Homework Assignment 8 Solution Set
PHYCS 4420
22 March, 2004
Problem 1 (Griffiths 6.12(b))
~ = χm H
~ because
~ = ksẑ we cannot say that M
For frozen-in magnitization M
~ is not due to H.
~ However, we can still write
M
I
~ · d~l = If reeencl. = 0
H
~ = 0.
=⇒ H
Thus,
~ = µ0 H
~ + µ0 M
~
B
~
=⇒ Binside = µ0 ksẑ.
Outside the cylinder the field is zero because the magnetization there is zero.
This is pretty easy to understand if you compare the bound currents that are
produced to this magnetization to the currents in the long solenoids we’ve already seen.
Problem 2 (Griffiths 6.13(b) & (c))
~ must be continuous. So, inside a long,
b The parallel component of H
skinny cavity we have
~ =H
~0
H
and
~ = µ0 H
~0 = B
~ 0 − µ0 M
~.
B
~ is continuous. So, inside a wide, flat
c The perpendicular component of B
cavity we have
~ =B
~0
B
and
~ = 1B
~0 = H
~0 + M
~.
H
µ0
1
Problem 3 (Griffiths 6.16)
Let the current on the inner cylinder flow in the ẑ direction (in cylindrical
~ in the region between the conductors.
coordinates). We start by finding H
I
~ · d~l = If reeencl.
H
~ = I θ̂.
=⇒ H
2πr
Thus,
~ = µH
~ = µ0 (1 + χm ) I θ̂.
B
2πr
~ = χm H
~ =
Now, where are the bound currents? We know that M
is no volume current density since
χm I
2πr θ̂.
There
~ ×M
~ =0
∇
everywhere between the conductors. However, at the inner conductor we have
a bound surface current density
~b = M
~ × n̂
K
r=a
χm I
(θ̂ × (−r̂))
=
2πa
χm I
ẑ.
=
2πa
mI
The bound current, then, is just Ib = χ2πa
ẑ2πa = χm I. The total current
enclosed by an Amperian loop anywhere between the two conductors is the sum
of the free current and the bound current. Therefore, we can write
I
~ · d~l = µ0 Itotal
B
~ = µ0 I + χm I θ̂
=⇒ B
2πr
µI
=
θ̂
2πr
~ =
which certainly agrees with our previous result (i.e., from this field we get H
I
2πr θ̂).
Problem 4 (Griffiths 6.17)
~ since it depends only on the free
As always, we use ampere’s law to find H
current. Inside the cylinder we have (with the current flowing in the ẑ direction,
as usual)
2
~ = I r θ̂
H
2πr a2
Ir
=
θ̂
2πa2
2
and outside we have
~ = I θ̂.
H
2πr
Thus,
~ in = µ0 (1 + χm ) Ir θ̂
B
2πa2
I
~ out = µ0
B
.
2πr
~ = χm H
~ = χm Ir 2 θ̂. So, in this
The magnetization inside the cylinder is just M
2πa
case there is a bound volume current density
~ ×M
~
J~b = ∇
1
∂Mθ
r
ẑ
=
r
∂r
χm I
=
ẑ
πa2
as well as a bound surface current density at the edge of the cylinder (r = a)
~b = M
~ × n̂
K
r=a
I = χm
θ̂ × r̂
2πa
χm I
=
(−ẑ).
2πa
Clearly these bound currents cancel each other out, leaving the total current
as just the original free current I. This must be true, since the field outside the
cylinder is the same as that for an ideal, one-dimensional current.
Problem 5 (Griffiths 6.26 - optional)
The components of the magnetic field lines that are perpendicular to the
interface must be continuous. Parallel to the interface the magnetic field lines
are discontinuous by an amount proportional to the total bound surface current
~ field, which depends only on free current, must
density there. However, the H
be continuous parallel to the interface. Thus,
tan θ2
=
tan θ1
B2k
B2⊥
B1k
B1⊥
µ2 H2k
µ1 H1k
µ2
=
.
µ1
=
3
Problem 6 (Griffiths 7.8)
~ = µ0 I . Thus, if we define out of the page
a The field from a long wire is B
2πr
to be the positive direction the flux through the square loop in Figure 7.17 is
Z s+a
µ0 I
ΦB =
adr
2πr
s
µ0 Ia s + a
=
ln
.
2π
s
b
∂ΦB
∂t
∂ΦB ∂s
=
∂s ∂t
µ0 I
a2
=
v.
2π (s + a)s
ε=
The flux is positive (out of the page), but decreasing. The induced current will
want to maintain the positive flux. Thus, the induced current flows counter
clockwise.
c Moving the loop to the right doesn’t change the flux, so there is no
induced current.
Problem 7 (Grifffiths 7.10 - optional)
Let’s start our loop perpendicular to the field, so that at time t = 0 we get
ΦB = Ba2 . Now, at any time t after this the flux will be
ΦB (t) = Ba2 cos θ = Ba2 cos(ωt).
So,
∂ΦB
∂t
= Ba2 ω sin(ωt)
ε(t) = −
which alternates sinusoidally, producing an AC current. One way to engineer
this in order to get such a current out of this device is shown in Figure 7.18 in
the text. Can you think of a simple way to get a (approximately) DC current
with only a slight modification?
4
Problem 8 (Griffiths 7.12)
ε
R
1
∂ΦB
= (−
)
R
∂t
a 2
1
B0 ω sin(ωt)
= π
R
2
πa2 ωB0
=
sin(ωt)
4R
I =
where a positive current indicates current flowing in the θ̂ direction.
Problem 9 (Griffiths 7.15)
~ = µ0 nI ẑ inside the solenoid and
The magnetic field from the solenoid is B
zero outside. Thus, the flux through a circular (to preserve symmetry) loop
centered along the axis of the solenoid is
ΦB(in) (t) = πr2 µ0 nI(t)
ΦB(out) (t) = πa2 µ0 nI(t)
where the ẑ direction is the direction of positive flux. For such a loop we know
that
I
~ · d~l = − ∂ΦB
E
∂t
~ = − 1 ∂ΦB θ̂.
=⇒ E
2πr ∂t
Thus,
~ in = − 1 ∂ πr2 µ0 nI(t)θ̂
E
2πr ∂t
µ0 nr ∂I(t)
=−
θ̂
2
∂t
~ out = − 1 ∂ πa2 µ0 nI(t)θ̂
E
2πr ∂t
µo na2 ∂I(t)
=−
θ̂.
2r
∂t
5