PHY4324/Fall 09: EM II
HOMEWORK ASSIGNMENT #1:SOLUTIONS
due by 11:45 p.m. Wed 09/02
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly. Every (algebraic) final result
must be supplemented by a check of units. Without such a check, no more
than 75% of the credit will be given even for an otherwise correct solution.
1. G6.8 [25 points]
Inside the cylinder, the bulk bound current density Jb = ∇ × M. Since there are no conduction currents,
∇ × B =µ0 Jb = µ0 ∇ × M. Obviously, B has only the φ̂ component. Applying the Stokes theorem, we obtain
then 2πsBφ (s) = 2πsµ0 Mφ (s) →
Bφ (s) = µ0 Mφ (s) = µ0 ks2 .
Outside the cylinder, M = 0 → Jb = 0 → ∇ × B = 0. This does not necessarily means that B = 0 because, for
example, B can be a constant. To show that the only constant consistent with the boundary conditions at the
surface of the cylinder, we recall that the tangential component of B is discontinuous: B||out − B||in = µ0 K. The
surface current of bound currents K = M × n = (0, 0, −Mφ ). The tangential component of B is Bφ . Therefore,
B||out = B||in + µ0 K = µ0 (Mφ − Mφ ) = 0 and B = 0 anywhere outside the cylinder.
2. G6.15 [25 points]
Hint: notice (and explain why!) that W is continuous at the surface whereas its radial derivative satisfies a
boundary condition
−
∂Win ∂Wout +
= M ⊥ = M cos θ.
∂r r=R
∂r r=R
(1)
The continuity of W follows from the gradient theorem. Integrate H = −∇W over any contour starting at point
RB
RB
A inside the sphere and ending at point B outside the sphere: A H·dl = − A (∇W ) · dl =W (A) − W (B).
RB
Since the points A and B can be made infinitesimally close, A H·dl is infitesimally small and W (A) = W (B).
Condition (1) follows from the continuity of the normal component of B in the absence of free currents and
out
in
the relation between H, B, and B : B =µ0 (H + M) . Since B⊥ is continuous, so is H + M, or H⊥
− H⊥
=
out
in
out
in
in
−(M⊥ − M⊥ ). Outside the sphere, M = 0, therefore, H⊥ − H⊥ = M⊥ . Since normal components are along
in,out
in
the radial line for a sphere, H⊥
= −∂W in,out /∂r and M⊥
= M cos θ, where θ is the azimuthal angle of the
spherical system. Both inside and outside the sphere, W satisfies the Laplace’s equation ∇2 W = 0. Recalling
the separation of variables in spherical geometry (Sec. 3.3.2), we write W as
X
W =
aℓ rℓ Pℓ (cos θ) , for r < R
ℓ
X bℓ
W =
Pℓ (cos θ) , for r > R.
rℓ+1
ℓ
The boundary condition for W yields bℓ = aℓ R2ℓ+1 whereas that for ∂W/∂r reads
X (ℓ + 1) bℓ
ℓ−1
Pℓ (cos θ) = M cos θ = M P1 (cos θ) .
+
ℓa
R
ℓ
Rℓ+2
(2)
ℓ
Since the Legendre polynomials form a complete basis, the sum on the l.h.s. has only one term ℓ = 1 and (2)
reduces to
2b1
+ a1 = M.
R3
2
Substituting b1 = a1 R3 into the last equation, we obtain
3a1 = M
or
W =
M
M
M
rP1 (cos θ) =
r cos θ =
z, r < R.
3
3
3
Therefore, H has only the z-component equal to
Hz = −
M
∂W
=− .
∂z
3
The magnetic field inside the sphere is also along the z-axis
Bz = µ0 (Hz + M ) =
2
µ0 M.
3
3. G6.17 [25 points]
Stokes theorem for the H field
I
H·dl =Ienc .
Inside the wire,
2πsHφ = I
s
s2
→ Hφ = I
.
a2
2πa2
Outside the wire,
2πsHφ = I → Hφ = I
1
.
2πs
The magnetic field Bφ = µ0 (1 + χm )Hφ inside the wire and Bφ = µ0 Hφ inside the wire. Magnetization
Mφ = χm Hφ (inside) and Mφ = 0 outside. The bulk bound current density inside the wire
Jb = ∇ × M = χm ∇ × H = χm Jf = χm (I/πa2 )ẑ.
The bulk bound current flows along the free current for χm > 0 and opposite to it for χm < 0.
The surface current density
K = M × n =χm (H × n) = −χm Hφ (s = a)ẑ = −χm
I
ẑ.
2πs
The surface current is always opposite to the direction of the bulk bound current. The total bound current
Ib = Jb πa2 + K2πa = 0.
4. G6.26 [25 points]
Because ∇ · B = 0, the normal component of B is always continuous:
B1⊥ = B2⊥ → µ1 H1⊥ = µ2 H2⊥ .
In the absence of free currents, we also have ∇ × H = 0 which means that the tangential component of H is
also continuous
H1|| = H2||
Since tan θ = H|| /H⊥ , dividing the two equations by each other we obtain the required result.
3
5. Bonus: G6.18 [25 points]
As in Problem G6.15, we will use the ”magnetic potential” defined by H = −∇W. As before, W itself is
continuous. However, now the magnetization is not fixed but is equal to χm H. Instead of Eq. (1), we now get
−
∂Win ∂Win
∂Wout +
= M ⊥ = χm H ⊥ = −χm
→
∂r r=R
∂r r=R
∂r
∂Win
∂Wout = (1 + χm )
.
∂r r=R
∂r
Far away from the sphere, the magnetic field must approach the external field B0 , therefore, the solution for W
outside the sphere must have the form
W =−
X Pℓ (cos θ)
B0
, for r > R.
r cos θ +
bℓ
µ0
rℓ+1
ℓ
Inside the sphere
W =
X
aℓ rl Pℓ (cos θ) ,
ℓ
as in 6.15. Again, it is obvious that the sum contains only the ℓ = 1 terms. Matching the boundary conditions
for these terms, we obtain
a1 R = −
B0
b1
R+ 2
µ0
R
and
2b1
B0
− 3
(1 + χm )a1 = −
µ0
R
From here, we find a1
a1 = −
3B0
1
.
µ0 3 + χm
The H field inside the sphere
H=
B0
3
ẑ.
µ0 3 + χm
Magnetization
M =χm H =
B0 3χm
ẑ
µ0 3 + χm
and the magnetic field
B = µ0 (1 + χm )H =
(1 + χm )
3(1 + χm )
B0 =
B0
3 + χm
1 + χm /3