PHY 641: EM-1
Problems and Solutions
Gulmammad Mammadov
Physics Department, Syracuse University,
Syracuse, NY 13244-1130, USA
Email: gmammado@syr.edu or gulmammad@gmail.com
February 21, 2008
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Problem 6.18 (Jackson)
Consider the Dirac expression
A(x ) =
g
4π
Z
L
dI0 × (x − x 0 )
|x − x 0 |3
(1)
for a vector potential of a magnetic monopole and its associated string L.
Suppose for definiteness that the monopole is located at the origin and the
string along the negative z axis.
(a) Calculate A explicitly and show that in spherical coordinates it has
components
Ar = 0,
Aθ = 0,
Aϕ =
g(1 − cos θ)
g
θ
=
tan
4πr sin θ
4πr
2
(2)
(b) Verify that B = ∇ × A is the Coulomb-like field of a point charge, except
perhaps at θ = π.
(c) With the B determined in part b, evaluate the total magnetic flux
passing through the circular loop of radius R sin θ shown in the figure (see
in Jackson). Consider θ < π/2 and π/2 < θ separately, but always calculate
upward flux.
H
(d) From
A · dI around the loop, determine the total magnetic flux
through the loop. Compare the result with that found in part c. Show that
they are equal for 0 < θ < π/2, but have a constant difference for π/2 < θ < π.
Interpret this difference.
Solution:
Since the string is along the negative z axis, then x 0 being a position vector from
the origin to a given point on the string, has components (0, 0, −z) = (0, 0, −r0 ),
while x with |x | = r is a position vector from the origin to an observation point.
And also for an element of length on the string we have dI0 = dx 0 .
(a) From the figure we see that dI0 × (x − x 0 ) has non-zero component only in
the ϕ̂ direction, and so does A.
|x − x 0 | = [(r cos(π − θ) − r0 )2 + r2 sin2 (π − θ)]1/2 = [(r cos θ − r0 )2 + r2 sin2 θ]1/2
(3)
And
dI0 × (x − x 0 ) = dx 0 × (x − x 0 ) = dr0 |x − x 0 |
2
r sin θ
ϕ̂ = r sin θdr0 ϕ̂
|x − x 0 |
(4)
If we substitute (3) and (4) into (1), we get
Z ∞
Z ∞
r sin θdr0
g
ady
g
=
Aϕ =
4π 0 [(r cos θ − r0 )2 + r2 sin2 θ]3/2
4π −r cos θ [y 2 + a2 ]3/2
with a = r sin θ and y = r0 − r cos θ. The last integral gives
s
∞
g(1 − cos θ)
g
y 2 =
Aϕ =
2
2
4πa
y +a 4πr sin θ
(5)
(6)
−r cos θ
which is in agreement with (2).
(b) Using (2) we can compute components of B = ∇ × A in the spherical
polar coordinates (θ 6= π),
∂
∂Aθ
g
1
(Aϕ sin θ) −
=
(7)
Br =
r sin θ ∂θ
∂ϕ
4πr2
1 ∂
∂Ar
Bϕ =
(rAθ ) −
=0
(8)
r ∂r
∂θ
1
1 ∂Ar
∂
Bθ =
−
(rAϕ ) = 0
(9)
r sin θ ∂ϕ
∂r
Therefore we see that B is the Coulomb-like field of a point charge.
(c) If θ < π/2, the magnetic flux passing trough the circle with the radius
R sin θ is
Z
Z
Z 2π
Z θ
g
g
R2 sin θ0 dθ0 = (1 − cos θ) (10)
Φ1 = B · dS = Br dS =
dϕ
2
4πR
2
0
0
And if θ > π/2
Z
Z
Φ2 = B · dS = −
0
2π
Z
dϕ
θ
π
g
g
R2 sin θ0 dθ0 = − (1 + cos θ)
2
4πR
2
(11)
the minus sign comes because we consider upward flux. Then the total flux
g
g
Φ1 + (−Φ2 ) = (1 − cos θ) + (1 + cos θ) = g
(12)
2
2
(d) Since Aϕ is the only non-vanishing component of the vector potential, using
(6) we have
I
I
g(1 − cos θ)
g
Φ=
A · dI = Aϕ dI =
2πR sin θ = (1 − cos θ)
(13)
4πR sin θ
2
If we compare (13) to (10) and (11), we see that Φ = Φ1 while Φ = −Φ2 and
Φ − Φ2 = g. This shows that there is a flux coming up along the z axis from
the string.
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