TUTORIAL 5
Problem 1
Consider the loop consisting of radial lines and segments of circles with center
P.
~ at P in terms of a, b, θ and I.
Determine the magnetic field H
Proof. Use Bio-Savart Law, for a current element this implies
~
Id~l × R
4πR3
As this is a two-dimensional problem and there is polar symmetry, we work with
polar coordinates with the origin at P.
~ = ρ~aρ and Id~l = Id~l = Idρ~aρ , hence
Along φ = 0 and φ = θ, R
~ =
dH
~ = 0.
Id~l × R
~ at P due to the radial segments.
There is no contribution to H
Next consider the circular segments. For r = a, Id~l = −Iadφ~aφ where the negative sign arises from the current direction moving in a counter-clockwise manner.
~ = a~aρ then
Setting R
~1 =
dH
−Ia2 dφ(~aφ × ~aρ )
4πa2
integrating this we have
~1 = − I
H
4πa
Z
θ
dφ(−~az ) =
0
Iθ
~az
4πa
Repeating this for r = b, with Id~l = Ibdφ~aφ and R = b~aρ then
H2 = −
Iθ
~az
4πb
Adding these two together we find
~ =H
~1 + H
~ 2 = Iθ(b − a) ~az
H
4πab
1
2
TUTORIAL 5
Problem 2
Given the magnetic field in free space
~ = xy 2~ax + x2 z~ay − y 2 z~az A/m
H
Determine
a) The current density at P (2, −1, 3)
b) The time rate of change of the electric charge at P .
~
Proof. a) Use the Maxwell equation J~ = ∇ × H
~ax
~ay
~az
~ = ∂x
∂
∂z
H
y
xy 2 x2 z −y 2 z
= −(x2 + 2yz)~ax + 2x(z − y)~az
At P (2, −1, 3) then
J~ = −2~ax + 16~az .
b) Using the Continuity equation
∂t ρv = −∇ · J~ = −∂x Jx − ∂y Jy − ∂z Jz = 2x − 2x = 0
Problem 3
An infinitely long cylinder of radius a and permeability µ is placed such that its
axis coincides with the z-axis. The cylinder carries a current I uniformly distributed
across it. The current is in the z-direction as well.
Find
a) Magnetization inside the cylinder.
b) Volume magnetization current density inside the cylinder.
Proof. a) From the symmetry of the problem, we assume the direction of the magnetic field to be azimuthal (i.e. proportional to ~aφ ). Apply Ampère’s Law in
H
~ · d~l = Ienc to any circular Ampèrian path of radius ρ ≤ a
the integral form L H
then
I
~ = Iρ ~aφ
Hφ × 2πρ =
× πρ2 → H
2
πa
2πa2
µIρ
~ = µH
~ =
Therefore B
aφ and the magnetization is
2πa2 ~
~
B
µ
Iρ
~
~
M=
−H =
−1
~aφ
µ0
µ0
2πa2
b) The magnetization current density is then
~aρ
ρ~aφ
~az
1
∂
∂
∂z
~ =
J~m = ∇ × M
φ
ρ
ρ
Iρ2 µµ −1
0
0
0
2πa2


µ
2
1  Iρ µ0 − 1 
=
~az
ρ
2πa2
,ρ
TUTORIAL 5
3
and so
I
J~m =
πa2
µ
− 1 ~az
µ0
Problem 4
~ depends only on the distance from
A circularly symmetrical magnetic field (B
the axis) pointing perpendicular to the page occupies the shaded region
R
~ a) is zero, show that a charged particle that starts at the center
If the total flux ( Bd~
will emerge from the field region on a radial path (assuming it has the velocity to
escape at all).
Proof. The angular momentum acquired by the particle as it moves out from the
center to the edge of the field is
Z ~
Z
Z
dL
~ =
L
dt − T~ dt = (~r × F~ )dt
dt
Z
Z
~
~
=
~r × q(~v × B)dt
= q ~r × (d~l × B)
Z
Z
~
~
~ r · d~l)]
= q[ (~r · B)dl − B(~
~ and so ~r · B
~ = 0 and ~r · d~l = ~r · d~r = 1 d(~r · ~r) = 1 dr2 =
As ~r is perpendicular to B
2
2
1
rdr = 2π (2πrdr) and so if we say the max radius of the field is R, then
Z R
Z
−q
~ =− q
~
~ a
L
B2πrdr
=−
Bd~
2π 0
2π
R
~ = B
~ · d~a then
Writing the total flux as Φ
~ =− q Φ
~
L
2π
4
TUTORIAL 5
~ = 0 then L
~ = 0 and the charge emerges with zero angular momentum so it is
If Φ
moving along a radial line.
Problem 5
~ vanishes and we can express H
~ as the
If J~f = 0 everywhere, the curl of H
gradient of a scalar potential W :
~ = −∇W.
H
~ = −∇ · M now gives
The equation ∇ · H
∇2 W = −∇ · M,
and W obets Poisson’s equation with −∇ · M as the ”source”. Use separation of
variables to determine the field inside a uniformly magnetized sphere.
Proof. In this context separation of variables ensures we may write the potential
on the interior and exterior of the sphere as the series
k
= ∇∞
k=1 Ak r Pk (cos θ), r > R
Bk
Wout (r, θ) = ∇∞
k=1 k+1 Pk (cos θ), r < R
r
To get conditions on the form of these series, we impose boundary conditions. In
this case, continuity of W follows from the gradient theorem
Z ~b
Z ~b
~ · d~l
W (~b) − W (~a) =
∇W · d~l = −
H
Win (r, θ)
~
a
~
a
and so the following conditions are sensible:
(1) Win (R, θ) = Wout (R, θ)
in
~ r = M cosθ when r = R.
(2) − ∂W∂rout + ∂W
∂r = M · ~
As the sphere is magnetized we can choose that the z axis is aligned with it, and so
~ = M~az in the second boundary condition. Here, ~r denotes the position vector
M
and so θ represents the angle between the z-axis and ~r,
The first condition implies
Bk
→ Bk = R2k+1 Ak
Rk+1
and substituting this into the second boundary condition we find
X
X
Bk
M cos θ =
(k + 1) k+2 Pk (cos θ) +
kAk Rk−2 Pk (cos θ)
R
X
=
(2k + 1)Rk−1 Ak Pk (cos θ)
Ak R k =
this implies that Ak = 0 if k 6= 1 and 3A1 = M → A1 =
of the form
M
M
Win (r, θ) =
r cos θ =
z
3
3
~ in = −∇Win
from this we can compute H
~ in = −M ~az = − 1 M
~
H
3
3
M
3 .
The potential is now
TUTORIAL 5
5
and finally the field on the interior of a uniformly magnetized sphere is
~ +M
~ ) = µ0 (H
~ +M
~ ) = µ0 − 1 M
~ +M
~ = 2 µ0 M
~
B = µ0 (H
3
3
Problem 6
A sphere of linear magnetic material is placed in an otherwise uniform magnetic
~ 0 . Find the new field inside the sphere using separation of variables.
field B
~ θ) → B
~0 = B
~ 0~az and so H
~ =
Proof. For large r we would like B(r,
and W → − −1
µ0 B0 r cos θ. Choosing the potentials:
1 ~
µ0 B
=
B0
az
µ0 ~
k
∇∞
k=1 Ak r Pk (cos θ), r > R
B0
Bk
Wout (r, θ) = − r cos θ + ∇∞
k=1 k+1 Pk (cos θ), r < R
µ0
r
the boundary conditions in this case are
(1) Win (R, θ) = Wout (R, θ)
in
(2) −µ0 ∂W∂rout + µ ∂W
∂r = 0 when r = R.
Expanding the second condition this is
X
X
Bk
B0
kAk Rk−1 Pk (cos θ) = 0
µ0
cos θ +
(k + 1) k+2 Pk (cos θ) + µ
µ0
R
Win (r, θ)
=
For k 6= 1, the first condition implies Bk = R2k+1 Ak and so
[µ0 (k + 1) + µk]Ak Rk−1 = 0 → Ak = 0.
For k = 1 we find from the first condition
B0 R B1
A1 R = −
+ 2
µ0
R
the second condition gives
B1
B0 + 2µ0 3 + µA1 = 0
R
combining these equations, we can write A1 as
3B0
A1 = −
.
2µ0 + µ
The potential on the interior is thus
−3B0 z
−3B0
Win (r, θ) =
r cos θ =
2µ0 + µ
2µ0 + µ
~ in = −∇Win :
and so H
~ in = − 3B0 ~az = −3 B
~ 0.
H
2µ0 + µ
2µ0 + µ
~ is
Putting this together, B
~
~ = µH
~ = 3µB0 =
B
2µ0 + µ
1 + χm
1 + χ3m
~0
B