PHYS 100B (Prof. Congjun Wu)
Solution to HW 3
January 10, 2011
Problem 1 (Griths 5.24)
If B is uniform, show that A(r) = 1 (r B) works. That is, check that A = 0 and A = B. Is this
2
result unique, or are there other functions with the same divergence and
HW8 solution
Wei-Ting Kuo
March 1, 2016
Problem 7.22
(a) From eq.5.41, the magnetic field from a loop is
B(z) =
0 I
R2
.
2
2 (R + z 2 )3/2
Here, I assume that the small loop is so small that the magnetic field from the large loop is constant
across the w
HW1 solution
Wei-Ting Kuo
January 13, 2016
Problem 5.2
We can use the general solution derived in Ex. 5.2,
y(t) = C1 cos t + C2 sin t +
E
t + C3
B
and
z(t) = C2 cos t C1 sin t + C4 .
To determine these four constants, I will utilize the initial condition
HW2 solution
Wei-Ting Kuo
January 18, 2016
Problem 5.3
(a) Zero deflection means the total force exerted on the electrons is zero.
~ + ~v B)
~ =0
F = q(E
~ = |~v B|
~ |~v | = |E|/|
~ B|
~
|E|
(b) The only remaining force is magnetic force.
~ 2
|E|
~ 2R
HW3 solution
Wei-Ting Kuo
January 22, 2016
Problem 5.19
Since we are talking about steady current here, which means J~ = 0, the current density can be written
~ Input this into the integration and we get
as the curl of a vector, J = A.
Z
Z
I
~
~
~ d~l.
J
HW7 solution
Wei-Ting Kuo
February 22, 2016
Problem 7.14
As the magnet passes the pipe, the magnetic flux of the pipe keeps changing. We can divide the pipe into
two parts. The first part is above the magnet and the second part is below the magnet. As th
HW6 solution
Wei-Ting Kuo
February 12, 2016
Problem 7.2
(a) From Kirchhoffs law,
q
dq
t
q
+ qR
=0
= R
= ln q + A.
C
C
dt
RC
At t = 0, q = Q = CV0 A = ln Q q = CV0 et/RC . The current is I = dq/dt = VR0 et/RC .
(b)Original energy is 21 CV02 . Integration
HW5 solution
Wei-Ting Kuo
February 5, 2016
Problem 6.12
(a) The relations between magnetization and bound currents are
~ and K
~b = M
~ n
~ = k and K
~b = M
~ n
J~b = M
J~b = M
= kR.
Based on Amperes law, the magnetic field inside the cylinder is
~ = 0
HW4 solution
Wei-Ting Kuo
January 31, 2016
Problem 5.34
The trick here is to use spherical coordinates to represent the magnetic dipole moment. To simplify the
calculation, we can choose the direction of magnetic dipole moment is along z direction. Thus,
52
CHAPTER 3. SPECIAL TECHNIQUES
(for r ~ R). Meanwhile, Bl = A1R21+1(Eq. 3.81-this followsfrom the continuity of V at R). Therefore 8kR4/5, So
-3kR2 1 8kR4 1 V(r,O) = 5 r2PI(COSO) + r4P3(COSO) k
I
if 1= 3
Bl = cfw_ -3kR2/5, if 1= 1
(zero otherwise).
s-
PHYS 100B (Prof. Congjun Wu) Solution to Midterm Problems
February 4, 2011
Problem 2 (Helmholtz coil) Two identical circular loops with radius R are put together along the same z-axis, at a distance d apart. The upper one is put at z = d/2, and the lower
PHYS 100B (Prof. Congjun Wu) Solution to HW 6
March 4, 2011
Problem 1 (Griffiths 8.2) (Griffiths 7.31) A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w a, forms a parall
PHYS 100B (Prof. Congjun Wu) Solution to HW 5
February 21, 2011
Problem 1 (Griffiths 6.26) At the interface between two linear magnetic materials, the magnetic field lines band. Show that tan 2 / tan 1 = 2 /1 , assuming there is no free current at the bou
PHYS 100B (Prof. Congjun Wu)
Solution to HW 2
January 8, 2011
Problem 1 (Griths 5.8)
(a) Find the magnetic eld at the center of a square loop, which carries a steady current I . Let R be the distance
from center to side (Fig. ?).
Solution: (B = 20 I/(R).)
PHYS 100B (Prof. Congjun Wu)
Solution to HW 1
January 8, 2011
Problem 1 (Griths 5.3) In 1897 J. J. Thomson discovered the electron by measuring the charge-to-mass
ratio of cathode rays (actually, streams of electrons, with charge q and mass m) as follows:
HW9 solution
Wei-Ting Kuo
March 6, 2016
Problem 7.34
Magnetic field can come from the current and the change of electric field. In this question, since the
charges are being accumulated on the two plates, the electric field inside the gap is increasing.