Chapter 8. The Steady Magnetic Field 8.1 Biot-Savart Law

Transcription

1 hapter 8. The teady Magnetic Field 8. Bit-avart Law The surce f steady magnetic field a permanent magnet, a time varying electric field, a direct current. Hayt; /9/009; 8- The magnetic field intensity frm a differential current element IdL aˆ IdL dh = = 3 ince the current always flws in a clsed circuit IdL aˆ H = urface current density, J It is defined as a current per unit length alng the directin perpendicular t the current. Therefre in Bit-avart law IdL= J d s and H J ˆ a = d Bit-avart law fr current density J J aˆ H = dv v

2 Eample The magnetic field frm an infinite line current Hayt; /9/009; 8- The magnetic field shuld be independent f and due t symmetry. The field pint : r = ρ ˆ a ρ The surce pint : r ' = aˆ ' Therefre ˆ ˆ = r r ' = ρar ' a, aˆ ρaˆ ' aˆ = ρ + ' r Using dl = d aˆ ' ( ρ ρ ' ) ( + ) 3/ Id' aˆ aˆ aˆ dh = 4 π ρ ' ( ρ ρ ) I ρ ' ( ) π + ( ρ + ) Id' aˆ aˆ ' aˆ d aˆ I H = = = aˆ 3/ 3/ 4 π ρ ' 4 ' πρ Eample The magnetic field frm a finite line current Frm the abve result α I ρtanα ρ d ' aˆ I I H = sinθ sinα ' tan 3/ sinα = = = ρ α ρ + ' ρ ρ α ( ) ( ) ( ) Let ' = ρ tan θ, then d ' = ρcs θdθ

3 Hayt; /9/009; 8-3

4 8. Ampere s ircuital Law The clsed path integral f H is equal t the current enclsed by the path HdL i = I : ight hand rule between and I Hayt; /9/009; 8-4 Eample An infinitely lng current carrying wire alng -ais The symmetry predicts n variatin alng and. Frm Bit-avart law H dl Only H eists. H r Path hse the integratin path such that ( ) ( ) and ( H) is cnstant n In this prblem the path shuld be a circle centered n -ais H dl π H d H π i d H I 0 ρ ρ 0 ρ π = I H = πρ Eample A caial cable The symmetry shws that H is independent f and. The symmetry als shws that there eists nly H, Fig (b). () In the regin a < ρ < b I H =. πρ () In the regin ρ < a The enclsed current is Ienclsed πρ = I. πa Frm Ampere s circuital law ρ πρ H = I a ρ H = I π a (3) In the regin b < ρ < c π ρ πρh = I I π I c ρ H = πρ c b (4) In the regin ρ > c H = 0 ( b ) ( c b )

5 Eample A sheet current Hayt; /9/009; 8-5 The surface current density is given by K = K aˆ y y The symmetry shw that n change f H alng r y directin. ubdivide the sheet int a number f filaments. N H cmpnent y H prduced by ne filament is canceled by a symmetrically lcated filament. H = 0 hse a path Frm Ampere s circuital law H L + H L = K L ( ) y H H Ky = () hse a path H H = K () 3 y mbine () and () H = H 3 By symmetry H = H (3) Insert (3) int () and () H = Ky ( > 0) H = Ky ( < 0)

6 Eample lenid Hayt; /9/009; 8-6 The magnetic field well inside the slenid NI H = aˆ d Gd apprimatin when nt clser than a t the ends nr clser than d/n t the surface Eample Trid NI H = ˆ πρ H = 0 a (Inside trid) (Outside trid)

7 8.3 url Assume the surface nrmal -ais The directin f traverse : (ight-hand rule) Find HdL i alng the lp Hayt; /9/009; 8-7 H y HdL = H Δy Hy + Δ Δy ( i ) y, ( HdL i ) = H, 3 ( Δ) H + Δy ( Δ ) and s n. 3 H y The clse path integral f H is H y H HdL i ΔΔy y Ampere s circuital law H y H HdL i H y H HdL i ΔΔy JΔΔy J y n y plane ΔΔy y HdL i H y H lim = = J () Δ, Δy0 ΔΔy y imilarly lim Δy, Δ0 HdL i H H y = = J and ΔyΔ y lim Δ, Δ0 HdL i H H = = J y () ΔΔ mbine (), () and (3) H H y H H H y H aˆ ˆ ˆ + ay a J y + = y The definitin f curl ( curl H ) = lim Δ 0 N N HdL i Δ N : The curl f H in N directin lsed path integral n a surface nrmal t N directin. The area enclsed by the path. Nte that curl f H is a vectr curl H = curl H aˆ + curl H aˆ + curl H aˆ ( ) ( ) ( ) ( ) y N y

8 In rectangular crdinates H H y H H H y H curl H = aˆ + aˆ aˆ y + y y Hayt; /9/009; 8-8 Fr easy evaluatin aˆ aˆy aˆ curl H = y H H H Anther ntatin curl H = H y In general, haˆ haˆ haˆ A = hhh 3 u u u 3 ha ha ha u u 3 u3 y 3 where h = h = h3 = : ectangular crdinates h =, h = ρ, h3 = : ylindrical crdinates h =, h = r, h = r sin θ : pherical crdinates 3 The pint frm f Ampere s circuital law H = J The pint frm f the circulatin f E. E = 0 lsed path integral is als called as circulatin The curl f a vectr is the circulatin per unit area. Nn-er curl makes a paddle wheel rtate in the field. Nn-er curl Zer curl

9 Hayt; /9/009; 8-9

10 8.4 tkes Therem Hayt; /9/009; 8-0 The surface is subdivided int a number f Δ. Apply the curl t ne f Δ s HdL i Δ ( H) i aˆ N : a ˆN is the right hand nrmal Δ HdL H Δ i Δ i ( ) um all ver the subdivisin HdL ( H ) Δ i Δ i HdL i = ( H) iδ : tkes therem Ampere s circuital law H = J H id = Ji d ( ) HdL i I Eample Prve i A = 0 Let i A = T Take vlume integral n bth sides i Adv= T dv v v Tdv= 0 v T = 0 Divergence therem v is arbitrary ( A) i d AdL= i 0, Because the clsed surface has n pening. tkes therem. i A = 0 ummary IdL aˆ H =, HdL = I i, H = J

11 Hayt; /9/009; 8-

12 8.5 Magnetic Flu and Magnetic Flu Density In free space the magnetic flu density is defined as B = μ H Hayt; /9/009; 8- where the permeability in free space is given by 7 μ = 0 [ H / m] The magnetic flu is Φ= Bd i The electric flu lines begin and end n psitive and negative charge, respectively. Ψ= Dd i = Q N such surce fr the magnetic flu lines. The magnetic flu lines are clsed and d nt terminate n a magnetic charge Bd= i 0 i B = 0 Fur Mawell s equatins fr static fields id = ρv E = 0 ib = 0 H = J Divergence therem tkes' therem Dd i = Q= ρ vl vdv EdL i = 0 Bd i = 0 HdL i = I= Jd i The cnstitutive relatins D = εe B = μ H The electric ptential E = V

13 8.6 The calar and Vectr Magnetic Ptentials The scalar magnetic ptential may be defined as H = V m Hayt; /9/009; 8-3 But V m is multi-valued and therefre used little. tart frm the vectr identity i A = 0 ince i B = 0, we define the vectr magnetic ptential as B = A Take curl f bth sides and use B = μh H A = J μ It can be prved by using Bit-avart law (sectin 8.7) μidl A = + G : G is an arbitrary scalar functin It is custmary t set G=0 Eample The vectr magnetic ptential abut a differential filament Identifying dl = d a μ ˆ Ida da = ρ + ˆ The magnetic field is btained as da Id ρ dh = da aˆ aˆ μ 4 μ ρ π ρ + ( ) 3/ : The same result as Bit-avart A fr a current density and a surface current density. IdL is replaced by Jdv and J d, respectively. μjdv A = vl A = μj d

14 Hayt; /9/009; Derivatin f the teady-magnetic-field Laws Fur basic equatins in the magnetic field IdL aˆ H =, H = J, B = μh, B = A Derive Bit-avart law frm A μjdv A = : is surce pint and is field pint. vl μjdv J H = A vl dv μ 4 4 vl μ π π Use ( V ) = ( ) V + ( V ) H = J ( J) dv + vl =0 aˆ = + y y + J aˆ ˆ I dl a H = dv vl =, where ( ) ( ) ( ) Jdv is replaced by IdL Derive differential equatin f A frm Ampere s circuital H A ( A) A J μ μ = i The first term μ J μ ia = dv J ( J ) dv i + vl i i vl =0 i( V ) = V i( ) + ( i V ) Use 3 = = μ μ J ia = J dv ( J) dv i vl i i vl =0 frm static cntinuity eq. Frm the vectr identity Frm the divergence therem μ J ia = d 0 i, since enclses all the current and n current n the surface Therefre, t satisfy the Ampere s circuital law A = μ J

15 Eample A caial cable with radii f a and b, and current I in directin In the gap J = 0 A = 0 Hayt; /9/009; 8-5 ince I is in directin, nly = 0 A A eists, In cylindrical crdinates A A A A = i A = ρ + + = 0 ρ ρ ρ ρ ince A is a functin nly f ρ A ρ = 0 ρ ρ ρ A = ln ρ + ln ρ A =, : a er reference at ρ = b b The magnetic flu density A B = A aˆ aˆ ρ ρ H = aˆ μρ Apply Ampere s circuital law π π HdL i aˆ d aˆ I 0 iρ = = μρ μ μ I = π Therefre μi b A = ln π ρ I H = πρ