Magnetic Vector Potential When We Derived the Scalar Electric Potential

Home , Electric potential, Scalar potential, Vector potential

Magnetic vector potential When we derived the scalar electric potential we started with the relation ~ E~ = 0 to conclude that E~ could be written as the gradient of a scalar r £ 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~ = ~ A~ ~ B~ = ~ ~ A~ = 0 r £ ) r ¢ r ¢ r £ ³ ´ Then the second equation becomes

~ B~ = ~ ~ A~ = ~ ~ A~ 2A~ = µ ~j r £ r £ r £ r r ¢ ¡ r 0 ³ ´ ³ ´ We had some ﬂexibility in choosing the scalar potential V because E~ = ~ V is ¡r not changed if we add a constant to V, since ~ (constant) = 0. Similarly here, r if we add to A~ the gradient of a scalar function, A~ = A~ + ~ χ, we have 2 1 r B~ = ~ A~ = ~ A~ + ~ χ = ~ A~ = B~ 2 r £ 2 r £ 1 r r £ 1 1 ³ ´ With this ﬂexibility, we may choose ~ A~ = 0. For suppose this is not true. r ¢ Then ~ A~ + ~ χ = ~ A~ + 2χ = 0 r ¢ 1 r r ¢ 1 r So we have an equation fo³r the func´tion χ

2χ = ~ A~ r ¡r ¢ 1

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 ~ A~ = 0 from the start. This is r ¢ called the Coulomb gauge condition. With this choice, the equation for A~ is

2A~ = µ ~j (1) r ¡ 0 We may look at this equation one component at a time (provided that we use Cartesian components.) Thus, for the x component ¡ 2A = µ j r x ¡ 0 x This equation has the same form as the equation for V ρ 2V = r ¡ ε0 and thus the solution will also have the same form: µ j (~r ) A (~r) = 0 x 0 dτ x 4π R 0 Z

1 and since we have an identical relation for each component, then ~ µ0 j (~r0) A~ = dτ0 (2) 4π R Z Now remember that ~j dτ corresponds to Id`~, so if the current is confined in wires, the result is µ Id`~ A~ = 0 0 (3) 4π R Z At this point we may stop and consider if there is any rule for magnetic field analagous to our RULE 1 for electric fields. Since there is no magetic charge, there is no "point charge" field. But we can use our expansion

l 1 1 (r0) = Pl cos θ0 R rl+1 l=0 X ¡ ¢ where ~r is on the polar axis. Then

µ0 I l A~ (~r = rz^) = (r ) P cos θ0 d`~ 4π rl+1 0 l 0 Z X ¡ ¢ The l = 0 term is µ0I A~ = d`~0 0 4π r Z ~ Since the current ﬂows in closed loops, d`0 = 0. (This result is actually more general, because in a static situation ~ ~j = 0, and the lines of ~j also form closed r¢ R loops.) This is the result we expected. The next term is

µ0I µ0I A~ = r0 cos θ0d`~0 = (~r0 z^) d`~0 1 4πr2 4πr2 ¢ Z Z We can use Stokes theorem to evaluate this integral.

~u d~l = ~ ~u ^n dA ¢ r £ ¢ Z Z ³ ´ Let ~u = ~cχ where ~c is a constant vector and χ is a scalar function. Then

~c χd~l = ~ ~cχ n^dA ¢ r £ ¢ Z Z ³ ´ = ~ χ ~c n^ dA r £ ¢ Z h³ ´ i We may re-arrange the triple scalar product

~c χd~l = ~c ~ χ n^ dA ¢ ¡ ¢ r £ Z Z ³ ´

2 This is true for an arbitrary constant vector ~c, so, with χ = (~r z^) 0 ¢

(~r0 z^) d~l0 = ~ 0 (~r0 z^) n^0dA0 ¢ ¡ r ¢ £ Z Z h i = ^z ~ 0 ~r0 + z^ ~ 0 ~r0 ^n0dA0 ¡ £ r £ ¢ r £ Z h ³ ´ ³ ´ i = (0 + z^ n^ ) dA ¡ £ 0 0 Z = ^z n^ dA = ^n dA z^ ¡ £ 0 0 0 0 £ Z Z Note that z^ can come out of the integral because it is a constant. So

µ0I µ0 µ0 A~ = n^0dA0 z^ = m~ z^ = m~ r^ 1 4πr2 £ 4π r2 £ 4πr2 £ Z where

m~ = I n^0dA0 Z is the magnetic moment of the loop. The corresponding magnetic ﬁeld is µ B~ = ~ 0 m~ ^r 1 r £ 4πr2 £ µ h 3 i 1 = 0 ^r (m~ ~r) + ~ (m~ ~r) 4π ¡ r4 £ £ r3 r £ £ µ ¶ µ 3 1 = 0 [m~ ^r (m~ ^r)] + m~ ~ ~r + m~ ~ ~r 4π ¡ r3 ¡ ¢ r3 ¡ ¢ r r ¢ µ µ ³ ³ ´ ³ ´´¶ = 0 ( 3m~ + 3~r (m~ r^) m~ + 3m~ ) 4πr3 ¡ ¢ ¡ µ = 0 [3~r (m~ ^r) m~ ] 4πr3 ¢ ¡ This is a dipole ﬁeld. Thus the magnetic equivalent of RULE 1 is :

At a great distance from a current distribution, the magnetic ﬁeld is a dipole ﬁeld

Here is another useful result:

~ ~ ~ ~ ~ A d` = A n^ da = B ^n da = ©B (4) ¢ S r £ ¢ S ¢ IC Z ³ ´ Z

Thus the circulation of A~ around a curve C equals the magnetic ﬂux through any surface S spanning the curve. Boundary conditions for B~ We start with the Maxwell equations. Remember, if the equation has a divergence we integrate over a small volume (pillbox) that crosses the boundary.

3 But if the equation has a curl, we integrate over a rectangular surface that lies perpendicular to the surface. So we start with ~ B~ = 0 r ¢ ~ B~dτ = 0 = B~ dA~ r ¢ ¢ Z IS

But because we chose h d, the integral over the sides is negligible, and on ¿ the bottom side dA~ = ^ndA, so we have 2 ¡ B~ B~ n^ = 0 (5) 1 ¡ 2 ¢ ³ ´ The normal component of B~ is continuous. For the curl equation, we use the rectangle shown:

Then ~ ~ ~ ~ ~ B dA = µ0j dA S r £ ¢ S ¢ Z ³ ´ Z B~ d`~ = µ ~j N^ dh w ¢ 0 ¢ IC Z

B~ B~ t^ w = µ µ ~jdh N^ w 1 ¡ 2 ¢ ¡ 0 0 ¢ µZ ¶ ³ ´ ¡ ¢ B~ B~ n^ N^ = µ K~ N^ 1 ¡ 2 ¢ ¡ £ 0 ¢ ³ ´ ³ ´ 4 Rearrange the triple scalar product on the left to get

B~ B~ n^ N^ = µ K~ N^ ¡ 1 ¡ 2 £ ¢ 0 ¢ h³ ´ i Since we may orient the rectangle so that N^ is any vector in the surface, we have n^ B~ B~ = µ K~ (6) £ 1 ¡ 2 0 ³ ´ 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:

n^ B~ B~ n^ = µ K~ ^n £ 1 ¡ 2 £ 0 £ h ³ ´i B~ B~ ^n n^ B~ B~ = µ K~ ^n 1 ¡ 2 ¡ ¢ 1 ¡ 2 0 £ ³ ´ h ³ ´i But now we may make use of (5) to obtain

B~ B~ = µ K~ ^n (7) 1 ¡ 2 0 £ ³ ´ 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 ~ A~ = 0 r ¢ 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,

A~ d`~ = © = B~ N^ wh ¢ B ¢ IC A~ A~ t^ w = B~ N^wh 0 as h 0 1 ¡ 2 ¢ ¡ ¢ ! ! ³ ´ ¡ ¢ 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):

~ A~ A~ = µ K~ n^ r £ 1 ¡ 2 0 £ ³ ´

5 So the derivatives of A~ have a discontinuity. But which ones? Let’s expand

n^ B~ = n^ ~ A~ = n ~ A n^ ~ A~ £ £ r £ ir i ¡ ¢ r ³ ´ ³ ´ Then n^ B~ B~ = n ~ (A A ) n^ ~ A~ A~ = µ K~ (10) £ 1 ¡ 2 ir i,1 ¡ i,2 ¡ ¢ r 1 ¡ 2 0 ³ ´ ³ ´ ³ ´ But we have shown that each component of A~ is continuous at the surface. So the components of ~ (A A ) r i,1 ¡ i,2 parallel to the surface must be zero. Thus only the normal derivatives remain. Then the normal component of equation (10) is identically zero, and the only non-zero components of the boundary condition are the tangential components ~ ~ ~ ~ n^ A1 A2 = µ0K (11) ¢ r ¡ tan ¡ ³ ´ ³ ´ 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, ~ B~ = 0 and we may use a ´ r £ 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 ~ B~ = 0 2© = 0 (12) r ¢ ) r m ag B continuous ^n ~ © is continuous (13) normal ) ¢ r m ag n^ B~ B~ = µ K~ n^ ~ (© © ) = µ K~ (14) £ 1 ¡ 2 0 ) £ r m ag1 ¡ m ag2 ¡ 0 Let’s use th³ese boun´dary 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 = σωa sin θφ^ £ 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 (1 with ~j = 0) 2© = 0 r mag and because we have azimuthal symmetry, the solution is of the form

6 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 do we know? At the boundary, from (13) ∂© ∂© m ag,out m ag,in = 0 ∂r ¡ ∂r ¯r=a ¯r=a ¯ ¯ ¯ 1 l 1 ¯ 1 Dl ¯ lCl a ¡ Pl (co¯s θ) = (l + 1) Pl (cos θ) ¡ al+2 l=1 l=1 X X D l + 1 C = l l > 0 (15) l ¡ a2l+1 l and from (14). 1 ∂© ∂© m ag,out m ag,in = µ σaω sin θ a ∂θ ¡ ∂θ ¡ 0 µ ¯r=a ¯r=a¶ 1 ∂© ¯ ∂© ¯ m ag,out ¯ m ag,in ¯ = 0 a sin θ ∂φ ¯ ¡ ∂φ ¯ µ ¯r=a ¯r=a¶ ¯ ¯ ¯ ¯ The last equation is automa¯tically satisfied. ¯ Thus the final condition we need to satisfy is

1 Dl ∂ 1 l 1 ∂ P (cos θ) C a ¡ P (cos θ) = µ σaω sin θ al+2 ∂θ l ¡ l ∂θ l ¡ 0 Xl=1 Xl=1 Now since P (cos θ) = cos θ and ∂ cos θ = sin θ, the first term in the sum is 1 ∂θ ¡ D 1 C sin θ ¡ a3 ¡ 1 µ ¶ so we may satisfy the boundary conditons by taking D 1 C = µ σaω a3 ¡ 1 0 and all the other Cl, Dl = 0. Then equation (15) gives D D 2 µ σa4ω 1 + 1 = µ σ aω D = 0 a3 a3 1 0 ) 1 3 and then µ σaω C = 2 0 1 ¡ 3 So 2 3 µ0σaωr cos θ inside © = ¡ 2 m ag 1 µ σa2ω a cos θ outside ½ 3 0 r 2 ¾ giving a field 2 µ σaωz^ inside ~ 3 0 B = a3 ^ ( µ0σaω 3r 3 2cos θ r^ + sin θ θ outside ) ³ ´ 7 Thus the field inside is uniform and the field outside is a pure dipole field. The dipole moment is 4π m = σa4ω 3 The dimensions of m are

charge (length)4 charge = (length)2 = current area area time time £ £ which is correct. You should verify that you get the same m by summing current loops. Compare this solution with Gri¢ths’ example 5.11. Which method do you think is easier?