15. Gauss's Law for Magnetic Field and Ampere's Law
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1 Univesity of Rhode sland PHY 204: Elementay Physics Physics Couse Mateials Gauss's Law fo Magnetic Field and Ampee's Law Gehad Mülle Univesity of Rhode sland, Ceative Commons License This wok is licensed unde a Ceative Commons Attibution-Noncommecial-Shae Alike 4.0 License. Follow this and additional woks at: Abstact Pat fifteen of couse mateials fo Elementay Physics (PHY 204), taught by Gehad Mülle at the Univesity of Rhode sland. Documents will be updated peiodically as moe enties become pesentable. Some of the slides contain figues fom the textbook, Paul A. Tiple and Gene Mosca. Physics fo Scientists and Enginees, 5 th /6 th editions. The copyight to these figues is owned by W.H. Feeman. We acknowledge pemission fom W.H. Feeman to use them on this couse web page. The textbook figues ae not to be used o copied fo any pupose outside this class without diect pemission fom W.H. Feeman. Recommended Citation Mülle, Gehad, "15. Gauss's Law fo Magnetic Field and Ampee's Law" (2015). PHY 204: Elementay Physics. Pape This Couse Mateial is bought to you fo fee and open access by the Physics Couse Mateials at DigitalCommons@UR. t has been accepted fo inclusion in PHY 204: Elementay Physics by an authoized administato of DigitalCommons@UR. Fo moe infomation, please contact digitalcommons@etal.ui.edu.
2 Gauss s Law fo Electic Field The net electic flux Φ E though any closed suface is equal to the net chage Q in inside divided by the pemittivity constant ǫ 0 : E d A = 4πkQ in = Q in ǫ 0 i.e. Φ E = Q in ǫ 0 The closed suface can be eal o fictitious. t is called Gaussian suface. The symbol H denotes an integal ove a closed suface in this context. with ǫ 0 = C 2 N 1 m 2 Gauss s law is a geneal elation between electic chage and electic field. n electostatics: Gauss s law is equivalent to Coulomb s law. Gauss s law is one of fou Maxwell s equations that goven cause and effect in electicity and magnetism. 27/10/2015 [tsl44 1/14]
3 Gauss s Law fo Magnetic Field The net magnetic flux Φ B though any closed suface is equal to zeo: B d A = 0. Thee ae no magnetic chages. Magnetic field lines always close in themselves. No matte how the (closed) Gaussian suface is chosen, the net magnetic flux though it always vanishes. The figues below illustate Gauss s laws fo the electic and magnetic fields in the context of an electic dipole (left) and a magnetic dipole (ight). 27/10/2015 [tsl236 2/14]
4 Ampèe s Law (Resticted Vesion) The ciculation integal of the magnetic field B aound any closed cuve (loop) C is equal to the net electic cuent C flowing though the loop: B d l = µ 0 C, with µ 0 = 4π 10 7 Tm/A The symbol H denotes an integal ove a closed cuve in this context. Note: Only the component of B tangential to the loop contibutes to the integal. The positive cuent diection though the loop is detemined by the ight-hand ule. 27/10/2015 [tsl237 3/14]
5 Ampèe s Law: Application (1) The line integals H B d s along the thee Ampeian loops ae as indicated. Find the diection ( J, N ) and the magnitude of the cuents 1, 2, µ 0 (2Α) 3 µ 0 (4Α) µ 0 (3Α) 27/10/2015 [tsl239 4/14]
6 Ampèe s Law: Application (2) An electic cuent flows though the wie in the diection indicated. Detemine fo each of the five Ampeian loops whethe the line integal H B d s is positive, negative, o zeo. (1) (2) (3) (4) (5) 27/10/2015 [tsl240 5/14]
7 Magnetic Field on the Axis of a Solenoid Numbe of tuns pe unit length: n = N/L Cuent ciculating in ing of width dx : ndx Magnetic field on axis of ing: db x = µ 0(ndx ) 2 Magnetic field on axis of solenoid: B x = µ Z x2 0n dx 2 R2 x 1 [(x x ) 2 + R 2 ] 3/2 = µ 0n 2 R 2 [(x x ) 2 + R 2 ] 3/2! x x 1 p (x x1 ) 2 + R x x 2 p 2 (x x2 ) 2 + R 2 27/10/2015 [tsl215 6/14]
8 Ampèe s Law: Magnetic Field nside a Long Solenoid Apply Ampèe s law, B d l = µ 0 C, to the ectangula Ampeian loop shown. Magnetic field inside: stong, unifom, diected along axis. Magnetic field outside: negligibly weak. Numbe of tuns pe unit length: n. Total cuent though Ampeian loop: C = na ( is the cuent in the wie). Ampèe s law applied to ectangula loop: Ba = µ 0 na. Magnetic field inside: B = µ 0 n. 27/10/2015 [tsl241 7/14]
9 Ampèe s Law: Magnetic Field nside a Tooid Apply Ampèe s law, B d l = µ 0 C, to the cicula Ampeian loop shown. Magnetic field inside: diected tangentially with magnitude depending on R only. Magnetic field outside: negligibly weak. Numbe of tuns: N. Total cuent though Ampeian loop: C = N ( is the cuent in the wie). Ampèe s law applied to cicula loop: B(2πR) = µ 0 N. Magnetic field inside: B = µ 0N 2πR. 27/10/2015 [tsl242 8/14]
10 Ampèe s Law: Magnetic Field nside a Wie Conside a long, staight wie of adius R. The cuent is distibuted unifomly ove the coss section. Apply Ampèe s law, B d l = µ 0 C, to the cicula loop of adius < R. The symmety dictates that the magnetic field B is diected tangentially with magnitude B depending on R only. Line integal: B d l = B(2π). Faction of cuent inside loop: C = π2 πr 2. Magnetic field at adius < R: B = µ 0 C 2π = µ 0 2πR 2. B inceases linealy with fom zeo at the cente. Magnetic field at the peimete: B = µ 0 2πR. 27/10/2015 [tsl243 9/14]
11 Ampèe s Law: Magnetic Field Outside a Wie Conside a long, staight wie of adius R with cuent. Apply Ampèe s law, B d l = µ 0 C, to the cicula loop of adius > R. The symmety dictates that the magnetic field B is diected tangentially with magnitude B depending on R only. Cuent inside loop: C =. Ampèe s law applied: B(2π) = µ 0. Magnetic field at adius > R: B = µ 0 2π. 27/10/2015 [tsl244 10/14]
12 Ampèe s Law: Coaxial Cable Conside a long coaxial cable, consisting of two cylindical conductos sepaated by an insulato as shown in a coss-sectional view. Thee is a cuent flowing out of the plane in the inne conducto and a cuent of equal magnitude flowing into the plane in the oute conducto. Calculate the magnetic field B as a function of the adial coodinate. b c (out) (in) a 27/10/2015 [tsl341 11/14]
13 Unit Exam : Poblem #3 (Sping 12) The coaxial cable shown in coss section has sufaces at adii 1mm, 3mm, and 5mm. Equal cuents flow though both conductos: int = ext = 0.03A (out). Find diection (, ) and magnitude (B 1, B 3, B 5, B 7 ) of the magnetic field at the fou adii indicated ( ). ext int 1mm 3mm 5mm 7mm 27/10/2015 [tsl437 12/14]
14 Unit Exam : Poblem #3 (Sping 12) The coaxial cable shown in coss section has sufaces at adii 1mm, 3mm, and 5mm. Equal cuents flow though both conductos: int = ext = 0.03A (out). Find diection (, ) and magnitude (B 1, B 3, B 5, B 7 ) of the magnetic field at the fou adii indicated ( ). ext int 1mm 3mm 5mm 7mm Solution: 2π(1mm)B 1 = µ 0 (0.03A) B 1 = 6µT 27/10/2015 [tsl437 12/14]
15 Unit Exam : Poblem #3 (Sping 12) The coaxial cable shown in coss section has sufaces at adii 1mm, 3mm, and 5mm. Equal cuents flow though both conductos: int = ext = 0.03A (out). Find diection (, ) and magnitude (B 1, B 3, B 5, B 7 ) of the magnetic field at the fou adii indicated ( ). ext int 1mm 3mm 5mm 7mm Solution: 2π(1mm)B 1 = µ 0 (0.03A) 2π(3mm)B 1 = µ 0 (0.03A) B 1 = 6µT B 1 = 2µT 27/10/2015 [tsl437 12/14]
16 Unit Exam : Poblem #3 (Sping 12) The coaxial cable shown in coss section has sufaces at adii 1mm, 3mm, and 5mm. Equal cuents flow though both conductos: int = ext = 0.03A (out). Find diection (, ) and magnitude (B 1, B 3, B 5, B 7 ) of the magnetic field at the fou adii indicated ( ). ext int 1mm 3mm 5mm 7mm Solution: 2π(1mm)B 1 = µ 0 (0.03A) 2π(3mm)B 1 = µ 0 (0.03A) 2π(5mm)B 1 = µ 0 (0.06A) B 1 = 6µT B 1 = 2µT B 1 = 2.4µT 27/10/2015 [tsl437 12/14]
17 Unit Exam : Poblem #3 (Sping 12) The coaxial cable shown in coss section has sufaces at adii 1mm, 3mm, and 5mm. Equal cuents flow though both conductos: int = ext = 0.03A (out). Find diection (, ) and magnitude (B 1, B 3, B 5, B 7 ) of the magnetic field at the fou adii indicated ( ). ext int 1mm 3mm 5mm 7mm Solution: 2π(1mm)B 1 = µ 0 (0.03A) 2π(3mm)B 1 = µ 0 (0.03A) 2π(5mm)B 1 = µ 0 (0.06A) 2π(7mm)B 1 = µ 0 (0.06A) B 1 = 6µT B 1 = 2µT B 1 = 2.4µT B 1 = 1.71µT 27/10/2015 [tsl437 12/14]
18 Unit Exam : Poblem #2 (Sping 08) (a) Conside a solid wie of adius R = 3mm. Find magnitude and diection (in/out) that poduces a magnetic field B = 7µT at adius = 8mm. (b) Conside a hollow cable with inne adius R int = 3mm and oute adius R ext = 5mm. A cuent out = 0.9A is diected out of the plane. Find diection (up/down) and magnitude B 2, B 6 of the magnetic field at adius 2 = 2mm and 6 = 6mm, espectively. (a) (b) out 8mm B 0mm 2mm 6mm 27/10/2015 [tsl382 13/14]
19 Unit Exam : Poblem #2 (Sping 08) (a) Conside a solid wie of adius R = 3mm. Find magnitude and diection (in/out) that poduces a magnetic field B = 7µT at adius = 8mm. (b) Conside a hollow cable with inne adius R int = 3mm and oute adius R ext = 5mm. A cuent out = 0.9A is diected out of the plane. Find diection (up/down) and magnitude B 2, B 6 of the magnetic field at adius 2 = 2mm and 6 = 6mm, espectively. Solution: (a) (b) (a) 7µT = µ 0 2π(8mm) = 0.28A (out). 8mm B out 0mm 2mm 6mm 27/10/2015 [tsl382 13/14]
20 Unit Exam : Poblem #2 (Sping 08) (a) Conside a solid wie of adius R = 3mm. Find magnitude and diection (in/out) that poduces a magnetic field B = 7µT at adius = 8mm. (b) Conside a hollow cable with inne adius R int = 3mm and oute adius R ext = 5mm. A cuent out = 0.9A is diected out of the plane. Find diection (up/down) and magnitude B 2, B 6 of the magnetic field at adius 2 = 2mm and 6 = 6mm, espectively. Solution: (a) (b) (a) 7µT = (b) B 2 = 0, µ 0 2π(8mm) B 6 = µ 0(0.9A) 2π(6mm) = 0.28A (out). = 30µT (up). 8mm B out 0mm 2mm 6mm 27/10/2015 [tsl382 13/14]
21 Unit Exam : Poblem #2 (Sping 11) The coaxial cable shown has sufaces at adii 1mm, 3mm, and 5mm. The magnetic field is the same at adii 2mm and 6mm, namely B = 7µT in the diection shown. (a) Find magnitude (in S units) and diection (in/out) of the cuent int flowing though the inne conducto. (b) Find magnitude (in S units) and diection (in/out) of the cuent ext flowing though the oute conducto. ext B B int 2mm 6mm 27/10/2015 [tsl416 14/14]
22 Unit Exam : Poblem #2 (Sping 11) The coaxial cable shown has sufaces at adii 1mm, 3mm, and 5mm. The magnetic field is the same at adii 2mm and 6mm, namely B = 7µT in the diection shown. (a) Find magnitude (in S units) and diection (in/out) of the cuent int flowing though the inne conducto. (b) Find magnitude (in S units) and diection (in/out) of the cuent ext flowing though the oute conducto. ext B B int 2mm 6mm Solution: (a) (7µT)(2π)(0.002m) = µ 0 int int = 0.07A (out) 27/10/2015 [tsl416 14/14]
23 Unit Exam : Poblem #2 (Sping 11) The coaxial cable shown has sufaces at adii 1mm, 3mm, and 5mm. The magnetic field is the same at adii 2mm and 6mm, namely B = 7µT in the diection shown. (a) Find magnitude (in S units) and diection (in/out) of the cuent int flowing though the inne conducto. (b) Find magnitude (in S units) and diection (in/out) of the cuent ext flowing though the oute conducto. ext B B int 2mm 6mm Solution: (a) (7µT)(2π)(0.002m) = µ 0 int int = 0.07A (out) (b) (7µT)(2π)(0.006m) = µ 0 ( int + ext ) int + ext = 0.21A (out) ext = 0.14A (out) 27/10/2015 [tsl416 14/14]
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