Massachusetts Institute of Technology
Department of Physics
Physics 8.07
Fall 2005
Problem Set 7 Solutions
Problem 1: Griffiths Problem 6.13 (p. 272)
~ 0 the magnetic field of a uniformly magnetized
This problem is done by adding to B
~ . Given the total magnetic field B,
~ the H-field in the cavity
object of magnetization −M
~
~
~
~
~.
is then H = B/µ0 . The H-field with no cavity is H0 ≡ B0 /µ0 − M
a) From Griffiths Example 6.1, the magnetic field in a sphere of uniform mag~ is − 2 µ0 M
~ giving B
~ =B
~ 0 − 2 µ0 M
~ and H
~ =B
~ 0 /µ0 − 2 M
~ =
netization −M
3
3
3
~.
~ 0 + 1M
H
3
~ parallel to the needle is equivalent to a
b) A long needle of magnetization M
~b = M
~ and internal magnetic field B
~ = µ0 M
~
solenoid with surface current K
~
~
~
~
(Griffiths Example 5.9). Thus, in the cavity B = B0 − µ0 M and H =
~ 0 /µ0 − M
~ =H
~ 0.
B
~ , radius R and thickness h R is
c) A wafer of uniform magnetization M
equivalent to a current loop of current I = hM . From Griffiths Example
~ =
5.6, the magnetic field in the center of the loop is B = µ0 I/(2R) or B
~
~
~
~
~
~ =
(µ0 h)/(2R)M . Thus, in the cavity B = B0 − (µ0 h)/(2R)M ≈ B0 and H
~ 0 /µ0 − (h/2R)M
~ ≈H
~0 + M
~.
B
Problem 2: Griffiths Problem 6.18 (p. 277)
~ = −∇W
~
~ M
~ vanishes everywhere except
Because J~f = 0 everywhere, H
where ∇2 W = ∇·
at r = R, where the boundary conditions give [Hθ ] = 0 and [Hr ] = −[Mr ]. Borrowing
the techniques of electrostatics, we write
W =
Ar cos θ ,
r<R
(B/r2 ) cos θ − H0 r cos θ , r > R ,
where H0 ≡ B0 /µ0 . (Note that we excluded a point multipole at r = 0, and we also
~ → B0~ez for r R.) Imposing the boundary condiimposed the boundary condition B
tion R[Hθ ] = −[∂W/∂θ] = 0 gives B = (A + H0 )R3 . For the other boundary condition,
~ = M~ez inside the sphere and zero outside so
[Hr ] = −∂W/∂r(R+ ) + ∂W/∂r(R− ) and M
~ ) = M cos θ, giving 3(A + H0 ) = M . Thus, A = −H0 + 1 M , or
that [Mr ] = −~er · (0 − M
3
1
~ = −∇W
~
H
= H0 − M ~ez for r < R .
3
1
Using M = χm H, we solve for H and M for r < R in terms of H0 ,
H=
H0
χm H0
, M=
.
1 + χm /3
1 + χm /3
~ = µ0 (H
~ +M
~ ) and H0 = B0 /µ0 give
Finally, B
~ =
B
!
1 + χm
~0 .
B
1 + χm /3
Problem 3: Griffiths Problem 7.7 (p. 299)
a) The current is I = E/R with E = Blv, and flows in the counter-clockwise
direction.
~ F = IlB = (Bl)2 v/R. The direction is opposite the
b) Using dF~ = Id~l × B,
direction of ~v .
c)
d~v
(Bl)2
1
mR
.
= F~ = −
~v ⇒ v = v0 e−t/τ , τ =
dt
m
mR
(Bl)2
d) The energy dissipated in the resistor is
Z
0
∞
Pjoule (t) dt =
Z
∞
0
(BL)2 v02 τ
E2
(Bl)2 2 Z ∞ −2t/τ
1
e
dt =
dt =
v0
= mv02 .
R
R
R
2
2
0
Problem 4: Griffiths Problem 7.16 (p. 309)
Assuming that the wire is very thin, and that the displacement current may be neglected
(quasi-static approximation), the magnetic field is
~ =
B
(µ0 I)/(2πs)~eφ , 0 < s < a ,
0,
s>a.
~
~ E
~ ∝
a) The magnetic field is in the ~eφ direction, and so too is ∂ B/∂t.
Thus, ∇×
~
~
~eφ . From the curl in cylindrical coordinates we deduce ∇ × E = (∂Es /∂z −
∂Ez /∂s)~eφ . Now, for a long wire there must be translational symmetry for
~ = Ez (s, t)~ez .
which ∂/∂z = 0, thus we conclude E
~ = 0 for s > a, we must have E
~ = 0 for s ≥ a. Integrating
b) Clearly, since B
∂Ez /∂s = ∂Bφ /∂t with I = I0 cos ωt, we find
~ = − µ0 I0 ω ln s sin ωt ~ez
E
2π
a
for s < a.
2