Magnetic field due to a current loop.

Transcription

1 Example using spheical hamonics Sp 18 Magnetic field due to a cuent loop. A cicula loop of adius a caies cuent I. We place the oigin at the cente of the loop, with pola axis pependicula to the plane of the loop. Then the cuent density is δ ( a) j = I δ (µ) a ˆφ (You can get this most easily by stating with the expession in cylindical coodinates j = Iδ (z) δ ( a) ˆφ and using z = cos θ. See also Lea pg 315 Example 6.7.) Then the magnetic vecto potential is (Notes 1 eqn 1) A (x) = µ j (x ) 4π x x d3 x (1) = µ δ ( 4πa I a) δ (µ ) ˆφ φ x x d 3 x We must take cae hee, because the unit vecto ˆφ is not a constant. We must e-expess it in tems of the constant Catesian unit vectos, and thus: ˆφ φ = sin φ ˆx +cosφ ŷ A (x) = µ δ ( 4πa I a) δ (µ ) sin φˆx x x +cosφ ŷ d 3 x Now we use ou most useful esult (J eqn 3.7, sphepobnotes eqn 5 ) to expand the 1/R in the integand. With < =min(, ) and similaly fo, 1

2 A (x) = µ I 4πa = µ I 4πa δ ( a) δ (µ ) l= l l= m= l π +1 = µ ai 1 l l= m= l l < l+1 l m= l 4π l +1 Y lm (θ, φ) δ ( a) δ (µ ) l < Y lm (θ, φ) l +1 4π l +1 Y lm (θ, φ) Ylm θ, φ sin φ ˆx +cosφ ŷ d 3 x l < l+1 l+1 π Ylm θ, φ sin φ ˆx +cosφ ŷ ( ) d dµ dφ π Ylm, φ sin φˆx +cosφ ŷ dφ wheeweusedthesiftingpopetytoevaluatetheintegalsove and µ. We now intepet < as the lesse of and a, and similaly fo. To do the integal ove φ, we ewite the sines and cosines in tems of exponentials: I lm = = π π π Ylm, φ sin φˆx +cosφ ŷ dφ ŷ l +1(l m)! + iˆx 4π (l + m)! P l m () e imφ e iφ ŷ iˆx + e iφ dφ The integal is zeo unless m = ±1. (Be alet hee if you use the manta "axisymmety so m =" you will get into big touble!) With m =+1we get: ŷ l +1(l 1)! + iˆx I l1 = 4π (l +1)! P l 1 () π and with m = 1 I l, 1 = ŷ l +1(l +1)! iˆx 4π (l 1)! P 1 l () = π l +1(l +1)! 4π (l 1)! = π l +1(l 1)! 4π (l +1)! P l 1 () (ŷ iˆx) So A (x) =µ ai π l +1 < l l+1 π ( 1) (l 1)! (l +1)! P 1 l () (ŷ iˆx) (Lea 8.54) l +1(l 1)! 4π (l +1)! P l 1 () [(ŷ + iˆx) Y l1 (θ, φ) (ŷ iˆx) Y l, 1 ]

3 The sum ove l stats at 1 now because with l =thee is no m = ±1 tem. Then, using Y l, m =( 1) m Ylm, (sphepobnotes eqn 4) we get π < A l l +1(l 1)! (x) = µ ai l +1 l+1 4π (l +1)! P l 1 () Pl 1 (µ) (ŷ + iˆx) e iφ +(ŷ iˆx) e iφ = µ ai < l 1 4 l+1 l (l +1) P l 1 () Pl 1 (µ)(ŷcos φ ˆx sin φ) = µ ai l < l+1 P 1 l () P 1 l (µ) l (l +1) ˆφ () We might have expected to find that A is in the φ diection, paallel to j. Check dimensions: A is cuent times µ, which is consistent with (1). We can simplify a bit by inseting the value of Pl 1 (). Fist note that P l is even if l is even, and odd if l is odd. Since Pl m is popotional to the mth deivative of P l, Pl m will be odd if l + m is odd and even if l + m is even. (Lea pg 385) So Pl m () = unless l + m is even, o, in this case, l is odd. Now we can use the ecusion elation (J3.9 o Lea 8.37) lp l (µ) = µpl (µ) Pl 1 (µ) lp l () = Pl 1 () = Pl 1 1 () (Lea 8.53) Thus, using Lea 8.47, with l =n +1, Pl 1 () = (l +1)P l+1 () = (l +1)( 1) (l+1)/ l!! (l +1)!! Pn+1 1 () = ( 1) n+1 (n +1)!! (n)!! = ( 1) n+1 (n +1)!! n n! Thus () becomes A (x) = µ ai = µ ai < n+1 n+ n= < n= 1 (n +1)!! (n +1)(n +) ( 1)n+1 n Pn+1 1 (µ) n! ˆφ n+1 (n +1)!! ( 1) n+1 (n +1)(n +1) n Pn+1 1 (µ) n! ˆφ Sepaating out the n =tem, we have A φ = µ ai < n+1 P1 1 (µ)+ ( 1) n < (n 1)!! 4 n (n +1)! P n+1 1 (µ) and (Lea 8.53) P1 1 = sin θ d (µ) = sin θ dµ 3 (3)

4 Outside the loop, a,and A (x) = µ ai a 4 sin θ Fo a, n =( =1) is the dominant tem: ( 1) n a n+1 (n 1)!! n (n +1)! P n+1 1 (µ) A (x) µ I 4 a sin θ ˆφ ˆφ (4) so A (x) µ a I 4 sin θ ˆφ = µ m 4π sin θ ˆφ = µ m 4π 3 (5) whee m = πa I ẑ is the magnetic moment of the loop. Compae with Jackson equation Then, in this limit B = A = ˆ µ m sin θ θ 4π sin θ ˆ µ = m ˆθ sinθcos θ + sin θ 4π = µ m 4π 3 ˆ cosθ + ˆθ sin θ This is a dipole field, as expected. Inside the loop, <aand we have: A (x) = µ ai 4a a sin θ ˆθ µ m 4π sin θ µ m 4π sin θ ( 1) n n+1 (n 1)!! a n (n +1)! P n+1 1 (µ) Nea the cente, a, the n =tem dominates again, and we have: and B (x) µ 4a I A µ I 4 ˆ sin θ sin θ ˆφ a sin θ ˆθ θ = µ a I ˆ cos θ ˆθ sin θ = µ a I ẑ a unifom field, as expected. z =. sin θ ˆφ (6) Compae with Lea and Buke equation 8.7 with 4

5 Field on axis: Fom LB 8.7, the fieldonthepola(z ) axisis µ B (z) = Ia (z + a ) 3/ ẑ So fo zawe can do a binomial expansion to get B (z) = µ Ia z a z + a 4 z 4 + = µ Ia z a z + 15 a 4 8 z 4 + We can also find B fom the solution (4) fo a: B = A = µ ai ˆ sin θ a 4 sin θ θ P 1 1 (µ)+ ˆθ a P 1 1 (µ)+ ẑ ẑ (7) ( 1) n+1 a n+1 (n 1)!! n+ n (n +1)! P n+1 1 (µ) ( 1) n+1 a n+1 (n 1)!! n (n +1)! P 1 n+1 (µ) Fo µ =1(θ =), the theta component is µ ai 4 times a P 1 1 (1) + ( 1) n+1 (n +1) a n+1 (n 1)!! n (n +1)! P n+1 1 (1) But Pn+1 1 (1) = 1 µ d dµ P n+1 (µ) = µ=1 So the theta component on axis is zeo, as we would expect fom the symmety (azimuthal symmety about the axis and eflection anti-symmety about the plane of the loop). The component is B = µ ai ˆ a sin θ 4 sin θ θ P 1 1 (µ)+ ( 1) n a n+1 (n 1)!! n+ n (n +1)! P n+1 1 (µ) = µ a I 4 3 ˆ 1 µ P1 1 (µ)+ ( 1) n a n (n 1)!! µ n n (n +1)! P n+1 1 (µ) Fom the definition of Pl 1 (Lea 8.53) and the diffeential equation fo P n+1 (Lea 8.19), we find d 1 µ Pn+1 (µ) 1 = d 1 µ d dµ dµ dµ P n+1 (µ) = (n +1)(n +)P n+1 (µ) 5

6 Then we evaluate at µ =1whee P n+1 (1) = 1 B (, ) = µ a I ˆ ( 1) n a n (n 1)!! n n (n +1)(n +) (n +1)! Changing to the z coodinate, we get B z (z) = µ a I z 3 ẑ 1+ 1 ( 1) n a n (n +1)!! z n n 1 n! = µ a I z 3 ẑ 1 3 a z + a4 5 3 z 4 + = µ a I z 3 ẑ 1 3 a 15 a z 8 z which agees with (7). 6