( π µ n) n N π µ n π µ Where N is the totl numer of turns in the solenoid. 3. Trnsformer i A c V A VB i A Coil A primry Coil B d secondry di V M A B nd di V A A V B M VA f the coils re nerly coincident so tht the flux for ech turn is the sme then

11 EMF Notes 9; Electromgnetic nduction N N M π µ nd N µ π A B N N M V V

12 EMF Notes 9; Electromgnetic nduction Energy storge in inductors. ecll: Putting chrge Doing Storing on cpcitor work potentil energy U elec CV Similrly: Mking current Doing Storing flow in n inductor work potentil U mg?? energy Becuse chnging current chnging B induced emf Which opposes ttempt to chnge current (clled ck emf) To find the energy stored in n inductor: et current in the inductor i initilly Apply externl emf current increses t rte This leds to ck emf E To mke current flow, the required externl emf is E ext di di di The externlly pplied power is therefore P E ext i The work done to increse the current from to is therefore di i t U P idi U Energy stored in n inductor chrged with current Where is the stored energy? t is in the MAGNETC FED. ecll: For solenoid of length, rdius, n turns per unit length:

13 EMF Notes 9; Electromgnetic nduction 3 n π ni B ( ) ( ) π π B n B n U Therefore ( ) solenoid of Volume B U The ENEGY DENSTY OF THE MAGNETC FED, u, is therefore; B u This holds generlly, for ny mgnetic field