Week 11 - Inductance

Transcription

1 Week - Inductnce November 6, 202 Exercise.: Discussion Questions ) A trnsformer consists bsiclly of two coils in close proximity but not in electricl contct. A current in one coil mgneticlly induces n emf in the second coil, with properties tht cn be controlled by djusting the geometry of the two coils. Such device will work only with lternting current, however, nd not with direct current. Explin. A trnsformer only works with lternting current becuse it s dependent upon the phenomenon of mutul inductnce. Here the emf in the second coil due to the first is given by ε 2 = M di dt, () nd thus, for there to be n emf, it hs to be chnging current in the first coil. How much emf for given current is determined by the mutul inductnce M, which is only reflection of the geometry of the coils. b) Suppose there is stedy current in n inductor. If you ttemt to reduce the current to zero instntneously by quickly opening switch, n rc cn pper t the switch contcts. Why? Is it physiclly possible to stop the current instntneously? Explin. The rc is formed becuse of the self inductnce in the circuit. When you open the switch the rte of chnge of the current become extremely high nd therefore the emf generted become enormous. This emf forces the current cross the cross the gp between the switch. Mthemticlly this is seen with the eqution ε = L di dt (2) where di/dt becomes very lrge nd negtive s you open the switch, nd consequently ε becomes huges nd forces the current cross the gp. Since ll circuits hve some self conductnce it is not possible to stop ny current instntneously. An electric rc is luminous dischrge of current tht is formed when current jumps gp in circuit between two electrodes.

2 Exercise.2: Inductnce of Solenoid A long, stright solenoid hs N turns, uniform cross-sectionl re A, nd length l. Show tht the inductnce of this solenoid is given by the eqution L = µ 0 AN 2 /l. Assume tht the mgnetic field is uniform inside the solenoid nd zero outside. (Your nswer is pproximte becuse B is ctully smller t the ends thn t the center. For this reson your nswer is ctully n upper limit on the inductnce.) For the cse of n infinitely long solenoid the mgnetic field is is B = µ 0 ni where n is the number of turns per unit length. Now we use this field to pproximte the inductnce of finite solenoid. A finte solenoid hs finite length l nd finite number of turns N so we cn write B = µ 0 ni = µ 0NI. (3) l For self inductnce the flux trough the re enclosed by the turns of wire in the solenoid is φ B = B(NA) = µ 0AN 2 I (4) l nd here we see the clim tht the flux φ B is lwys proportionl to the current I. The proportionlity constnt is the self inductnce L L = µ 0AN 2. (5) l Figure Exercise.3: A RL-Circuit In the circuit shown in figure, E = 50.0 V, R = 40.0 Ω,R 2 = 30.0 Ω nd L = 0.70 H. Switch S is closed t t = 0. Just fter the switch is closed.. 2

3 ) Wht is the potentil difference v b cross the resistor R? Just fter the switch is closed the current trough the inductor is zero becuse it hs nt hd ny time to increse (Recll how the current grows in simple RL-circuit). Therefore by pplying the loop rule to the upper loop E v b = 0 v b = E. (6) v b = E. (7) b) Which point, or b, is t higher potentil. As the current trvels trough R the voltge drops by IR, therefore b is t lower potentil thn. b is t the lower potentil. c) Wht is the potentil difference v cd cross the inductor L? There is no current trough R 2 so by the loop rule E i 2 R 2 v cd = E v cd = 0 v cd = E. (8) v cd = E. (9) d) Which point, c or d, is t higher potentil? The voltge increses when we go from b to trough the emf, so it must therefore drop s we go trough the inductor from c to d. Thus c is t the higher potentil. 3

4 c is t the higher potentil. The switch is left closed long time nd then opened. Just fter the switch is opened... Repet () to (d). () Since the switch hs been open for long time, i 2 is not chnging nymore (di 2 /dt = 0), so v cd = 0. Therefore by the loop rule E = v R2 = i 2 R 2 i 2 = E R 2. (0) (b) Now when the switch suddenly opens the current trough the inductor hs not hd time to chnge so the current trough the lower prt of the circuit (with S open) is i 2. This current trvels from b to, so b is therefore t higher potentil. i 2 = E R 2 () is t the higher potentil. Exercise.4: An Electromgnetic Cr Alrm Your ltes invention is cr lrm tht produces sound prticulrly nnoying frequency of 3500 Hz. To do this, the cr-lrm circuitry must produce n lternting electric current of the sme frequency. Tht s why your design includes n inductor nd cpcitor in series. The mximum voltge cross the cpcitor is to be 2.0 V (the sme voltge s tht of the cr bttery). To produce sufficiently loud sound, the cpcitor must store J of energy. Wht vlues of cpcitnce nd inductnce should you chose for your cr-lrm circuit? Here one is fter the reltion between frequency of LC-circuit nd it s cpcitnce nd inductne ω = /LC. If your not very good t remembering fromuls I would recommend remembering the method for the result. The volge round the loop hs to be zero, for cpcitor nd n inductor of respective voltges q/c nd Ldi/dt this mens tht L di dt + q C = 0. (2) Now the chrge on the cpcitor is relted to the current by I = dq/dt, therefore in terms of q we cn write d 2 q dt 2 + q = 0. (3) LC 4

5 A good guess for this differentil eqution is q(t) = Q cos (ωt + φ). Substitution yields ( Q ω 2 cos (ωt + φ) + ) cos (ωt + φ) = 0, (4) LC our guess is only solution ω = LC which is our desired result. Now for our cr-lrm frequency; ω is the ngulr frequency nd this is relted to the ctul frequency ω = 2πf. Therefore we get f = ω 2π = 2π LC. (5) We hve to find the pproprite vlues of C nd L. We know the mximum voltge cross the cpcitor s well s it s energy. Therefore we cn use the reltion for the energy of cpcitor U E = Q 2 /2C = V 2 C/2 to find it s cpcitnce. We get. C = 2U E V 2 = (2.0) 2 F = 222 µf (6) Now we cn determine the necessry inductnce by our eqution for the frequency. Rerrnging we get L = ( ) 2 ( 2πf C = 2π 3500 ) 2 H = 9.3 µh. (7) ( L = 2π 3500 ) 2 H = 9.3 µh. (8) Exercise.5: Solr Mgnetic Energy Mgnetic fields within sunspot cn be s strong s 0.4 T. (By comprison, the erth s mgnetic field is bout /0000 s strong.) Sunspots cn be s lrge s km in rdius. The mteril in sunspot hs density of bout kg/m 3. Assume µ for the sunspot mteril to be µ 0. If 00% of the mgnetic-field energy stored in sunspot could be used to eject the sunspot s mteril wy from the sun s surfce, t wht speed would tht mteril be ejected? Compre to the sun s escpe speed, which is bout m/s. Hint: Clculte the kinetic energ the mgnetic field could supply to m 3 if sunspot mteril. Under the ssumption tht the permebility is µ 0 the mgnetic field energy density is given by u B = B2 (9) 5

6 nd therefore energy in m 3 is U = B2 m 3. (20) Now if this energy is 00% converted to kinetic energy of the sunspot mteril, we cn sett up conservtion of energy eqution for m 3 of this mteril. When ll the energy is converted to kinetic energy we hve or equivlently B 2 m 3 = 2 mv2 (2) v = B µ0 m. (22) m 3 of sunspot mteril hve mss m = kg/m 3 m 3. We then get velocity v = B µ0 m = 0.4 m/s = m/s. (23) 4π This is only 3.4% of the escpe velocity of the sun, so the mgnetic field energy lone is not enough cuse coronl mss ejections which is mssive burst of prticles being ejected into spce. It remins mystery exctly how the prticles get tht much energy. 2 v == m/s. (24) This is only 3.4% of the escpe velocity of the sun, so the mgnetic field energy lone is not enough cuse coronl mss ejections which is mssive burst of prticles being ejected into spce. It remins mystery exctly how the prticles get tht much energy. 3 Exercise.6: Inductnce of Coxil Cble A smll solid conductor of with rdius is supported by insulting, nonmgnetic disks on the xis of thin-wlled tube with inner rdius b. The inner nd outer conductors crry equl currents i in opposite directions. ) Use Ampere s lw to find the mgnetic field t n point in the volume between the two conductors. 2 Solr flre Wikipedi. 3 Solr flre Wikipedi. 6

7 Figure 2 We construct circulr integrtion pth with rdius r centered t the xis of the cble. Then by Ampere s lw B dl = B2πr = µ 0 i (25) or equivlently B = µ 0i 2πr. (26) B = µ 0i 2πr. (27) b) Write the expression for the flux dφ B through nrrow strip of length l prllel to the xis, of width dr, t sitnce r from the xis of the cble nd lying in plne contining the xis. The nrrow strip is shown in figure 3. The flux trough this strip will be the mgnetic field (which is perpendiculr to the strip t every point) times the re of the strip. I.e. it will be dφ B = B (ldr) = l µ 0i dr. (28) 2πr dφ B = B (ldr) = l µ 0i dr. (29) 2πr 7

8 Figure 3 c) Integrte your expression from prt (b) over the volue between the two conductors to find the totl flux produced by the current i in the centrl conductor. To clculte the entire flux we hve to integrte this expression from to b. φ B = b dφ = l µ 0i 2π b dr r = l µ ( ) 0i b 2π ln. (30) φ B = b dφ = l µ ( ) 0i b 2π ln. (3) d) Show tht the inductnce of length l of the cble is L = l µ ( ) 0 b 2π ln. (32) The inductnce is the constnt of proportionlity between φ B nd i. Looking t our expression for the flux we see tht φ B = l µ ( ) 0 b 2π ln i = Li (33) 8

9 L = l µ ( ) 0 b 2π ln. (34) e) Use eqution 32 to clculte the energy stored in the mgnetic field for lentgh l of the cble. We know tht the energy stored in the mgnetic field of n inductor is 2 Li2. Therefore we get U = l µ 0i 2 ( ) b 4π ln. (35) U = l µ 0i 2 ( ) b 4π ln. (36) f) Verify tht this is the correct expression energy by lso clculting th e totl stored energy through the energy density formul u B = B2. Hint: Clculte the energy of thin, cylindricl, shell of length l nd rdius r plced in between the two conductors. We found the mgnetic field in between the two conductors in (). The result ws The corresponding energy density is therefore given by B = µ 0i 2πr. (37) u B = B2 = ( µ0i ) 2 2πr = µ 0i 2 8π 2 r 2. (38) Now to find the energy of thin, cylindricl shell of length l nd rdius r we must multiply the energy density with the volume of the cylidricl shell. The volume of such shell is given by dv = l2πrdr nd therefore du = u B dv = µ 0i 2 8π 2 r 2 2πrdr = l µ 0i 2 dr. (39) 4πr such tht the totl energy contined in the mgnetic field within the conductors is l µ 0i 2 4π b dr r = l µ 0i 2 4π ln b (40) which indeed grees with the expression from (e). 9