Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 36 Inductance Capacitance C ++++++- q CV dU dqV qdq C U q 0 0 U dU qdq / C di dB d iB dt dt dt di Inductance L L dt unit of L is henry (H) di dU dq idt L idt dt u U dU 1 2 1 1 2 q / C C (V ) qV 2 2 2 Electric energy 0 1 2 Lidi Li 0 2 i Magnetic energy Calculating the capacitance Procedure: 1. Suppose that the capacitor is charged, with ±q on the two plates respectively. 2. Find the electric field E in the region between the plates. 3. Evaluate the potential difference between the positive and negative plates, by using the formula: V V V E ds 4.The expected capacitance is then: C q V Calculating the Inductance Procedure: 1. Suppose i 2. Find the magnetic field B, FB 3. Evaluate the EMF by using the formula: d B di L dt dt d Ldi Li L i Calculating the Inductance B 0 ni ΦB μ 0 niA NΦB μ0 NniA N nl NΦB 0 NnAi 2 L 0 n lA i i L is independent of i and depends only on the geometry of the device. Calculating the Inductance ΦB BdA L i i b a 0i ldr l b 2r 0 ln i 2 a Calculating the capacitance q C V 20l b b a 20 rl dr ln a q q a a b b Inductance of a Toroid L? 0iN B 0 ni 2r b iN 0iNh b dr b 0 ΦB B dA a Bhdr a 2r hdr 2 a r NΦB 0 N h b ln L 2 a i 2 0iNh b ln 2 a Inductors with Magnetic Materials 0 N Nh h b b LL m ln ln 22 a a B m B0 L m L0 2 0 m0 2 L i C 0 A C d C' e 0 A d A d Ferromagnetic cores (κm >>1, κm =103 - 104) provide the means to obtain large inductances. RC Circuits Combine Resistor and Capacitor in Series a Switch at position (a) (b) b R C LR Circuits q(t ) Ce(1t/ RCet / RC ) q(tt)/ RC t / RC Vc (t ) e (1 e ) C dq t /RCt / RC i (t ) ee dt R R RC Circuits Combine Resistor and Capacitor in Series at→∞, i=ε/R. Switch at position (a) di di iR L iR L 0 R b dt dt C di dt di R dt iR L L t=0, i=0 i R LR Circuits i t di R ΔVR=-iR dt t =L/R inductive 0 i time constant 0 L tR R t L t ) i ( 1 e ) i R R t R ln i L R di i L dt RC Circuits Combine Resistor and Capacitor in Series at→∞, i=ε/R. Switch at position (b) R t d i L R b iR L 0 i e dt R C t=0, i=0 LR Circuits t =L/R i R e t 0, i R ΔVR=iR i di i L dt t , i 0 t t 22 1 ddqqq dq22 q 2 0 L q 2 2 2 LC C dt dt CL dt q q0 cos(t ) L-C circuit i ? L C dq i q0 sin( t ) Energy conservation dt q i q di L 0 C dt dq i dt t 1 q02 1 2 2 2 2 UE cos (t ) U B Lq0 sin (t ) 2 2C Electric energy Magnetic energy Damped and Forced oscillations d 2 x b dx 2 x0 2 dt m dt x(t ) xm e bt / 2 m cos(t ) If there are resistances in circuit, the U is no longer constant. m cos t Resonance 1 2 b R , m L, L LC C Rt / 2 L R q(t ) q0e cos(t ) 2 ( R/ 2 L) 2 di q dq L iR 0 i dt C dt d 2 q R dq 1 q 0m cos t 2 dt L dt LC Energy Storage in a Magnetic Field K a di iR L 0 dt di 2 i i R Li dt ΔVR=-iR i L i= (dq/dt)= (dq)/dt, the power by the emf device. i2R, the power consuming in the resistor. Li(di/dt), the rate at which energy is stored in the space of the inductor, it can be put out, when switch to b dU di Li dt dt dU Lidi 0 dU U 0 i Lidi 1 2 U Li 2 di dt 2 q 1q UE 2 C q energy is stored in the electric field 1 2 U B Li 2 L 0 n 2lA inductance UB i energy is stored in the magnetic field i magnetic field 1 uE 0 E 2 2 UB UB 2 B U B 0 n 2 uB i 2 0 V 2 2 0 n 2lA 2 i2 B 0 ni Analogy to Simple Harmonic Motion x xm sin( t ) 2 d x 2 x 0 2 dt k m 2 dx v xm cos(t ) dt Us 1 2 kx 2 K 1 2 mv 2 q q0 sin( t ) 2 d q 1 2 2 q 0 2 LC dt q x Lm i v 1 C k dq i q0 cos(t ) dt 2 1q UE 2C UB 1 2 Li 2 Electric Field E q 40 r F qE 2 rˆ Magnetic Field 0 qv rˆ B 2 4r 0ids rˆ dB 4r 2 F qv B dF ids B Electric Field P qd 1 p E 40 x 3 t P E Magnetic Field iA 0 B 3 2z t B Electric Field E E0 E ' Magnetic Field B B0 0 M 1 E E0 ( e 1) B m B0 ( m 1, m 1) e 0 m 0 e Es dl 0 q E ds 0 B dl 0i Ampere Law B ds 0 Gauss Law Electric Field Magnetic Field Induction d B Ei dl i dt t dS d (v B) ds (v B) ds Electric Field Magnetic Field N B di L L dt i q q C V C V q(t ) C (1 e t / RC 2 1q UE 2C 1 2 uE 0 E 2 ) i R (1 e R t L 1 2 U B Li 2 1 B2 uB 2 0 ) Electric Field Magnetic Field 1 LC 2 2 2 0 q q0 sin( t ) dq i q0 cos(t ) dt q 1q 2 UE sin (t ) 2C 2C 1 2 1 2 2 2 U B Li Lq cos (t ) 2 2 0 qx iv Lm 1 C k Example UE ? 2 q 1 q 1 ) 2 2 2 2 uE 0 E 2 0 ( 8 0l r 2 20lr 2 q 2 dr q2 dU B uB dV 2 2 2 (2rl )(dr ) 40l r 8 0l r 2 b q q2 b dr ln UB a 4 l r 40l a 0 20l q q b C q b V a 20 rl dr ln a a b q2 UE 2C q2 b ln 40l a Example UB ? 2 1 B2 i 1 0i 2 uB ( ) 02 2 b 2 0 8 r 2 0 2r a 0i 2 0i 2l dr dU B uB dV 2 2 (2rl )(dr ) 8 r 4 r 2 2 b i l dr i l b 0 0 UB ln a 4 r 4 a ΦB BdA L 2 2 Li i l b 0 i i UB ln b i 2 4 a 0 a 2r i ldr 0l b ln 2 a Exercises P839-841 9, 10, 23, 41 Problems P842 3, 5