Then the second equation becomes ³ j

Transcription

1 Magnetic vector potential When we derived the scalar electric potential we started with the relation r E = 0 to conclude that E could be written as the gradient of a scalar potential. That won t work for the magnetic field (except where j = 0), because the curl of B is not zero in general. Instead, the divergence of B is zero. That means that B may be written as the curl of a vector that we shall call A. B = r A ) r B = r r A = 0 Then the second equation becomes r B = r r A = r r A r 2 A j We had some flexibility in choosing the scalar potential V because E = rv is not changed if we add a constant to V, since r (constant) = 0. imilarly here, if we add to A the gradient of a scalar function, A 2 = A + rχ, we have B 2 = r A 2 = r A + rχ = r A = B With this flexibility, we may choose r A = 0. For suppose this is not true. Then r A + rχ = r A + r 2 χ = 0 o we have an equation for the function χ r 2 χ = r A Once we solve this we will have a vector A 2 whose divergence is zero. Once we know that we can do this, we may just set r A = 0 from the start. This is called the Coulomb gauge condition. With this choice, the equation for A is r 2 A = µ 0 j () We may look at this equation one component at a time (provided that we use Cartesian components.) Thus, for the x component r 2 A x = µ 0 j x This equation has the same form as the equation for V r 2 V = ρ ε 0 and thus the solution will also have the same form: A x ( r) = µ 0 jx ( r 0 ) 4π R dτ0

2 and since we have an identical relation for each component, then A = µ 0 j ( r 0 ) 4π R dτ0 (2) Now remember that jdτ corresponds to Id l, so if the current is confined in wires, the result is A = µ 0 Id l 0 (3) 4π R At this point we may stop and consider if there is any rule for magnetic field analagous to our RULE for electric fields. ince there is no magetic charge, there is no "point charge" field. But we can use our expansion R = X (r 0 ) l r P l+ l cos θ 0 l=0 where r is on the polar axis. Then A ( r = r^z) = µ X 0 I 4π r l+ (r 0 ) l P l cos θ 0 d l 0 The l = 0 term is A 0 = µ 0I 4πr ince the current flows in closed loops, R d l 0 = 0. (This result is actually more general, because in a static situation r j = 0, and the lines of j also form closed loops.) This is the result we expected. The next term is A I 4πr 2 d l 0 r 0 cos θ 0 d l 0 I 4πr 2 We can use tokes theorem to evaluate this integral. u d l = r u ^n da ( r 0 ^z)d l 0 Let u = cχ where c is a constant vector and χ is a scalar function.then c χd l = r cχ ^nda h i = rχ c ^n da We may re-arrange the triple scalar product c χd l = c rχ ^n da 2

3 This is true for an arbitrary constant vector c, so, with χ = ( r 0 ^z) ( r 0 ^z) d h i l 0 = r 0 ( r 0 ^z) ^n 0 da 0 = h^z r 0 r 0 + ^z r 0 r 0i ^n 0 da 0 = (0 + ^z ^n 0 ) da 0 = ^z ^n 0 da 0 = ^n 0 da 0 ^z Note that ^z can come out of the integral because it is a constant. o A = µ 0I 4πr 2 ^n 0 da 0 ^z 4πr 2 m ^z m ^r 4πr2 where m = I ^n 0 da 0 is the magnetic moment of the loop. The corresponding magnetic field is h µ0 i B = r 4πr 2 m ^r = µ µ 0 3 ^r ( m r) + 4π r4 r 3 r ( m r) 4π = = µ 0 µ 3 r 3 [ m ^r( m ^r)] + r 3 ( 3 m + 3 r ( m ^r) m + 3 m) 4πr3 µ 0 [3 r( m ^r) m] 4πr3 m r r + m r r This is a dipole field. Thus the magnetic equivalent of RULE is : At a great distance from a current distribution, the magnetic field is a dipole field Here is another useful result: I A d l = r A ^n da = C B ^n da = B (4) Thus the circulation of A around a curve C equals the magnetic flux through any surface spanning the curve. Boundary conditions for B We start with the Maxwell equations. Remember, if the equation has a divergenceweintegrateover asmall volume (pillbox) that crosses the boundary. 3

4 But if the equation has a curl, we integrate over a rectangular surface that lies perpendicular to the surface. o we start with r B = 0 I r Bdτ = 0 = B da But because we chose h d, the integral over the sides is negligible, and on the bottom side d A 2 = ^nda, so we have B B 2 ^n = 0 (5) The normal component of B is continuous. For the curl equation, we use the rectangle shown: Then r B d A = I B d l = µ 0 j d A µ 0 j ^Ndh w C B B 2 ^t w = µ µ 0 µ 0 B B 2 ^n ^N K ^N jdh ^N w 4

5 Rearrange the triple scalar product on the left to get h i B B 2 ^n ^N K ^N ince we may orient the rectangle so that ^N is any vector in the surface, we have ^n B B 2 K (6) Thus the tangential component of B has a discontinuity that depends on the surface current density K. Crossing both sides with ^n, we get an alternate version: h B B 2 i ^n B B 2 ^n h ^n But now we may make use of (5) to obtain ^n = µ 0K ^n B B 2 i K ^n B B 2 K ^n (7) What about the vector potential? Remember that for the scalar potential V we were able to show that V is continuous across the surface (in most cases). When we find A we first choose a gauge condition. The Coulomb gauge condition is r A = 0 and then we can use our usual pillbox trick to show that A ^n is continuous (8) For the tangential component, we make use of equation (4). Then, using the rectangle, I A d l = B = B ^N wh C A A 2 ^t w = B ^Nwh! 0 as h! 0 Thus we have A ^t is continuous (9) These two result taken together show that the vector potential as a whole is also continuous across the boundary. Finally let s put A into equation (6): r A A 2 K ^n 5

6 o the derivatives of A have a discontinuity. But which ones? Let s expand ^n B = ^n r A = n i ra i ^n r A Then ^n B B 2 = n i r(a i, A i,2 ) ^n r A A 2 K (0) But we have shown that each component of A is continuous at the surface. o the components of r(a i, A i,2 ) parallel to the surface must be zero. Thus only the normal derivatives remain. Then the normal component of equation (0) is identically zero, and the only non-zero components of the boundary condition are the tangential components ^n r A A 2 = µ 0K () tan Now this is neat. Each component of A satisfies Laplace s equation with Neumann boundary conditions, and so it must have a unique solution, as we already proved for V. Magnetic scalar potential When we have the special case of j 0, r B = 0 and we may use a magnetic scalar potenial mag. This can be useful if the current is confined to lines or sheets, because we can create a nice boundary-value problem for m ag. B = r m ag r B = 0 ) r 2 m ag = 0 (2) B normal continuous ) ^n r m ag is continuous (3) ^n B B 2 = µ 0K ) ^n r( m ag m ag2 ) = µ 0K (4) Let s use these boundary conditions to find the potential due to a spinning spherical shell of charge. The current is confined to the surface and has the value K = σ v = σ ω r = σωasinθ^φ where in the last expression put the z axis along the rotation axis. We will take σ to be a constant. The equation for m ag in the region entirely inside (or entirely outside) the sphere is ( with j = 0) r 2 mag = 0 and because we have azimuthal symmetry, the solution is of the form X in = C l r l P l (cosθ) out = l= X l= 6 D l r l+p l (cosθ)

7 We have omitted the l = 0 term because it contributes zero field inside, and we know there can be no monopole term outside. What else dowe know? At the boundary, from (3) m ag,out r and from (4). a asinθ m ag,in = 0 r X X lc l a l P l (cosθ) = l= µ m ag,out θ µ m ag,out φ l= C l = D l l + a 2l+ l m ag,in θ m ag,in φ (l + ) D l a l+2p l (cosθ) l > 0 (5) = µ 0 σaω sinθ = 0 The last equation is automatically satisfied. Thus the final condition we need to satisfy is X l= D l a l+2 θ P l (cos θ) X l= Now since P (cosθ) = cosθ and θ C l a l θ P l (cosθ) = µ 0 σaω sinθ cosθ = sin θ, the first term in the sum is sin θ µ D a C 3 so we may satisfy the boundary conditons by taking D a 3 C σaω and all the other C l, D l = 0. Then equation (5) gives and then o giving a field B = D a 3 + D 2 a 3 σaω ) D σa 4 ω 3 C = 2 µ 0σaω 3 ½ 2 3 m ag = µ ¾ 0σaωrcosθ inside 3 µ 0σa 2 ω a2 r cos θ outside 2 ( 2 3 µ 0σaω^z inside µ 0 σaω 2cosθ a3 ^r + sin θ ^θ outside 3r 3 ) 7

8 Thus the field inside is uniform and the field outside is apure dipole field. The dipole moment is The dimensions of m are m = 4π 3 σa4 ω charge (length) 4 = charge area time time (length)2 = current area which is correct. You should verify that you get the same m by summing current loops. Compare this solution with Gri ths example 5.. Which method do you think is easier? 8