PHY4324/Fall 09: EM II
Test preparation: Solutions
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. Energy stored in a two-loop configuration is
W =
L1 I12
L2 I22
+
+ M I1 I2 .
2
2
The energy must be non-negative. This requirement imposes some relation on the values of L1,2 and M. To see
this, diagonalize the quadratic form for W, defining new variables i1,2 as
p I1 = 1/ L1 (i1 cos θ + i2 sin θ)
p I2 = 1/ L2 (i1 sin θ + i2 cos θ) ,
where θ is to be determined from the condition that W is diagonal in variables i1,2 . Substituting these equations
into W, we obtain
1
1
M
2
2
(i1 cos θ + i2 sin θ) + (i1 sin θ + i2 cos θ) + √
(i1 cos θ + i2 sin θ) (i1 sin θ + i2 cos θ)
2
2
L1 L2
1
M
1
12
i1 + i22 sin (2θ) .
= i21 + i22 + i1 i2 sin (2θ) + √
i1 i2 +
2
2
2
L1 L2
W =
The cross-terms disappear if one chooses
sin (2θ) + √
M
= 0.
L1 L2
Then W is simplified to
M2
12
2
i + i2 1 −
.
W =
2 1
L1 L2
Obviously, W ≥ 0 if
M≤
p
L1 L2 .
In our case, M = 10 µH and L1 = 5 µH. Therefore,
L2 ≥
M2
100
=
µH = 20µH.
L1
5
a) no; b) yes; c) yes.
2. The magnetic field inside the solenoid
B = µ0 N I.
Energy stored
1
W =
2µ0
Z
B 2 dτ =
µ0 2 2 2
N I πR ℓ.
2
2
3. If N1 (N2 ) is the number of turns in the inner (outer) solenoid, the magnetic field inside the inner solenoid is
B1 = (N1 + N2 ) µ0 I while the field in the space in between two solenoids is B2 = N2 µ0 I. Energy stored is
h
Z
Z
iµ
1
0
2
2
2
W =
πℓ,
B1 dτ1 + B2 dτ2 = (N1 + N2 ) R12 + N22 (R22 − R12 )
2µ0
2
where R1 and R2 are the radii of inner and outer solenoids, correspondingly, and ℓ is their length.
1. Answer: M =
a2 b2
π
2 µ0 (b2 +z 2 )3/2
cos α
2. Displacement current through the surface of a cylinder of radius r and height h
∂D
2πrh
∂t
Q
D =
2πrh
dQ
Id =
dt
Id =
where Q is the charge on the inner cylinder. The potential difference between the cylinders is
Z a1
Z b
D
D
1
Q
a1
1
b
dr +
V = −
dr = −
ln
+
ln
ε1
ε2
2πh ε1
a
ε2 a1
a
a1
V 2πh
i
Q = −h
a1
1
b
1
ε1 ln a + ε2 ln a1
Id = −
dV
dt
1
ε1
2πh
ln aa1 + ε12 ln ab1
3. See solutions for Homework # 4.
7. Laser beam: The momentum density carried by the incoming beam of intensity I is I/c2 . Consider a slanted
2
prizm with a base of unit area as shown in the figure. The volume
of the prizm is cdt cos φ×1 m . All light within
2
this prizm falls on thesurface transporting momentum I/c cdt cos φ per unit
area. Momentum reflected per
unit area is −R I/c2 cdt cos φ. The change in momentum is [1 + R] I/c2 cdt cos φ. The pressure is equal to
the normal component of the force, that is, P = [1 + R] I/c2 c cos2 φ.