Today in Physics 217: Ampère’s Law Magnetic field in a solenoid, calculated with the Biot-Savart law The divergence and curl of the magnetic field Ampère’s law Magnetic field in a solenoid, calculated with Ampère’s law Summary of electrostatics and magnetostatics so far 6 November 2002 dA C A J v∫ B ⋅ dA = C Physics 217, Fall 2002 4π c ∫ A J ⋅ da = 4π I encl c 1 Another Biot-Savart law example: the solenoid Griffiths problem 5.11: find the magnetic field at point P on the axis of a tightly-wound solenoid (helical coil) consisting of n turns per unit length wrapped around a cylindrical tube of radius a and carrying current I. Express your answer in terms of θ 1 and θ 2 (it’s easiest that way). Consider the turns to be essentially circular, and use the result of example 5.6. What is the magnetic field on the axis of an infinite solenoid? a θ1 θ2 P I 6 November 2002 Physics 217, Fall 2002 2 Reminder of the result of Example 5.6 Magnetic field a distance z along the axis of a circular loop with radius R and current I: 2π I ˆ B=z c (z R2 2 +R 2 dBz dB dBz dB dBs ) 2 32 dBs dBz r z r dA R dA 6 November 2002 Physics 217, Fall 2002 3 The solenoid (continued) a θ1 P θ2 z I Suppose that n is so large that we can consider the loops in the coil to be displaced infinitesimally; then the number of loops in a length dz is ndz, and 2π Indz ˆ dB = z c 6 November 2002 (z a2 2 +a Physics 217, Fall 2002 ) 2 32 4 The solenoid (continued) a θ1 θ2 P z I Take so a tan θ = z 2 tan θ 2 1 + tan θ dθ = dz = − 2 dz = − 2 a cos θ z a ⇒ dz = − sin 2 θ ( 6 November 2002 ) dθ a Physics 217, Fall 2002 5 The solenoid (continued) a θ1 θ2 I 2π Indz ˆ dB = z c ( a2 P z 2π In adθ sin 3 θ ˆ = z − 3 2 2 c a 2 2 θ sin z +a ) 2π In sin θ dθ ; c θ2 2π In 2π In B = −zˆ sin θ dθ = zˆ ( cosθ 2 − cosθ 1 ) . c c = −zˆ ∫ θ1 6 November 2002 Physics 217, Fall 2002 6 The solenoid (continued) a θ1 θ2 P z I For an infinite solenoid, θ 2 = 0 and θ 1 = π , so B = zˆ 6 November 2002 2π In 4π In zˆ = µ0 Inzˆ in MKS . cos 0 cos − π = ( ) c c Physics 217, Fall 2002 7 The divergence and curl of B Any vector field is uniquely specified by its divergence and curl. What are the divergence and curl of B? Consider a volume V to contain current I, current density J ( r ′ ) : 1 J ( r ′ ) × rˆ ′ B (r ) = d τ c r2 S ∫ P Denote gradient with respect to the components of r and r’ by — and —′. Now note that 1 1 — = −—′ (because r = r − r ′), r r rˆ 1 and — = − 2 . r r 6 November 2002 V Physics 217, Fall 2002 J ( r ′) r r r’ dτ ′ 8 The divergence and curl of B (continued) With these, 1 B (r ) = − c rˆ 1 1 — × J ( r ′ ) dτ ′ × J ( r ′ ) dτ ′ = 2 c r r V V J ( r ′) 1 dτ ′ ( remember, J ≠ f ( r ) ) . = —× c r ∫ ∫ ∫ V This is a useful form for B, which we will use a lot next lecture too (the integral turns out to be the magnetic vector potential, A). Take its divergence: J ( r ′) 1 — ⋅ B (r ) = — ⋅ — × ∫ dτ ′ = 0 . r c V 6 November 2002 Physics 217, Fall 2002 The divergence of any curl is zero, remember. 9 The divergence and curl of B (continued) Integrate this last expression over any volume: ∫ — ⋅ B ( r ) dτ = v∫ B ⋅ da = 0 . Compare these to the expressions for E in electrostatics, and we see that magnetostatics involves no counterpart of charge: there’s no “magnetic charge.” Now for the curl: J ( r ′) 1 — × B (r ) = — × — × dτ ′ . r c V Use Product Rule #10: ∫ — × — × A = — ( — ⋅ A) − ∇2 A : 6 November 2002 Physics 217, Fall 2002 10 The divergence and curl of B (continued) 1 J ( r ′) J ( r ′) 1 2 dτ ′ − ∇ dτ ′ — × B (r ) = — — ⋅ c c r r V V ∫ ∫ 1 J ( r ′) 1 1 = — —⋅ J ( r ′ ) ∇ 2 dτ ′ . dτ ′ − c c r r V V Now use your old friend Product Rule #5, ∫ — ⋅ ( fA ) = f ( — ⋅ A ) + A ⋅ ( — f ) , to write —⋅ J ( r ′) 6 November 2002 ∫ r =0 (J independent of r) 1 1 1 dτ ′ = — ⋅ J ( r ′ ) + J ( r ′ ) ⋅ — = J ( r ′ ) ⋅ — r r r Physics 217, Fall 2002 11 The divergence and curl of B (continued) rˆ 1 ∇ = — ⋅— = — ⋅ 2 r r r 2 1 Also, 3 = πδ 4 (r) , so — × B ( r ) = 1 — J ( r ′ ) ⋅ — 1 dτ ′ + 4π c c r ∫ V ∫ J ( r ′ ) δ 3 ( r − r ′ ) dτ ′ V 1 4π 1 = − — J ( r ′ ) ⋅ —′ dτ ′ + J (r ) . c c r ∫ V Use Product Rule #5 again, on the first term: J ( r ′) 1 J ( r ′) 1 J ( r ′ ) ⋅ —′ = —′ ⋅ − —′ ⋅ J ( r ′ ) = —′ ⋅ r r r r =0 in magnetostatics 6 November 2002 Physics 217, Fall 2002 12 The divergence and curl of B (continued) So, J ( r ′ ) 4π 1 J (r ) — × B ( r ) = − — —′ ⋅ dτ ′ + c r c V ∫ 4π 1 J ( r ′) J (r ) . = − — ⋅ da′ + c r c S But by definition J = 0 on the surface, so the integral vanishes: v∫ — × B (r ) = 4π J (r ) . c Ampère’s Law This can be put into integral form by choosing an area that some current flows through: 6 November 2002 Physics 217, Fall 2002 13 The divergence and curl of B (continued) ∫ A 4π ( — × B ) ⋅ da = c v∫ B ⋅ dA = C dA ∫ J ⋅ da C A 4π I enclosed c . A J Ampère’s law is to magnetostatics what Gauss’ law is to electrostatics, except that one uses an Ampèrean loop to enclose current, instead of a Gaussian surface to enclose charge. The same tricks we learned with Gauss’ law and superposition have analogues in magnetostatics. 6 November 2002 Physics 217, Fall 2002 14 Example: field in an infinite solenoid The symmetry of the coil dictates that the field must be along z, and must be a lot stronger inside than out, so if the number of turns per unit length is n, and the current is I, v∫ C 4π B ⋅ dA = c ∫ A ∆z C A 4π J ⋅ da = I enclosed c 4π 4π nI zˆ B∆z = In∆z ⇒ B = c c [ = µ0 nI in MKS ] . Same as before! 6 November 2002 Physics 217, Fall 2002 15 Maxwell’s equations for electrostatics and magnetostatics Note the similarities and differences: — ⋅ Ε = 4πρ —⋅B = 0 —× E = 0 4π J —×B = c v∫ E ⋅ da = 4π Qencl v∫ B ⋅ da = 0 v∫ E ⋅ dA = 0 6 November 2002 4π v∫ B ⋅ dA = c Iencl Physics 217, Fall 2002 16 Maxwell’s equations for electrostatics and magnetostatics Note the similarities and differences (MKS): ρ —⋅Ε = ε0 —× E = 0 6 November 2002 —⋅B = 0 — × B = µ0 J 1 v∫ E ⋅ da = ε 0 Qencl v∫ B ⋅ da = 0 v∫ E ⋅ dA = 0 v∫ B ⋅ dA = µ0 Iencl Physics 217, Fall 2002 17