Lecture 20 Ampère s Law

Transcription

1 Lecture 20 Ampère s Law Sections: 7.2, partially 7.7 Homework: See homework file

2 Ampère s Law in ntegral Form 1 the field of a straight wire with current (Lecture 19) B H = = a a φ φ µ, T 2πρ, A/m 2πρ for this cylindrically symmetric problem with circular field lines the result suggests the relation H d L = encl magnetic B field lines Ampère s law can be derived from the Biot-Savart law (section 7.7 in textbook, Hayt/Buck 8 th ed.) LETURE 20 slide 2 a z a φ

3 Ampère s Law in ntegral Form 2 [ S ] H dl= = J ds encl contour can be any does not need to be a field line is total current flowing through area S bounded by positive relates to the direction of by the right-hand rule S magnetic field lines LETURE 20 slide 3

4 Ampère s Law in ntegral Form 3 examples on the sign of current contributions H d L = + H d L = H d L = 1 2 LETURE 20 slide 4

5 Applications of Ampère s Law problems where the magnetic field is circularly symmetric contours exist, along which the magnetic field is constant or zero Ampère s contours concept is similar to Gauss law in electrostatics LETURE 20 slide 5

6 Applications of Ampère s Law: Straight Thin Wire Example 1: Find the magnetic field of an infinitesimally thin straight infinitely long wire with current. a z 2π d = Hφ φ φ( ρdφ) = H L encl a a 0 dl Hφ2 πρ = Hφ =, A/m 2πρ encl [ S ] ρ a φ H same result obtained via Biot-Savart law in Lecture 15 LETURE 20 slide 6

7 Applications of Ampère s Law: Straight Thick Wire 1 Example 2: A straight conductor of circular cross-section and radius a carries current. Find the magnetic field inside and outside the conductor. The conductor has no magnetic properties. (a) outside case the same as in Example 1 ρp a total enclosed current is Hφ ( P) =, A/m πρ 2 P (b) inside ρp a J ρ P P ρ P P assume uniform current distribution 2 J = a, A/m 2 z π a a LETURE 20 slide 7

8 Applications of Ampère s Law: Straight Thick Wire 2 encl 2 P ρp 2 φ 2 2 ρ Hφ 2 πρp = J πρp = H ( P) =, A/m a 2π a 2π a 0 H φ ρ P a 1/ ρp ρ P LETURE 20 slide 8

9 Applications of Ampère s Law: Straight Thick Wire 3 LETURE 20 slide 9

10 Applications of Ampère s Law: oaxial able 1 Example 3: Find the magnetic field everywhere in the cross-section of a coaxial cable. (a) inside wire, ρ P a same as in Example 2-b H φ ρp =, A/m 2 2π a c b a H (b) between wire and shield, a ρ P b same as in Example 2-a H φ = 2πρP, A/m (c) inside shield, b ρ P c 2 πρ ( 2 2) ( 2 2 PHφ = J π ρp b = π ρ ) ( 2 2 P b π c b ) LETURE 20 slide 10

11 Applications of Ampère s Law: oaxial able 2 (c) inside shield, cont. ( c2 ρ 2 P ) 2 2 Hφ = 2 πρ ( c b ) P (d) outside cable, ρ P > c H φ =? LETURE 20 slide 11

12 Applications of Ampère s Law: Toroid Example 4: Find the magnetic field inside a toroid of N turns carrying current. H d L = H 2πρ = N φ encl N H φ =, ρ1 ρ ρ2 2πρ ρ 2 H ρ 1 ρ What is the field outside the toroid? a φ LETURE 20 slide 12

13 Applications of Ampère s Law: Surface urrent Sheet 1 Example 5: Find the magnetic field due to an infinite sheet of uniform current flowing in the y-direction. H field does not vary with x and y since the source does not vary with x and y LETURE 20 slide 13

14 Applications of Ampère s Law: Surface urrent Sheet 2 H y = 0 since current is along y (field is always to current) H z = 0 due to cancellation of contributions from two symmetrical current elements along x z y H k P H j # j # k H = H + H = H a j k x x x resultant field has only x-component and does not vary with x and y LETURE 20 slide 14

15 Applications of Ampère s Law: Surface urrent Sheet 3 apply Ampère s law along xz plane contour 1-1'-2'-2-1 H dl= encl field does not vary with x H 1 x1dx + ( H ) ( 1 z dz + H 2 x2) dx + H 2 zdz = K y L 0 zero contribution from segments 1'-2' and 2-1 (H z = 0) 0 H L H L= K L H H = K x1 x2 y x1 x2 y H 1 as follows from Biot-Savart s law, field is anti-symmetrical wrt the current sheet y z dla y x H = H H = K /2, H = K /2 x2 x1 x1 y x2 y H 2 LETURE 20 slide 15

16 Applications of Ampère s Law: Surface urrent Sheet 4 Ampère s law can be applied to another xz plane contour stretched in the z-direction, see 3-3'-2' H dx + ( H ) dx = K L H H = K 3 2 x 3 x 2 y x 3 x 2 y H = H = K x3 x1 y /2 field does not depend on distance from the infinite current sheet result analogous to D-field of an infinite charged sheet n H = K a, A/m 2 ρs D= an, /m 2 2 K H 2 P 1 a n2 a n1 P 2 H 1 LETURE 20 slide 16

17 Surface current sheet of current density K = 4a z A/m lies in the plane defined by y = 4 m. Find the magnetic field at the origin (0,0,0). Z-Axis Y-Axis 4 H X-Axis LETURE 20 slide 17

18 Applications of Ampère s Law: Solenoid 1 field is constant in z and ϕ because the source is constant in z and ϕ z H φ = 0 because current is along φ H ρ = 0 because the ρ contributions of two symmetrical turns along z cancel a φ a ρ H j # j K # k az solenoid axis P H = H + H = H a H k j k z z field has only a z-component and it is constant in z and ϕ LETURE 20 slide 18 K a = N d

19 H dl= z1 z2 d 1 1 N H z1 H z2 = d a b N repeat along 11'3'31: H z1 H z3 = Hz3 = Hz2 d outside the solenoid field is constant and zero (see contour a-b-c-d) antisymmetry H H = 0 2H = 0 H= H a = 0, ρ > a z, ab z, cd z, ab z H = H encl N H L H L= L z3 z2 inside solenoid Applications of Ampère s Law: Solenoid 2 = 0 N Hz1 = = Kφ = Ka, A/m d d L LETURE 20 slide 19 z c a z

20 You have learned: that the work integral of H along a closed path is not zero but is equal to the enclosed current (Ampère s law) how to apply Ampère s law to symmetric problems that the field of a current sheet is constant in space, it is tangential to the sheet but is orthogonal to the surface current (right-hand rule applies) what the field is in a toroid and in a solenoid (equal to the surface current density of the toroid/solenoid) what the field is inside wires with current (growing linearly inside the wire, decaying as 1/ρ outside) LETURE 20 slide 20