Jackson Classical Electrodynamics Solutions 3.16
Feb 11, 2001 ... Classical Electrodynamics, Third Edition ... Following Jackson, we take the observation point x on the x axis, so its coordi- nates are (ρ, φ = 0,z).
Solutions to Problems in Jackson, Classical Electrodynamics, Third Edition Homer Reid February 11, 2001
Chapter 5: Problems 10-18
Problem 5.10 A circular current loop of radius a carrying a current I lies in the x − y plane with its center at the origin. (a) Show that the only nonvanishing component of the vector potential is Z µ0 Ia ∞ Aφ (ρ, z) = dk cos kz I1 (kρ< )K1 (kρ> ) π 0 where ρ< (ρ> ) is the smaller (larger) of a and ρ. (b) Show that an alternative expression for Aφ is Z µ0 Ia ∞ dke−k|z| J1 (ka)J1 (kρ). Aφ (ρ, z) = 2 0 (c) Write down integral expressions for the components of magnetic induction, using the expressions of parts a and b. Evaluate explicitly the components of B on the z axis by performing the necessary integrations. (a) Translating Jackson's equation (5.33) into cylindrical coordinates, we have Jφ = Iδ(z)δ(ρ − a)
(1)
Following Jackson, we take the observation point x on the x axis, so its coordinates are (ρ, φ = 0, z). Since there is no current in the z direction, and since the 1
Homer Reid's Solutions to Jackson Problems: Chapter 5
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current density is cylindrically symmetric, there is no vector potential in the ρ or z directions. In the φ direction we have Aφ = −Ax sin φ + Ay cos φ = Ay Z µ0 Jy (x0 ) = dx0 4π |x − x0 | Z µ0 Jφ (x0 ) cos φ0 0 = dx 4π |x − x0 | Z 0 Jφ (x0 )eiφ µ0 Re dx0 = 4π |x − x0 | # " Z Z ∞ ∞ X 0 0 µ0 2 eim(φ−φ ) cos[k(z − z 0 )]Im (kρ< )Km (kρ> ) dk dx0 Re Jφ (x0 )eiφ = 4π π m=−∞ 0 where we substituted in Jackson's equation (3.148). Rearranging the order of integration and remembering that φ = 0, we have Z ∞ Z ∞ X µ0 0 i(1−m)φ0 0 0 Aφ = Re Jφ (x )e cos[k(z − z )]Im (kρ< )Km (kρ> )dx dk 2π 2 m=−∞ 0 If m = 1, the φ integral yields 2π; otherwise it vanishes. Thus Z Z ∞ Z ∞ µ0 ∞ Aφ = Jφ (r0 , z 0 ) cos[k(z − z 0 )]I1 (kρ< )K1 (kρ> )ρ0 dz 0 dr0 dk π 0 0 −∞ Substituting (1), we have Aφ =
Iaµ0 π
Z
∞
cos kz I1 (kρ< )K1 (kρ> ) dk. 0
(b) The procedure for obtaining this expression is identical to the one I just went through, but with the expression from Problem 3.16(b) used for the Green's function instead of equation (3.148). (c) Let's suppose that the observation point is in the interior region of the current loop, so ρ< = ρ, ρ> = a. Then ∂Aφ Bρ = [∇ × A]ρ = − ∂z Z Iaµ0 ∞ =− k sin kz I1 (kρ)K1 (ka) dk π 0 1 ∂Aφ Bz = [∇ × A]z = Aφ + ρ ∂ρ Z Iaµ0 ∞ I1 (kρ) 0 + kI1 (kρ) K1 (ka) dk = cos kz π ρ 0
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As ρ = 0, I1 (ρ) → 0, I1 (ρ)/ρ → 1/2, and I10 (ρ) → 1/2, so Bρ (ρ = 0) = 0 Z Iaµ0 ∞ k cos kzK1 (ka) dk Bz (ρ = 0) = π 0 Z ∞ Iaµ0 ∂ = sin kzK1 (ka)dk π ∂z 0 The integral may be done by parts: ∞ Z ∞ Z 1 z ∞ sin kzK1 (kz) dk = − sin kzK0 (ka) + cos kzK0 (ka) dk a a 0 0 0
K0 is finite at zero but sin vanishes there, and sin is finite at infinity but K0 vanishes there, so the first term vanishes. The integral in the second term is Jackson's equation (3.150). Plugging it in to the above, z Iµ0 ∂ 2 2 ∂z (z + a2 )1/2 a2 Iµ0 = . 2 (z 2 + a2 )3/2
Bz (ρ = 0) =
Problem 5.11 A circular loop of wire carrying a current I is located with its center at the origin of coordinates and the normal to its plane having spherical angles θ0 , φ0 . There is an applied magnetic field, Bx = B0 (1 + βy) and By = B0 (1 + βx). (a) Calculate the force acting on the loop without making any approximations. Compare your result with the approximate result (5.69). Comment. (b) Calculate the torque in lowest order. Can you deduce anything about the higher order contributions? Do they vanish for the circular loop? What about for other shapes? (a) Basically we're dealing with two different reference frames here. In the "lab" frame, R, the magnetic field exists only in the xy plane, and the normal to the current loop has angles θ0 , φ0 . We define the "rotated" frame R0 by aligning the z 0 axis with the normal to the current loop, so that in R0 the current loop exists only in the x0 y 0 plane, but the magnetic field now has a z 0 component. The force on the current loop is Z F = (J × B)dV. (2)
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Homer Reid's Solutions to Jackson Problems: Chapter 5 PSfrag replacements z1 = z
z1 z0
θ0 y 0 = y1
y1 x
φ0
x1
y
x1 x0 R1 → R 0
R → R1
Figure 1: Successive coordinate transformations in Problem 5.11.
The components of J are easy to express in R0 , but more complicated in R; the opposite is true for B. There are two ways to do the problem: we can work out the components of J in R and do the integral in R, or we can work out the components of B in R0 and do the integral in R0 , in which case we would have to transform the components of the force back to R to get the answer we desire. I think the former approach is easier. To derive the transformation matrix relating the coordinates of a point in R and R0 , I imagined that the transformation arose from two separate transformations, as depicted in figure (??). The first transformation is a rotation through φ0 around the z axis, which takes us from R to an intermediate frame R1 . Then we rotate through θ0 around the y1 axis, which takes us to R0 . Evidently, the coordinates of a point in the various frames are related by x1 cos φ0 sin φ0 0 x y1 = − sin φ0 cos φ0 0 y (3) z1 0 0 1 z 0 x1 cos θ0 0 − sin θ0 x y1 y0 = 0 1 0 (4) z1 sin θ0 0 cos θ0 z0 Multiplying matrices, 0 cos θ0 cos φ0 x y 0 = − sin φ0 sin θ0 cos φ0 z0
cos θ0 sin φ0 cos φ0 sin θ0 sin φ0
x − sin θ0 y . 0 z cos θ0
(5)
This matrix also gives us the transformation between unit vectors in the two
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Homer Reid's Solutions to Jackson Problems: Chapter 5
frames: ˆi0 cos θ0 cos φ0 ˆj0 = − sin φ0 ˆ0 sin θ0 cos φ0 k
cos θ0 sin φ0 cos φ0 sin θ0 sin φ0
− sin θ0 0 cos θ0
ˆi ˆj . ˆ k
(6)
We will also the inverse transformation, i.e. the expressions for coordinates in R in terms of coordinates in R0 : 0 x cos θ0 cos φ0 − sin φ0 sin θ0 cos φ0 x y = cos θ0 sin φ0 cos φ0 sin θ0 sin φ0 y 0 . (7) z − sin θ0 0 cos θ0 z0
To do the integral in (2) it's convenient to parameterize a point on the current loop by an angle φ0 reckoned from the x0 axis in R0 . If the loop radius is a, then the coordinates of a point on the loop are x0 = a cos φ0 , y 0 = a sin φ0 , and the current density/volume element product is J dV = Id l = (Ia dφ0 )φˆ0 = Ia dφ0 [− sin φ0ˆi0 + cos φ0ˆj0 ] h = Ia dφ0 (− sin φ0 cos θ0 cos φ0 − cos φ0 sin φ0 )ˆi
ˆ + (sin φ0 sin φ0 + cos φ0 cos φ0 )ˆj + (sin φ0 sin θ0 )k
i
We also need the components of the B field at a point on the current loop: B(φ0 ) = B0 [1 + βy(φ0 )]ˆi + B0 [1 + βx(φ0 )] = B0 [1 + aβ(cos φ0 cos θ0 sin φ0 + sin φ0 cos φ0 )]ˆi + B0 [1 + aβ(cos φ0 cos θ0 cos φ0 − sin φ0 sin φ0 )]ˆj The components of the cross product are [J × B]x dV = −Jz By dV = (· · · )βIa2 B0 dφ0 sin2 φ0 sin θ0 sin φ0 [J × B]y dV = Jz Bx dV
= (· · · ) + βIa2 B0 dφ0 sin2 φ0 sin θ0 cos φ0 [J × B]z dV = (Jx By − Jy Bx ) dV = (· · · ) + 0
where we only wrote out terms containing a factor of cos2 φ0 or sin2 φ0 , since only these terms survive after the integral around the current loop (we grouped all the remaining terms into (· · · )). In the surviving terms, cos2 φ0 and sin2 φ0 turn into factors of π after the integral around the loop. Then the force components are Fx = πβIa2 B0 sin θ0 sin φ0 Fy = πβIa2 B0 sin θ0 cos φ0 Fz = 0.
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To compare this with the first-order approximate result, note that the magnetic moment has magnitude πa2 I and is oriented along the z 0 axis: ˆ0 = πa2 I sin θ0 cos φ0ˆi + sin θ0 sin φ0ˆj + cos θ0 k ˆ m = πa2 I k so
∇ B · m = ∇ B0 (1 + βy)mx + B0 (1 + βx)my = B0 β myˆi + mxˆj
= πβIa2 B0 sin θ0 sin φ0ˆi + sin θ0 cos φ0ˆj)
in exact agreement with the result we calculated so laboriously above.
Problem 5.12 Two concentric circular loops of radii a, b and currents I, I 0 , respectively (b < a), have an angle α between their planes. Show that the torque on one of the loops is about the line of intersection of the two planes containing the loops and has the magnitude 2 2n ∞ µ0 πII 0 b2 X (n + 1) Γ(n + 3/2) b 1 N= P2n+1 (cos α). 2a (2n + 1) Γ(n + 2)Γ(3/2) a n=0 The torque on the smaller loop is Z N = r × Jb (r) × Ba (r) dr Z n o = r · Ba (r) Jb (r) − r · Jb (r) Ba (r) dr.
where Jb is the current density of the smaller loop and Ba is the magnetic field of the larger loop. But r · Jb vanishes, because the current flows in a circle around the origin—there is no current flowing toward or away from the origin. Thus Z N = rBr (r)Jb (r)dr (8) where Br is the radial component of the magnetic field of the larger current loop. As in the last problem, it's convenient to define two reference frames for this situation. Let R be the frame in which the smaller loop (radius b, current I) lies in the xy plane, and R0 the frame in which the larger loop lies in the x0 y 0 plane. We might as well take the line of intersection of the two planes to be the y axis, so y = y 0 . Then the z 0 axis has spherical coordinates (θ = α, φ = 0) in
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R, and for transforming back and forth between the two frames we may use the transformation matrices we derived in the last problem, with θ0 = α, φ0 = 0. If we choose to evaluate the integral (8) in frame R, the current density is Jb (r) = Iδ(r − b)δ(θ − π/2) − sin φˆi + cos φˆj
so the components of the torque are Z 2π 2 Br (r = b, θ = π/2, φ) sin φ dφ Nx = −Ib Ny = Ib2
Z
(9)
0 2π
Br (r = b, θ = π/2, φ) cos φ dφ
(10)
0
To do the integral in (8), we need an expression for the radial component Br of the field of the larger loop. Of course, we already have an expression for the field in R0 : in that frame the field is just that of a circular current loop in the x0 y 0 plane, Jackson's equation (5.48): ∞
Br0 (r0 , θ0 ) =
2l+1 µ0 I 0 a X (−1)l (2l + 1)!! r< P (cos θ0 ). 2l+2 2l+1 2r0 2l l! r> l=0
We are interested in evaluating this field at points along the smaller current loop, and for all such points r = b; then r< = b, r> = a and we have Br0 (r0 = b, θ0 ) =
2l ∞ µ0 I 0 X (−1)l (2l + 1)!! b P2l+1 (cos θ0 ). 2a 2l l! a
(11)
l=0
To transform this to frame R, we first note that, since the origins of R and R0 coincide, the unit vectors ˆ r and ˆ r0 coincide, so Br = Br0 . Next, (11) expresses 0 the field in terms of cos θ , the polar angle in frame R0 . How do we write this in terms of the angles θ and φ in frame R? Well, note that z0 r x sin α + z cos α = r r sin θ cos φ sin α + r cos θ cos α = r = sin θ sin α cos φ + cos θ cos α
cos θ0 =
(12)
where in the second line we used the transformation matrix from Problem 5.11 to write down z 0 in terms of x and z. Equation (12) is telling us what our coordinates in R0 are in terms of our coordinates in R; if a point has angular coordinates θ, φ in R, then (12) tells us what angle θ 0 it has in R0 . (We could also work out what the azimuthal angle φ0 would be, but we don't need to, because (11) doesn't depend on φ0 .)
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To express the Legendre function in (11) with the argument (12), we may make use of the addition theorem for associated Legendre polynomials: Pl (cos θ0 ) = Pl (cos θ cos α + sin θ sin α cos φ) = Pl (cos θ)Pl (cos α) + 2
l X
Plm (cos θ)Plm (cos α) cos mφ.
m=1
Of course, the smaller loop exists in the xy plane, so for all points on that loop we have θ = π/2, whence Pl (cos θ0 ) = Pl (0)Pl (cos α) + 2
l X
Plm (0)Plm (cos θ) cos mφ.
m=1
We may now write down an expression for the radial component of the magnetic field of the larger loop, evaluated at points on the smaller loop, in terms of the angle φ that goes from 0 to 2π around that loop: 2l ( ∞ µ0 I 0 X (−1)l (2l + 1)!! b P2l+1 (0)P2l+1 (cos α) Br (φ) = 2a 2l l! a l=0 ) 2l+1 X m m +2 P2l+1 (0)P2l+1 (cos α) cos mφ . m=1
This looks ugly, but in fact when we plug it into the integrals (9) and (10) the sin φ and cos φ terms beat against the cos mφ term, integrating to 0 in the former case and πδm1 in the latter. The torque is Nx = 0 2l ∞ πµ0 II 0 b2 X (−1)l (2l + 1)!! b 1 1 Ny = P2l+1 (0)P2l+1 (cos α). a 2l l! a l=0
To finish we just need to rewrite the numerical factor under the sum: (−1)l (2l + 1)!! 1 (2l + 1)!! Γ(l + 3/2) P2l+1 (0) = 2l l! 2l l! Γ(l + 1)Γ(3/2) (2l + 3 − 2)(2l + 3 − 4)(2l + 3 − 6) · · · (5)(3) Γ(l + 3/2) = 2l Γ(l + 1) Γ(l + 1)Γ(3/2) Γ(l + 3/2) (l + 3/2 − 1)(l + 3/2 − 2) · · · (5/2)(3/2) = Γ(l + 1) Γ(l + 1)Γ(3/2) 2 Γ(l + 3/2) = Γ(l + 1)Γ(3/2) 2 Γ(l + 3/2) 2 = (l + 1) Γ(l + 2)Γ(3/2)
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So my answer is 2 2l ∞ b Γ(l + 3/2) πµ0 II 0 b2 X 1 2 P2l+1 (cos α). (l + 1) Ny = a Γ(l + 2)Γ(3/2) a l=0
Evidently I'm off by a factor of 1/(l + 1)(2l + 1) under the sum, but I can't find where. Can anybody help?
Problem 5.13 A sphere of radius a carries a uniform surface-charge distribution σ. The sphere is rotated about a diameter with constant angular velocity ω. Find the vector potential and magnetic-flux density both inside and outside the sphere.
Problem 5.14 A long, hollow, right circular cylinder of inner (outer) radius a (b), and of relative permeability µr , is placed in a region of initially uniform magnetic-flux density B0 at right angles to the field. Find the flux density at all points in space, and sketch the logarithm of the ratio of the magnitudes of B on the cylinder axis to B0 as a function of log10 µr for a2 /b2 = 0.5, 0.1. Neglect end effects. We'll take the cylinder axis as the z axis of our coordinate system, and we'll take B0 along the x axis: B0 = B0ˆi. To the extent that we ignore end effects, we may imagine the fields to have no z dependence, so we effectively have a two dimensional problem. There are two distinct current distributions in this problem. The first is a current distribution Jfree giving rise to the uniform field B0 far away from the cylinder; this current distribution is only nonvanishing at points outside the cylinder. The second is a current distribution Jbound = ∇ × M existing only within the cylinder. Since there is no free current within the cylinder or in its inner region, the equations determining H in those regions are ∇ · B = ∇ · (µH) = 0,
∇ × H = Jfree = 0.
These imply that, within the cylinder and in its inner region, we may derive H from a scalar potential: H = −∇Φm , with Φm satisfying the Laplace equation. In the external region, there is free current, so things are not so simple. To proceed we may separate the H field in the external region into two components: one that arises from the free current, and one that arises from the bound currents within the cylinder. The former is just (1/µ0 )B0 and the second is again derivable from a scalar potential satisfying the Laplace equation. So, in the external region, H = (1/µ0 )B0 − ∇Φm .
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Homer Reid's Solutions to Jackson Problems: Chapter 5
So our task is to find expressions for Φm in the three regions such that the boundary conditions on B and H are satisfied at the borders of the regions. Writing down the solutions of the 2-D Laplace equation in the three regions, and excluding terms which blow up as ρ → 0 or ρ → ∞, we have P ∞ n n cos nφ + Bn sin nφ Pn=1 ρ n A o ∞ Φm (ρ, φ) = ρn Cn cos nφ + Dn sin nφ + ρ−n En cos nφ + Fn sin nφ n=1 P∞ ρ−n G cos nφ + H sin nφ n n n=1
r
Jackson Classical Electrodynamics Solutions 3.16
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