Problem Set 5 Solutions

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Massachusetts Institute of Technology
Department of Physics
Physics 8.07
Fall 2005
Problem Set 5 Solutions
Problem 1: Fields in a non-uniform material
a) Griffths Problems 4.33 (p. 198)
~ t ] = 0. With the
Boundary conditions at the interface are [Dn ] = 0 and [E
geometry of Griffiths Figure 4.34, the normal components are given by a
~ = E,
~
factor cos θ and the tangential components by a factor sin θ. With D
this gives
Dn = 1 E1 cos θ1 = 2 E2 cos θ2 , Et = E1 sin θ1 = E2 sin θ2 .
Combining these gives
tan θ2
2
=
.
tan θ1
1
Because the sense of deflection is similar to Snell’s law, a convex dielectric
“lens” will “focus” the electric field. It’s fascinating that the amount of
focusing is different for static field lines than for electromagnetic waves – EM
waves don’t necessarily travel along the static field lines!
~ ·D
~ = 0 or
b) We want to find Ex (z) and Ez (z). Instead of [Dn ] = 0 we have ∇
d
(Ez ) = 0 .
dz
~ t ] = 0 we have
The solution of this is (z)E(z) = constant. Instead of [E
~ ×E
~ = 0 or
∇
dEx
=0.
dz
The solution is Ex = constant. With the values given at z = 0, we get
0
Ex = E0 , Ez =
E0 .
(z)
These imply
Dx = (z)E0 , Dz = 0 E0 ,
and, since ρ = ρf + ρb = ρb ,
~ ·E
~ = −E0 0
ρb = 0 ∇
1
2
d
.
dz
Problem 2: Griffiths Problem 5.3 (p. 208)
~ + ~v × B
~ ) = 0. With ~v , E
~ and B
~ mutually perpena) No deflection means q(E
dicular, this implies E = vB or v = E/B.
~ = m~a. Charges will travel in circular
b) With only a magnetic field, F~ = q~v × B
~ and centripetal acceleration
orbits of radius R with ~v perpendicular to B
v 2 /R, implying mv 2 /R = qvB or q/m = v/(RB) hence
q
E
= 2 .
m
B R
Problem 3: Griffiths Problem 5.10 (p. 220)
This problem requires the magnetic field of a long, straight wire carrying steady current
~ = µ0 I/(2πs)~eφ where s is the distance from the wire and ~eφ is the azimuthal unit
I: B
vector whose direction is given by the right-hand rule. Also, the force on a current I
~ In this problem, we define the Cartesian coordinate
flowing along d~l is dF~ = Id~l × B.
system so that the loop lies in the x-y plane, with the long wire pointing along ~ex . The
magnetic field on the loop points in the direction ~ez .
a) The forces on the two segments perpendicular to the long wire are equal and
opposite and therefore cancel. The other two segments have d~l = ±~ex dx
giving forces in the direction ±~ex × ~ez = ∓~ey . Putting in the distances y = s
and y = s + a, and integrating over x from 0 to a, we get
"
a2
1
µ 0 I 2 a2
µ0 I 2 a 1
−
1
+
O
~ey =
F~ =
2π
s s+a
2πs2
s2
!#
~ey .
For a s, the magnetic field at the center of the loop is
~ =
B
µ0 I
µ0 I
~ez ≈
~ez .
2π(s + a/2)
2πs
For a s, the force on the loop may therefore be written
~ m
~), m
F~ = ∇(
~ ·B
~ = −Ia2~ez .
b) The geometry of the triangular loop is a bit more complicated. Label the
three sides 1 (left), 2 (right), and 3 (bottom, parallel to the long wire). Along
segments 1 and 2,
√ !
1
3
d~l = dl ~ex ±
~ey .
2
2
2
The forces arising from the y-component of d~l cancel in the sum. For a point
that is a distance ` from the bottom corner
√ of size 1 or 2, the y-value (needed
to compute the magnetic field) is s + ` 3/2. The net force contribution from
sides 1 and 2 is therefore
√ !
2 Z a
2
µ
I
3
(d`/2)
(~
e
×
~
e
)
µ
I
a
0
x
z
0
√
F~1+2 =
2
= − √ ln 1 +
~ey .
2π
2s
0
s + ` 3/2
π 3
Adding to this the force on the horizontal segment gives the total force:
√ !#
"
µ0 I 2 a
3a
2
~
F =
− √ ln 1 +
~ey .
2π s
2 s
3
For a small loop, a s, we can expand the logarithm using ln(1 + ) =
− 2 /2 + · · ·, yielding
√
√
√
!
3 µ 0 I 2 a2
3 µ 0 I 2 a2
3 2
~
~
~
~
F =
= ∇(m
~ ·B) , m
~ =−
~ey = ∇ −
Ia ~ez .
2
8π s
8π
s
4
√
The area of the equilateral triangle is ( 3/4)a2 , so that m = I ∗ Area, with
direction perpendicular to the loop chosen by the right-hand rule. The vector
m
~ is called the magnetic moment and is defined in Griffiths equation (5.84).
Problem 4: Griffiths Problem 5.38 (p. 247)
The current density is J~ = ρ−~v where ρ− < 0. Assuming J~ is uniform in the wire, the
magnetic field is circumferential with magnitude given by Ampère’s law,
1
1
B = µ0 Js = µ0 ρ− vs .
2
2
Now, in order that the negative charges be in a steady state, the net force on them must
~ + ~v × B
~ = 0 yielding
vanish, implying E
~ = −|v− B|~es = 1 µ0 ρ− v 2~s .
E
2
(Note the sign choice is needed because with ρ− < 0 the field points radially toward the
axis, so that negative charges feel an outwards electric force which cancels the inward
magnetic force.) This field is generated by a net charge density
~ ·E
~ = 0 µ 0 ρ− v 2 ,
ρ = ρ− + ρ+ = 0 ∇
giving ρ+ = −(1 − 0 µ0 v 2 )ρ− or, with c2 = 1/(0 µ0 ), ρ− = −γ 2 ρ+ , γ = (1 − v 2 /c2 )−1/2 .
The negative charges do not pinch because of the electric repulsion away from the axis.
The compensating positive charge resides on the outside of the wire.
3
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