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Phys 3327 Fall ’11
Solutions 2
Solutions: Homework 2
Ex 2.1: Particle moving in Magnetic Field
(a) If the initial velocity is perpendicular to the magnetic field B, the magnitude of the Lorentz Force (Eq
1.52) is
F =
1
|q|
|qu × B| →
uB
c
c
(1)
The direction of this force is perpendicular to the plane containing u and B. Thus the magnetic force
does no work on the particle, and its speed u = |u| and kinetic energy 1/2mu2 remain constant. Since
the force is always perpendicular to the velocity, the particle moves in a circle, and its acceleration has
the centripetal form u2 /R. Newton’s equation of motion is
u2
|q|
uB = m
c
R
(2)
The radius of the circular orbit is thus
mcu
|q|B
(3)
|q|B
u
=
R
mc
(4)
R=
(b) The angular velocity has the magnitude
ω=
The sense of gyration is such that for a positive particle the vector angular velocity ω is antiparallel to
B; hence,
ω=−
q
B
mc
(5)
(c) Any component of initial velocity parallel to B produces no net force and therefor remains constant.
This general motion has the form of a helix.
Source: Heald & Marion Solutions Manual
Ex. 2.2: Higher-Order Mulitpole Moments
(a) Consider a system of charges defined by a density ρ(r). (This may include delta functions for lowerdimensional or discrete distributions.) The dipole moment of this distribution is by definition
Z
p = d3 rρ(r)r ,
(6)
where we have implicity defined a coordinate system for r with origin of course r = 0.
A change of coordinates may be implemented by a simple shift r → r ′ = r − a, so that the new
coordinate system has its origin at r = a in the old coordinates: see the figure. The transformed
density is ρ′ (r ′ ) = ρ(r), so that explicitly e.g. ρ′ (0) = ρ(a). Note also d3 r ′ = d3 r.
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Phys 3327 Fall ’11
Solutions 2
In this new coordinate system, the transformed dipole moment is
Z
p′ = d3 r ′ ρ′ (r ′ )r ′
Z
= d3 rρ(r)(r − a)
Z
= p − a d3 rρ(r) .
(7)
R
But, clearly d3 rρ(r) = q, the total charge. For zero total charge, it follows that p′ = p, so dipole
moment is independent of choice of origin.
(b) For Q 6= 0, we have
p′ = p − qa .
(8)
Clearly, if we choose a = (1/q)p, then p′ = 0.
(c) The quadrupole tensor has components
Qij =
Z
ρ(r) 3ri rj − r2 δij d3 r ,
(9)
where we write the coordinates of r as (r1 , r2 , r3 ). Under the same coordinate shift r → r ′ = r − a as
above, we have
Z
′
Qij = ρ′ (r ′ ) 3ri′ rj′ − (r′ )2 δij d3 r ′
Z
= ρ(r) 3(ri − ai )(rj − aj ) − (r − a)2 δij d3 r
Z
= Qij + ρ(r) 3(−ri aj − rj ai + ai aj ) + (2r · a + a2 )δij d3 r
= Qij − 3(pi aj + pj ai ) + 2p · aδij + (3ai aj − a2 δij )q .
(10)
For p = 0 and q = 0, then Q′ij = Qij as required.
(d) Suppose the net charge q = 0, but p 6= 0. Then we have the relation
Q′ij = Qij − 3(pi aj + pj ai ) + 2p · aδij .
(11)
The tensor Q is symmetric and traceless, so in general it has five independent components. Since a
has only three independent components ai , from (11) we see that we can only set three of the five
independent components of Q to zero. Hence we cannot set Q′ = 0 under choice of origin alone.
Ex. 2.3: Multipole Expansion
2
Phys 3327 Fall ’11
Solutions 2
(a) From simple application of superposition, we have
2e
e
−
,
|r1 | |r1 − 1|
2e
e
Φ(r2 ) =
− 2
.
|r2 | (r2 + 1)1/2
Φ(r1 ) =
(b)
(12)
(i) The monopole moment is the net charge
q=
X
qα = e .
(13)
α
(ii) For a discrete distribution, the dipole moment is
p=
X
qα rα = −eex .
(14)
α
(iii) Finally, the quadrupole tensor has components
Qij =
X
qα (3xα,i xα,j − rα2 δij ) .
(15)
α
Since the charge distribution has symmetry about the x-axis, we expect that Q should be diagonal,
and that Q22 = Q33 (Note we use subscripts 1, 2 and 3 to denote the x, y and z coordinates
respectively). We can check this by directly calculating all the components of Q, which is simple
to do for this system. We have in matrix notation


−2e 0 0
Q =  0 e 0
(16)
0 0 e
as expected.
(c) In this case, the first three terms of the multipole expansion are
3xi xj − r2 δij
1X
q p · er
Qij
+
Φ(r) = +
r
r2
6 ij
r5
=
q p · er
1 Q11 2
+
(x − y 2 /2 − z 2 /2) ,
+
r
r2
2 r5
(17)
where r = (x, y, z). Hence, for r1 = r1 ex and r2 = r2 ey ,
e
e
e
−
,
−
|r1 | |r1 |2
|r1 |3
e
e
Φ(r2 ) =
.
+
|r2 | 2|r2 |3
Φ(r1 ) =
(18)
Clearly these are different from the potentials in (12): these potentials are expansions of the latter in
powers of (1/r). Another way to say this is that the potentials (18) converge asymptotically to those
in (12). A brief check of this is to note that in the limits r1,2 → ∞, we have from (12) Φ(r1 ) → e/|r1 |,
and Φ(r2 ) → e/|r2 |, which are the monopole terms. More carefully, note that for r1 > 1 we have a
geometric series
2e
1
e
e
1
1
2e
=
1+
−
−
+ 2 + ... ,
(19)
Φ(r1 ) =
r1
r1 1 − 1/r1
r1
r1
r1
r1
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Phys 3327 Fall ’11
Solutions 2
which matches the asymptotic expression above. Similarly, finding the Taylor expansion for Φ(r2 ) in
powers of 1/r2 directly yields
2e
e
1
p
Φ(r2 ) =
−
r2
r2
1 + (1/r2 )2
1
e
2e
1 − 2 + ... ,
−
(20)
=
r2
r2
2r2
as above.
Ex. 2.4: Charged Ring
a) It’s just a step function:
b) By definition of the density ρl = ρl (θ), we have dq = ρl (θ)adθ, and hence
Z 2π
Qij =
dθ ρl (θ)a 3xi xj − r2 δij .
(21)
0
Note that x3 = 0 for all the contributions, and (x1 , x2 ) = (a cos θ, a sin θ).
c) Since x3 = 0, it follows that Q32 , Q31 , Q23 , Q13 = 0. Further, Q33 is zero since
Z 2π
dθρl (θ)a2 = 0 .
Q33 = −
(22)
0
The 11 and 22 components are both zero because the integrand contains the step function multiplied
by an always-positive function of twice the periodicity. That is,
Z 2π
Q11 =
dθρl (θ)a3 3 cos2 θ − 1
0
Z
π/2
π/2
= 2a3
Z
π/2
= 6a3
Z
= 2a
3
dθρl (θ) 3 cos(2θ) + 2 + 2a
0
0
0
= 0.
3
Z
π
dθρl (θ) 3 cos(2θ) + 2
π/2
dθρl (θ) 3 cos(2θ) + 2 + 2a3 λ
Z
0
dθ ρl (θ) − ρl (θ) 3 cos(2θ) + 2
Since Q is traceless, then clearly Q22 = 0, too.
4
π/2
ρl (θ + π/2)dθ 3 cos(2θ + π) + 2
(23)
Phys 3327 Fall ’11
Solutions 2
Only the 12 elements are nonzero:
Q12 = Q21 =
Z
2π
dθ ρl (θ)a3 (3 cos θ sin θ)
(24)
0
3a3
=
2
Z
2π
dθ ρl (θ) (sin 2θ)
(25)
0
3a3
4λ
2
= 6a3 λ
=
Z
π/2
dθ (sin 2θ)
(26)
0
(27)
Hence the quadrupole moment is
Qαβ


0 6a3 λ 0
=  6a3 λ 0 0  .
0
0 0
(28)
Ex. 2.5: Force on a Magnetic Dipole
(a) Consider a magnetic dipole m in a uniform magnetic field B. The dipole moment by definition is
Z
1
dv ′ [r ′ × J(r ′ )] ,
(29)
m=
c V
and we showed in class that
Z
dv ′ J(r ′ ) = 0 .
V
5
(30)
Phys 3327 Fall ’11
Solutions 2
The Lorentz force acting on such an ensemble J(r ′ ) is simply
Z
Z
1
1
F =
dv ′ J(r ′ ) × B = 0
dv ′ J(r ′ ) × B =
c V
c
V
(31)
since B is uniform.
(b) A magnetic dipole moment produces a magnetic field which is not uniform: it behaves as ∼ 1/r3 at
large r. A second dipole moment will therefore experience a net force due to this field, which is the
interaction force between the dipoles.
An alternative way:
(a) Consider a magnetic dipole m and a magnetic field B. The torque acting on this dipole is (by definition)
τ = m × B = −mB sin(θ)em×B ,
(32)
where θ is defined as the angle from B to m, whence the minus sign. Noting θ is the canonical variable,
the potential energy of this configuration is found by rotating m such that
U (θ) − U (0) = −
Z
θ
τ dθ
0
= −mB(cos θ − 1)
= −m · B(θ) + m · B(0)
(33)
so we choose the gauge U = −m · B. It follows that the net force on the dipole
F = ∇(m · B) .
This a general result for the force on a dipole in any magnetic field.
6
(34)