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Solutions to Problem Set 10, Physics 370, Spring 2014
1
TOTAL POINTS POSSIBLE: 65 points.
1. Griffiths Problem 5.34 (Tweaked): Starting with equation 5.88 for
the magnetic field of a dipole:
ˆ
~ dip (~r) = µ0 m 2 cos θˆ
r
+
sin
θ
θ
B
4πr 3
show that the magnetic field of a dipole can be written in coordinatefree form:
~ dip (~r) = µ0 1 [3(m
B
~ · rˆ)ˆ
r − m]
~
4π r 3
HINT: This is a fairly straightforward problem if you recall that
ˆ θ.
ˆ Some people may find it helpful to work
m
~ = (m
~ · rˆ)ˆ
r + (m
~ · θ)
back from the desired result above back to equation 5.88. HINT #2:
You may find inspiration from re-examining Problem 3.36 on Problem
Set 6.
(10 points possible) The magnetic field of a “pure” magnetic dipole
which points along the z-axis is given by equation 5.88:
ˆ
~ dip (~r) = µ0 m 2 cos θˆ
B
r
+
sin
θ
θ
4πr 3
For a magnetic dipole along the z-axis, then we can envision the necessary dot products with this sketch
z
θ
m
rˆ
θˆ
y
where you should be able to see that
m
~ · rˆ = |m||ˆ
~ r| cos θ = m cos θ
(1)
where θ is the angle between m
~ and rˆ and also the angle between the
z axis and the position we are computing the magnetic field. For the
Solutions to Problem Set 10, Physics 370, Spring 2014
2
second dot product, I will use θ as the polar angle from the z axis, so
the angle between m
~ and θˆ is θ + π2 . As such
ˆ cos(θ + π ) = −m sin θ
m
~ · θˆ = |m||
~ θ|
2
(2)
where I used the trigonometric identity, cos(θ + π2 ) = − sin θ. Given
these dot products,the magnetic field can be expressed
µ0 ˆ
ˆ
~
B=
2(m
~ · rˆ)ˆ
r − (m
~ · θ)θ .
(3)
4πr 3
ˆ θˆ such that
As noted in my hint, we can write m
~ = (m
~ · rˆ)ˆ
r + (m
~ · θ)
ˆ θˆ = m−(
(m·
~ θ)
~
m·
~ rˆ)ˆ
r. Putting this into the expression for the magnetic
field from equation (3) gives
~ = µ0 (3(m
B
~ · rˆ)ˆ
r − m)
~
4πr 3
(4)
which is the desired result.
As I noted, I had previously assigned a very similar problem to this
one (Problem 3.36 of Griffiths), but involving the electric dipole staring
with equation 3.103:
~ dip (~r) =
E
p
ˆ
(2 cos θˆ
r + sin θθ).
4πǫ0 r 3
Notice these equations for the magnetic dipole field and electric dipole
field are identical except for the constant in front! So I was able to take
the same approach I used for Problem 3.36. Why is this important?
Just to show that a dipole field is a dipole field, whether electric or
magnetic, the description of the shape and strength of the field is the
same.
Solutions to Problem Set 10, Physics 370, Spring 2014
3
2. Griffiths Problem 5.35: A circular loop of wire, with radius R, lies
in the xy plane, centered at the origin, and carries a current I running
counterclockwise as viewed from the positive z axis.
(a) What is its magnetic dipole moment?
(b) What is the approximate magnetic field at points far from the
origin?
(c) Show that, for points on the z axis, your answer is consistent with
R2
the exact field B(z) = µ20 I (R2 +z
(equation 5.41) from Example
2 )3/2
5.6, when z ≫ R.
(a) (3 points possible) Given the circular loop is centered on the
origin, the area of the loop is πr 2 and the normal vector to the
area is in the zˆ direction, therefore via equation 5.86 the magnetic
dipole moment should be
m
~ = I~a = IπR2 zˆ
(5)
(b) (2 points possible) Since this magnetic dipole is oriented along
the z axis, the magnetic field far away from the magnetic dipole
is just given by equation 5.88
ˆ
~ dip (~r) = µ0 m (2 cos θˆ
r + sin θθ)
B
4πr 3
µ0 IπR2
ˆ
=
(2 cos θˆ
r + sin θθ)
4πr 3
µ0 IR2
ˆ
(2 cos θˆ
r + sin θθ)
=
4r 3
(6a)
(6b)
Just to be clear, the reason you as asked about the field “far
away” from the magnetic dipole is because equation 5.86 assumes
a “pure” dipole field, whereas near a current loop, the field will
look a little different. To see a visualization of this, examine Figure
5.55 on page 255 of your textbook.
4
Solutions to Problem Set 10, Physics 370, Spring 2014
(c) (5 points possible) On the z-axis, if z > 0, then θ = 0, r = z,
and rˆ = zˆ therefore equation 6b becomes:
2
µ0 IR2
~ z>0 (z) = µ0 IR (2ˆ
z
)
=
zˆ.
B
4z 3
2z 3
(7)
How about below the xy plane? For z < 0, we have θ = π, r = −z,
and rˆ = −ˆ
z therefore equation 6b becomes:
2
µ0 IR2
~ z<0 (z) = µ0 IR (2(−ˆ
z
))
=
zˆ.
B
4(−z)3
2z 3
(8)
So it turns out the same equation applies both above and below
the xy plane. So on the z axis, we simply have:
µ0 IR2
~
zˆ.
B(z)
=
2z 3
(9)
3/2
And in the limit where z ≫ R, then (R2 + z 2 )
→ z 3 , so this
expression is the same as equation 5.41 for z > 0.
Solutions to Problem Set 10, Physics 370, Spring 2014
5
3. Griffiths Problem 6.01 tweaked: Calculate the torque exerted on
the square loop shown in Fig. 6.6 (below), due to the circular loop
(assume r is much larger that a or b). If the square loop is free to
rotate, what will its equilibrium orientation be?
Figure 6.06
Clearly state, in English, which direction the square loop will turn in
response to this torque. HINT: There are several forms of the equation
for the magnetic field of a dipole given in the book. I would suggest
using the coordinate-free version in equation 5.89 in this problem to
reduce the complexity of the math.
(10 points possible) The torque on a current loop with dipole moment m
~ is given by equation 6.1
~ =m
~
N
~ × B.
(10)
In this case, if I declare up to be the z direction and right the y direction
(such that x is out of the page), I can use the dipole moment for the
square loop, m
~ sq = Ib2 yˆ for the dipole moment. The magnetic field the
square loop is experiencing is the dipole magnetic field of the circular
loop. As suggested in my hint, I will express this using equation 5.89:
~ dip (~r) = µ0 1 [3(m
B
~ · rˆ)ˆ
r − m]
~ ,
4π r 3
(11)
where the dipole moment of the circular loop is m
~ cir = Iπa2 zˆ and rˆ = yˆ
in this case, therefore equation 11 can be written:
~ dip (~r) = µ0 1 3(Iπa2 zˆ · yˆ)ˆ
y − Iπa2 zˆ
(12a)
B
3
4π r
µ0 I a2
=−
zˆ.
(12b)
4 r3
Solutions to Problem Set 10, Physics 370, Spring 2014
6
Using this result in the expression for torque in equation 10, we find
the torque on the square loop is:
2
µ
I
a
0
2
~ = Ib yˆ × −
N
zˆ
(13a)
4 r3
µ0 I 2 a2 b2
xˆ
(13b)
=−
4 r3
Since the torque is in the −ˆ
x direction, this implies (via the right hand
rule) that the square loop will want to turn “clockwise” as we see it.
Finally, given that the magnetic field due to the circular dipole loop
points downward at the location of the square loop, the equilbrium
orientation of the square dipole is for its dipole moment to point downward.
4. Griffiths Problem 6.07 (tweaked): An infinitely long circular cylin~ parallel to its axis. You’d like to
der carries a uniform magnetization M
~
find the magnetic field (due to M both inside and outside the cylinder).
~ b ) and volume (J~b ) currents for this
(a) Find the bound surface (K
cylinder.
~ b and J~b . Explain why
(b) Give the physical interpretation of both K
solving for these currents is critical to determining the magnetic
~ in the cylinder.
field produced by M
~ ) inside and outside the cylinder.
(c) Find the magnetic field (due to M
I’ll sort of answer part (b) here first, since all magnetic fields are created by currents, it is critical we determine the bound currents due
~ in order to solve for the magnetic field created by the uniformly
to M
magnetized object.
(a) (3 points possible) The bound surface current is given by equa~b = M
~ ×n
tion 6.14, K
ˆ , where in the case of the cylinder, the
tangential unit vector at the surface is radial, n
ˆ = sˆ. Since we are
~ is parallel to the axis of the cylinder
told the magnetization M
~ = M zˆ, we know:
such that M
ˆ
~b = M
~ ×n
K
ˆ = M zˆ × rˆ = M φ.
(14)
Solutions to Problem Set 10, Physics 370, Spring 2014
7
~ ×
The bound volume current is given by equation 6.13, J~ = ∇
~ , such that in cylindrical coordinates (noting the only non-zero
M
component of the magnetization is Mz = M):
~ ×M
~
J~b = ∇
(15a)
1 ∂Mz
∂Ms ∂Mz ˆ 1 ∂
∂Mφ
∂Ms
=
sˆ +
zˆ
−
−
φ+
(sMφ ) −
s ∂φ
∂z
∂z
∂s
s ∂s
∂φ
(15b)
J~b = 0.
(15c)
So all we have here is a circumferential bound surface current
which will be generating the magnetic field.
~ is the magnetic dipole per
(b) (4 points possible) Recall that M
unit volume. Magnetic dipoles are the result of current loops (a
la Figure 6.15 on p. 277 of Griffiths). The bound surface current
can be taken to be an ’edge’ effect of all the little current loops
making up the magnetic dipoles in an object. Adjoining current
loops normally cancel out, but at the edges of the object, they
don’t. The net result is a bound surface current. The bound volume current, like bound volume charge density, is the result of
a variation in the magnetization within the material (for bound
volume charge density, its variation in polarizations that lead to
bound volume charge). Since this cylinder has uniform magnteization, there should not be any bound volume current. Finally, since
all currents create magnetic fields, we can take these bound currents and compute the magnetic field they create. This is, in fact,
the only “easy” way to compute the magnetic field of a magnetized
object.
(c) (3 points possible) Recall this is just a circumferential bound
surface current creating the magnetic field. We’ve really seen this
before, in the case of a solenoid, which is just a circumferential current traveling about a cylinder. Therefore the magnetic field due
to this magnetized cylinder should be the same as the magnetic
field of a solenoid (via equation 5.59):
µ0 nI zˆ = µ0 K zˆ = µ0 M zˆ, inside cylinder
~
B=
(16)
0, outside cylinder
Solutions to Problem Set 10, Physics 370, Spring 2014
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5. Griffiths Problem 6.09 tweaked: A short circular cylinder of radius
~ parallel
a and length L carries a “frozen-in” uniform magnetization M
to its axis. Find the bound current, and sketch the magnetic field of
the cylinder. (Make three sketches: one for L ≫ a, one for L ≪ a, and
for L ≈ a.) HINT: Very few explicit calculations are involved here.
This is mostly a conceptual problem.
(5 points possible) The bound currents in this problem as the same
as in Problem 6.7 because we again have a cylinder magnetized parallel
~ b = M φˆ
to its axis. So all we have is a circumferential bound current, K
with a magnetic field that looks like that of a solenoid.
For the case where L ≫ a, we essentially have a solenoid of finite
length, therefore:
In the case where L ≪ a, we essentially get
something that looks like a current loop, so
the magnetic field should look like that of a
magnetic dipole (as shown to the right).
Finally, in the case where L ≈ a, we have an intermediate case, so we expect the fields should
look like something “in between” the two previous examples (as shown in the figure to the
left).
[NOTE: All the figures here are copied from the
Instructor’s Solution Manual which is Copyright
1999 Prentice-Hall.]
Solutions to Problem Set 10, Physics 370, Spring 2014
9
6. Griffiths Problem 6.10: An iron rod of length L and square
cross section (side a), is given a uniform longitudinal magneti~ , and then bent around into a circle with a narrow gap
zation M
(width w), as shown in Fig. 6.14 (below). Find the magnetic field
at the center of the gap, assuming w ≪ a ≪ L. [Hint: Treat it
as the superposition of a complete torus plus a square loop with
reversed current. You determined the magnetic field in a square
current loop in Problem 5.8 on Problem Set 8.]
Figure 6.14
VERY SUBTLE HINT AND A SHOT AT EXTRA CREDIT:
You might be inclined to state M is uniform and therefore there
~ = M φˆ
was no bound current. This is incorrect because while M
is uniform in magnitude, it is not uniform in direction since it is
circumferential. As such, it is not guaranteed to have no bound
current, and in fact, if you compute the curl of the magnetization, it is NOT zero!1 However, I had a discussion about this
with David Griffiths via email and it turns out that if you bend
a rod around to form a torus, that will make M a function of
radius and M will vary in precisely the way necessary to lead to
~ ×M
~ = J~b = 0. If you feel like taking a shot at extra credit,
∇
~ will vary in precisely this way. Only hint I will
explain why M
give is think of what would happen to mass density in this situation. Otherwise, you can, without justification, feel free to cite
Dr. Griffiths himself on this one!
ˆ then the bound volume current should be (using curl in
~ = M φ,
If magnetization M
cylindrical coordinates):
1
~ ×M
~ = − ∂ Mφ sˆ − 1 ∂ (sMφ )ˆ
z.
J~ = ∇
∂z
s ∂s
M
This means if M is constant that J~ = − sφ zˆ.
(17)
10
Solutions to Problem Set 10, Physics 370, Spring 2014
(10 points possible) I start by computing the bound currents.
If I assume the torus is centered at the origin, then using cylinˆ
~ = M φ,
drical coordinates I can take the magnetization to be M
then n
ˆ is either zˆ on the top/bottom of the torus, and sˆ in the
inner/outer sides of the torus. So the bound surface currents are
(via equation 6.14):
~b = M
~ ×n
K
ˆ
~ top = M φˆ × zˆ = M sˆ
K
b
bottom
~
K
b
(18a)
(18b)
= M φˆ × (−ˆ
z ) = −M sˆ
= M φˆ × (−ˆ
s) = M zˆ
(18c)
~ binner
K
~ outer = M φˆ × sˆ = −M zˆ
K
b
(18d)
(18e)
(Note: I don’t need to worry about the faces of the gaps since
~ ×n
there n
ˆ = ±φˆ so M
ˆ = 0). So, in essence, we have a bound
surface current that is moving up the inner surface, across the top
surface, down the outer surface, and across the bottom surface,
forming a toroidal bound current loop like the torus in Example
5.10 (p. 238-239).
~ ×M
~ = J~b = 0 is fine, but here
Citing Dr. Griffiths as evidence ∇
is why M varies in precisely the way necessary to have no curl.
[Start of Extra Credit Digression] If a real (initially uniform
density) iron rod is bent into a torus as suggested, it is easy to
see that the matter density will become inversely proportional to
the radius, such that if we take s0 to be the average radius of the
toroid and ρ0 to be the average mass density, then the total mass
in some segment of the toroid, covering angle θ, will be:
ρ(s)asdθds = ρ0 as0 dθds → ρ(s) = ρ0
s0
.
s
(19)
Since magnetization M is a measure of magnetic dipole moment
per volume, you can imagine that it will behave like mass per
volume, such that:
M(s) = M0
s0
.
s
(20)
Solutions to Problem Set 10, Physics 370, Spring 2014
11
~ outer must be lower
This does mean the surface current density K
b
~ inner , but since the surface area
than the surface current density K
b
of the inner/outer surfaces is A = 2πas, then the total bound
current through those surfaces will be equal:
Iinner = Kbinner A = M0
Iouter
s0
2πasinner = M0 s0 2πa
sinner
s0
= Kbouter A = M0
2πasouter = M0 s0 2πa
souter
(21a)
(21b)
~ = M φˆ as Griffiths actually states in the probIf you assume M
lem, then the smaller radius of the inner surface means:
Iinner = Kbinner A = M(2πasinner )
Iouter = Kbouter A = M(2πasouter )
(22a)
(22b)
which means the total current on these two surfaces are not equal!
Where’d the missing current go? It goes into the now non-zero
bound current J~b in this scenario. [End of Extra Credit Digression] [Thanks to Dr. Griffiths himself for helping me resolve
this subtlety via email.]
Another approach you could take, if you can’t buy Dr. Griffith’s
argument (or didn’t know about it without the hint I gave) is
to state because L ≫ a then the difference in the magnetization
across the object will be small enough to safely ignore the variation in bound surface currents and to assume the bound volume
current is zero.
Assuming only bound surface currents, we have a toroidal current
and we can compute the magnetic field inside a complete torus
(without a gap) using Amp`ere’s Law and a circumferential circular Amp´erian loop running through the inside of the torus (inside
the current loop) such that:
I
~ · d~ℓ = µ0 Ienc
B
(23a)
B(2πs) = µ0 (K(2πs))
~ torus = µ0 K φˆ = µ0 M
~.
B
(23b)
(23c)
Solutions to Problem Set 10, Physics 370, Spring 2014
12
Having determined the magnetic field in a complete torus given
by equation 23c, we can now use the hint. We can consider the
magnetic field in the gap, where there is no current, as the sum
of the magnetic field of a complete torus with current running in
one direction and a small square loop with the same dimensions
as the gap, but the current running in the opposite direction, so
the sum of the two currents is zero. From our previous solution
to Problem 5.8 (which was assigned on Problem Set 8) and the
orientation of the square loop’s axis with the φˆ direction, I have
the magnetic field
√
2µ0 I ˆ
~
Bsqr = −
φ
(24)
πR
(the minus sign is from the direction of the current). Here R =
a/2 and I = Kw = Mw and therefore:
√
√
2
2µ
Mw
2µ0 w ~
2
0
~ sqr = −
B
M
(25)
φˆ = −
πa
πa
So the net magnetic field in the gap should be:
"
√ #
√
2µ
2w
2
2
w
0
~ = µ0 M
~ 1−
~ gap = B
~ torus + B
~ sqr = µ0 M
~ −
. (26)
M
B
πa
πa
Solutions to Problem Set 10, Physics 370, Spring 2014
13
7. Griffiths Problem 6.12: An infinitely long cylinder, of radius
R, carries a “frozen-in” magnetization, parallel to the axis,
~ = ksˆ
M
z
(27)
where k is a constant and s is the distance from the axis; there
is no free current anywhere. Find the magnetic field inside and
outside the cylinder by two different methods:
(a) As in Section 6.2, locate all the bound currents, and calculate the
field they produce.
~ and
(b) Use Amp`eres law (in the form of Equation 6.20) to find H,
1
~ −M
~ (Equation 6.18). (Notice that the
~ from H
~ ≡ B
then get B
µ0
second method is much faster, and avoids any explicit reference
to the bound currents.)
(a) (6 points possible) The first method is to find the bound currents and then calculate the magnetic field they produce using
Amp`ere’s law. The bound volume current is found using equation
6.13 (working in cylindrical coordinates) :
~ ×M
~
J~b = ∇
(28a)
∂Ms ∂Mz ˆ 1 ∂
∂Mφ
∂Ms
1 ∂Mz
sˆ +
zˆ
φ+
−
−
(sMφ ) −
=
s ∂φ
∂z
∂z
∂s
s ∂s
∂φ
(28b)
∂Mz ˆ
ˆ
J~b = −
φ = −k φ.
(28c)
∂s
The bound surface current is found using equation 6.14 (noting
the surface is radial such that n
ˆ = sˆ)
ˆ
~b = M
~ ×n
K
ˆ = kR(ˆ
z × sˆ) = kRφ.
(29)
Since both bound currents are in the φ direction, the magnetic
field will be along the axis inside the cylinder. Furthermore, the
total bound current must be zero. Since these bound currents run
Solutions to Problem Set 10, Physics 370, Spring 2014
14
along the same axes, but in opposite directions, the total magnetic
~ out = 0. Applying Amp`ere’s
field outside the cylinder must be B
H
~ · d~ℓ = Bℓ
law using the loop shown in the figure below gives B
~ b.
while the current enclosed is due to both J~b and K
(This Figure is copied from the Instructor’s Solution Manual
and Copyright 1999 Prentice-Hall.)
Note that the integration is done clockwise around the loop so
that positive area points in the φˆ direction, into the page. The
current enclosed due to the bound volume current is
Z
Z R
IJ = J~b · d~a =
(−k)ℓds′ = −kℓ(R − s).
(30)
s
The current enclosed due to the bound surface current density is
Z
~ b · d~ℓ = Kb ℓ = kRℓ.
IK = K
(31)
The total current enclosed is
I = IJ + IK = kℓs;
(32)
Note that the contribution of the bound surface current K partially cancels the bound volume current J. Amp`ere’s law therefore
gives
I
~ · d~ℓ = µ0 Ienc
B
(33a)
B(ℓ) = µ0 (kℓs)
µ0 I
~ in = µ0 ksˆ
B
z=
zˆ;
ℓ
(33b)
(33c)
Solutions to Problem Set 10, Physics 370, Spring 2014
15
where I used the the fact that when B is positive the magnetic
field points in the same direction as the direction of integration in
the loop (the portion in the B field), which is +ˆ
z.
(b) (4 points possible) A much easier way to determine the field is
~ According to Amp`ere’s law in materials this auxiliary
to use H.
~ is determined by the free current via equation 6.12,
field H
I
~ · d~ℓ = If, enc .
H
(34)
~ = 0.
Because there is no free current here, H
We obtain the magnetic field from the definition of the auxiliary
field:
~ ≡ 1B
~ −M
~ →B
~ = µ 0 (H
~ + M)
~ = µ0 M.
~
H
(35)
µ0
Inside the cylinder the magnetization is given by equation 27, so
that
~ in = µ0 ksˆ
B
z,
(36)
the same result we obtained above. Outside the field the magne~ = 0 so the magnetic field B
~ out = 0 also.
tization M
[Thanks to Dr. Craig for providing the initial draft of this solution
in LATEX format.]