Physics 4B. Question and 3 tie (clockwise), then 2 and 5 tie (zero), then 4 and 6 tie (counterclockwise) B i. ( T / s) = 1.74 V.

Transcription

1 Physcs 4 Solutons to Chapter 3 HW Chapter 3: Questons:, 4, 1 Problems:, 15, 19, 7, 33, 41, 45, 54, 65 Queston 3-1 and 3 te (clockwse), then and 5 te (zero), then 4 and 6 te (counterclockwse) Queston 3-4 (a) nto;(b) counterclockwse;(c) larger Queston 3-1 c, b, a Problem 3- Usng Faraday s law, the nduced emf s ( ) da d( π r ) dφ d A dr ε = = = = =πr =π (.1m)(.8T)(.75m/s) =.45V. Problem 3-15 (a) Let L be the length of a sde of the square crcut. Then the magnetc flux through the crcut s Φ = L /, and the nduced emf s ε dφ = = Now =.4.87t and d/ =.87 T/s. Thus, L d. ε = (. m) (. 87 T / s) = 1.74 V. The magnetc feld s out of the page and decreasng so the nduced emf s counterclockwse around the crcut, n the same drecton as the emf of the battery. The total emf s ε + ε =. V V = 1.7 V. (b) The current s n the sense of the total emf (counterclockwse).

2 Problem 3-19 Frst we wrte Φ = A cos θ. We note that the angular poston θ of the rotatng col s measured from some reference lne or plane, and we are mplctly makng such a choce by wrtng the magnetc flux as A cos θ (as opposed to, say, A sn θ). Snce the col s rotatng steadly, θ ncreases lnearly wth tme. Thus, θ = ωt f θ s understood to be n radans (here, ω = πf s the angular velocty of the col n radans per second, and f = 1 rev/mn 16.7 rev/s s the frequency). Snce the area of the rectangular col s A = (.5 m) (.3 m) =.15 m, Faraday s law leads to b g b g b g θ ε = N da cos = NA d cos π ft = NAπf sn π ft whch means t has a voltage ampltude of ε π π ( )( )( )( ) 3 max = fna = 16.7 rev s 1 turns.15 m 3.5T = V. Problem 3-7 (a) Consder a (thn) strp of area of heght dy and wh =. m. The strp s located at some < y <. The element of flux through the strp s dφ = da = 4t y dy c hb g where SI unts (and sgnfcant fgures) are understood. To fnd the total flux through the square loop, we ntegrate: d ( 4 t y ) dy t 3. Φ = Φ = = Thus, Faraday s law yelds dφ ε = = t. 4 3 At t =.5 s, the magntude of the nduced emf s V. (b) Its drecton (or sense ) s clockwse, by Lenz s law. Problem 3-33 (a) Lettng x be the dstance from the rght end of the rals to the rod, we fnd an expresson for the magnetc flux through the area enclosed by the rod and rals. y Eq. 9-17, the feld s = μ /πr, where r s the dstance from the long straght wre. We consder an nfntesmal horzontal strp of length x and wh dr, parallel to the wre and a dstance r from t; t has area A = x dr and the flux s μ Φ = = dr. π r d da x

3 y Eq. 3-1, the total flux through the area enclosed by the rod and rals s μx a+ Ldr μx a + L Φ = ln. π = a r π Accordng to Faraday s law the emf nduced n the loop s μ μv dφ dx a+ L a+ L ε = = ln = ln π π 7 ( π )( )( ) 4 1 T m/a 1A 5.m/s 1.cm + 1.cm = = π 1.cm (b) y Ohm s law, the nduced current s 4 ln.4 1 V. ( ) ( ) = ε = Ω = 4 4 / R.4 1 V / A. Snce the flux s ncreasng, the magnetc feld produced by the nduced current must be nto the page n the regon enclosed by the rod and rals. Ths means the current s clockwse. (c) Thermal energy s beng generated at the rate P ( 6. 1 A) (.4 ) 4 = R= Ω = W. (d) Snce the rod moves wth constant velocty, the net force on t s zero. The force of the external agent must have the same magntude as the magnetc force and must be n the opposte drecton. The magntude of the magnetc force on an nfntesmal segment of the rod, wth length dr at a dstance r from the long straght wre, s ( μ π ) df = dr= / r dr. We ntegrate to fnd the magntude of the total magnetc force on the rod: F μ a+ L dr μ a + L = ln π = a r π 7 4 ( π )( )( ) 4 1 T m/a 6. 1 A 1 A 1. cm + 1. cm = ln π 1. cm = N. Snce the feld s out of the page and the current n the rod s upward n the dagram, the force assocated wth the magnetc feld s toward the rght. The external agent must therefore apply a force of N, to the left. (e) y Eq. 7-48, the external agent does work at the rate

4 P = Fv = ( N)(5. m/s) = W. Ths s the same as the rate at whch thermal energy s generated n the rod. All the energy suppled by the agent s converted to thermal energy. Problem 3-41 (a) We nterpret the queston as askng for N multpled by the flux through one turn: c h b gc hbgb g 3 3 Φturns = NΦ = NA= N πr = T π 1. m = Wb. (b) Equaton 3-33 leads to 3 NΦ L = = 38. A Wb = H. Problem 3-45 (a) Speakng anthropomorphcally, the col wants to fght the changes so f t wants to push current rghtward (when the current s already gong rghtward) then must be n the process of decreasng. (b) From Eq (n absolute value) we get ε 17 V L = = = H. d /.5kA / s Problem (a) The nductor prevents a fast buld-up of the current through t, so mmedately after the swtch s closed, the current n the nductor s zero. It follows that (b) = 1 = 3.33A. ε 1V 1 = = = 3.33A. R + R 1. Ω+.Ω 1 (c) After a sutably long tme, the current reaches steady state. Then, the emf across the nductor s zero, and we may magne t replaced by a wre. The current n R 3 s 1. Krchhoff s loop rule gves ε R R = 1 1 ( ) ε R R = We solve these smultaneously for 1 and, and fnd

5 (d) and 1 ( R R3) ( 1 V)(. 3.Ω) + + ( 1.Ω)(.Ω ) + ( 1.Ω)( 3.Ω ) + (.Ω)( 3.Ω) ε + Ω+ = = = 4.55A, 3 = = ( 1 V)( 3.Ω) ( 1. )(. ) ( 1. )( 3. ) (. )( 3. ) ε R + + Ω Ω + Ω Ω + Ω Ω =.73A. (e) The left-hand branch s now broken. We take the current (mmedately) as zero n that branch when the swtch s opened (that s, 1 = ). (f) The current n R 3 changes less rapdly because there s an nductor n ts branch. In fact, mmedately after the swtch s opened t has the same value that t had before the swtch was opened. That value s 4.55 A.73 A = 1.8 A. The current n R s the same but n the opposte drecton as that n R 3, that s, = 1.8 A. A long tme later after the swtch s reopened, there are no longer any sources of emf n the crcut, so all currents eventually drop to zero. Thus, (g) 1 =, and (h) =. Problem 3-65 (a) The energy delvered by the battery s the ntegral of Eq (where we use Eq for the current): ε ε L P = e = t e R R + R t t battery Rt L Rt L ( 1 ) ( 1) ( 1.V ) ( ) ( 6.7 Ω)(. s) 5.5 H ( e ) 5.5 H 1 =. s Ω 6.7 Ω = 18.7 J. (b) The energy stored n the magnetc feld s gven by Eq. 3-49: 1 1 ε Rt L 1 1. V ( 6.7 Ω)(. s) 5.5 H U = L () t = L ( 1 e ) = ( 5.5H) 1e R 6.7 Ω = 5.1 J. (c) The dfference of the prevous two results gves the amount lost n the resstor: 18.7 J 5.1 J = 13.6 J.