13 Current sheet, solenoid, vector potential and current loops Js =ˆzJs z In the following examples we will calculate the magnetic fields B = µoH established by some simple current configurations by using the integral form y of static Ampere’s law. B L C B =ˆyB(x) W x Example 1: Consider a uniform surface current density Js = Jszˆ A/m flowing on µ J B(x)= o s x =0plane (see figure in the margin) — the current sheet extends infinitely in 2 y and z directions. Determine B and H. Solution: Since the current sheet extends infinitely in y and z directions we expect B As shown in Example 1 mag- to depend only on coordinate x.Also,thefieldshouldbethesuperpositionofthe netic field of a current sheet fields of an infinite number of current filaments, which suggests, by right-hand- rule, B =ˆyB(x),whereB(x) is an odd function of x.TodetermineB(x),such is independent of distance that B( x)= B(x),weapplyAmpere’slawbycomputingthecirculationof x from the current sheet. − − | | B around the rectangular path C shown in the figure in the margin. We expand Also H changes discontinu- B dl = µ I ously across the current sheet · o C !C by an amount Js. as B(x)L +0 B( x)L +0=µ J L, − − o s from which we obtain µoJs µoJs Js B(x)= B =ˆy sgn(x) and H =ˆy sgn(x). 2 ⇒ 2 2 1 Example 2: Consider a slab of thickness W over W <x<W which extends in- − 2 2 finitely in y and z directions and conducts a uniform current density of J =ˆzJo A/m2.DetermineH if the current density is zero outside the slab. Solution: Given the geometric similarities between this problem and Example 1, we postulate that B =ˆyB(x),whereB(x) is an odd function of x,thatisB( x)= − B(x).TodetermineB(x) we apply Ampere’s law by computing the circulation − of B around the rectangular path C shown in the figure in the margin. For W x< 2 ,weexpand B dl = µ I · o C !C as B(x)L +0 B( x)L +0=µ J 2xL B(x)=µ J x. z − − o o ⇒ o o W y For x 2 ,theexpansiongives ≥ Jozˆ W B(x)L +0 B( x)L +0=µoJoWL B(x)=µoJo . B L B =ˆyB(x) − − ⇒ 2 C Hence, we find that W x W W W yJˆ ox, x < 2 − 2 2 H = W | | "yJˆ o 2 sgn(x), otherwise. By(x) µoJoW 2 Note that the solution plotted in the margin shows no discontinuity at x = W ± 2 or elsewhere. W W x − 2 2 The figure in the margin depicts a finite section of an infinite solenoid. Asolenoidcanbeconstructedinpracticebywindingalongwireintoa 2 multi loop coil as depicted. A solenoid with its loop carrying a current I B =ˆzB in φˆ direction (as shown), produces effectively a surface current density of Js = INφˆ A/m, where N is the number density (1/m) of current loops in C the solenoid. In Example 3 we compute the magnetic field of the infinite solenoid using Ampere’s law. L I Example 3: An infinite solenoid having N loops per unit length is stacked in z- direction, each loop carrying a current of I Aincounter-clockwisedirectionwhen Infinite solenoid viewed from the top (see margin). Determine H. with N loops per unit length carrying I amps per loop Solution: Assuming that B =0outside the solenoid, and also B is independent of z within the solenoid, we find that Ampere’s law indicates for the circulation C B = µoIN shown in the margin B dl = µ I LB = µ INL. · o C ⇒ o !C This leads to B = µoIN and H =ˆzIN for the field within the solenoid. The assumption of zero magnetic flux density B =0for the exterior region is justified because: (a) if the exterior field is non-zero, then it must be independent of x and y (follows from Ampere’s law applied to any exterior path C with IC =0), and 2 (b) the finite interior flux Ψ=µoINπa can only be matched with the flux of the infinitely extended exterior region when the constant exterior flux density (because of (a)) is vanishingly small. 3 Static electric fields: Curl-free and are governed by • E =0, D = ρ where D = $E ∇× ∇· with $ = $r$o. Static magnetic fields: Divergence-free and are governed by • B =0, H = J where B = µH ∇· ∇× with µ = µrµo —relativepermeabilities µr other than unity (for free space) will be explained later on. Mathematically, we can generate a divergence-free vector field B(x, y, z) as B = A ∇× by taking the curl of any vector field A = A(x, y, z) (just like we can generate acurl-freeE by taking the gradient of any scalar field V (x, y, z)). − Verification: Notice that ∂ ∂ ∂ ∂x ∂y ∂z ∂ ∂ ∂ ∂ ∂ ∂ A = ( A)x + ( A)y + ( A)z = # ∂x ∂y ∂z # ∇·∇× ∂x ∇× ∂y ∇× ∂z ∇× # # # Ax Ay Az # # # ∂ ∂A ∂A ∂ ∂A ∂A ∂ ∂A # ∂A # = ( z y ) ( z x )+ ( y# x )=0.# ∂x ∂y − ∂z − ∂y ∂x − ∂z ∂z ∂x #− ∂y # If B = A represents a magnetostatic field, then A is called mag- • ∇× netostatic potential or vector potential. 4 – Vector potential A can be used in magnetostatics in similar ways to how electrostatic potential V is used in electrostatics. In electrostatics we can assign V =0to any point in space ◦ that is convenient in a given problem. In magnetostatics we can assign A to any scalar that is ◦ ∇· convenient in a given problem. – For example, if we make the assignment1 A =0, ∇· then we find that B = A = ( A) 2A = 2A. ∇× ∇×∇× ∇ ∇· −∇ −∇ This is a nice and convenient outcome, because, when combined with H = J B = µ J, ∇× ⇒∇× o it produces 2A = µ J, ∇ − o which is the magnetostatic version of Poisson’s equation ρ 2V = . ∇ −$o 1With this assignment — known as Coulomb gauge — A acquires the physical meaning of “potential momentum per unit charge”, just as scalar potential V is “potential enegy per unit charge” (see Konopinski, Am. J. Phys., 46,499,1978). 5 – In analogy with solution ρ(r') 3 V (r)= d r' z J(r") 4π$o r r' r r" $ | − | − of Poisson’s equation, it has a solution r r" µoJ(r') 3 A(r)= d r'. 4π r r' $ | − | y 2 Given any static current density J(r),theaboveequationcanbeusedto O obtain the corresponding vector potential A that simultaneously satisfies x A =0 and A = B. ∇· ∇× Once A is available, obtaining B = A is then just a matter of taking a ∇× curl. Magnetic flux density B of a single current loop I can be calculated • after determining its vector potential as follows: z y – For a loop of radius a on z =0plane, we can express the corresponding current density as 2 2 ( y',x', 0) J(r')=Iδ(z')δ( x' + y' a) − a − 2 2 x x' + y' I % ˆ where the ratio on the right is the unit vector φ'.% – Inserting this into the general solution for vector potential, and performing 2 2 Jφ = Iδ(z)δ( x + y a) the integration over z',weobtain & − 2 Also, in quasi-statics we use J(r!,t) to obtain A(r,t) and B = A over regions small compared to ∇× λ = c/f,withf the highest frequency in J(r!,t). 6 µoI 2 2 ( y',x', 0) A(r)= δ( x' + y' a) − dx'dy' 2 2 2 2 2 4π − (x x') +(y y') + z x' + y' $ % − − µoI ( y',x', 0) = δ(r' a) %− r'dr'dφ%' 2 2 2 4π − (x x') +(y y') + z r' $ π − − µoI ( a sin φ',acos φ', 0) = %− dφ' xAˆ x(r)+ˆyAy(r). 2 2 2 4π π (x a cos φ') +(y a sin φ') + z ≡ $− − − % – Given that Az =0,itcanbeshownthatB = A leads to ∇× ∂A ∂A ∂A ∂A B = y ,B= x ,B= y x . x − ∂z y ∂z z ∂x − ∂y – From the expected azimuthal symmetry of B about the z-axis, it is sufficient to evaluate these on, say, y =0plane — after some algebra, and dropping the primes, we find, on y =0plane, π µoaI z cos φ Bx = 2 2 2 3/2 dφ, 4π π (x + a + z 2ax cos φ) $− − π µoaI z sin φ By = 2 2 2 3/2 dφ, 4π π (x + a + z 2ax cos φ) $− − and π µoaI a x cos φ Bz = 2 2 −2 3/2 dφ. 4π π (x + a + z 2ax cos φ) $− − – We note that By =0since the By integrand above is odd in φ and the integration limits are centered about the origin. Hence, the field on y =0 plane is given as B =ˆxBx +ˆzBz with Bx and Bz defined above. – There are no closed form expressions for the Bx and Bz integrals above for an arbitrary (x, z). 7 However, it can be easily seen that if x =0(i.e., along the z-axis), Bx =0 ◦ (as symmetry would dictate) and π 2 µoaI a µoIa Bz = 2 2 3/2 dφ = 2 2 3/2 . 4π π (a + z ) 2(a + z ) 3 $− For z a, 2 | |) 2 µoIa 1 Bz 3 , ≈ 2 z z | | 0 which is positive and varies with the inverse third power of distance z . | | !1 I B B – Also, x and z integrals can be performed numerically. Figure !2 in the margin depicts the pattern of Bˆ on y =0plane for a loop !3 !3 !2 !1 0 1 2 3 of radius a =1computed using Mathematica.