Along the wire segment parallel to the electric field between the plates,
which is assumed to be uniform,
E =
ó (ç) õ
E·dl = Eh
Since the electric field is conservative, the integral of the field around
a closed loop should be zero. Since we get a non-zero result, we have ignored
something. That something is the fringing field inside the end and outside the
capacitor plates. The component of the fringing field around the circuit is
just enough to cancel the effects of the uniform field inside the plates.
The answer to the question is
E = 0
Text: 7.8
SOLUTION
(a) Calculate the magnetic flux through the loop.
F
=
ó õ
B·da
=
ó õ
s+a
s
æ è
moI
2py
^
z
ö ø
·
æ è
a dy
^
z
ö ø
=
moIa
2p
ó õ
s+a
s
dy
y
=
moIa
2p
ln
æ è
s+a
s
ö ø
(b) Now, move the loop directly away from the wire with speed v.
v=v
^
y
v=dy/dt
E
=
-
dF
dt
=
-
moIa
2p
æ è
s
s+a
ö ø
é ë
1
s
-
s+a
s2
ù û
ds
dt
=
-
moIav
2p
æ è
1
s+a
-
1
s
ö ø
=
+
moIav
2ps(s+a)
The sign indicates that the current induced in the loop is CCW. Lenz's law
supports this.
(c) There is no induced voltage in the loop because the magnetic flux
through the loop is NOT changing.
Text: 7.13
SOLUTION
The magnetic flux through the loop is
E
=
ó õ
B·da
=
ó õ
æ è
ky3t2
^
z
ö ø
·
æ è
a dy
^
z
ö ø
=
kat2
ó õ
a
0
y3dy
=
1
4
kat2y4
ê ê
a
0
=
1
4
ka5t2
The induced emf is then
E
=
-
dF
dt
=
-
1
2
ka5t
Text: 7.16
SOLUTION
(a) We might answer this question by looking at
Ñ×E
=
-
¶B
¶t
=
-
mo
2ps
dI
dt
^
f
The curl of E must be azimuthal. Looking up the operation
æ è
¶Ef
¶z
-
¶Ez
¶s
ö ø
^
f
=
-
mo
2ps
dI
dt
^
f
The first partial derivative on the left-hand-side must be zero. Because of
symmetry, the electric field does not vary with z. Therefore, the electric
field cannot be azimuthal. The second partial derivative contains the
z-component of E. Therefore, I would conclude that E is
longitudinal.
(b) Compare the two sides of the above equation and solve it for
the current
I = Iocoswt
-
¶Ez
¶s
=
-
mo
2ps
dI
dt
¶Ez
¶s
=
mo
2ps
(-Iowsinwt)
=
-
mowIo
2ps
sinwt
Ez
=
-
mowIo
2p
sinwt
ó õ
s
a
ds
s
=
-
mowIo
2p
sinwt ln
æ è
s
a
ö ø
Since the above integration involved a partial derivative, a time-dependent
term could have been added to the result. I chose to set it to zero.
The final result is
E
=
é ë
-
mowIosinwt
sp
ln
æ è
s
a
ö ø
ù û
^
z
An alternate way to determine the electric field is to use
ó (ç) õ
E·da
=
-
dF
dt
Set up a rectangular loop in the radial plane with two sides parallel to
the wire at the center of the cylinder and two sides along the radius of the
cylinder. For the sides parallel to the wire, let one side be inside the
cylinder and the other wire be outside the cylinder. Assume that the
electric field outside the outer conductor is essentially zero. The value of
the integral is
ó (ç) õ
E·dl
=
El
where, l is the length of each parallel to the wire. The magnetic flux
passing through the rectangular loop is
F
=
ó õ
a
s
æ è
moI
2ps¢
^
f
ö ø
·
æ è
lds¢
^
f
ö ø
=
moIl
2p
ó õ
a
s
ds¢
s¢
=
moIl
2p
ln
æ è
a
s
ö ø
Take the time derivative using the expression given for the current.
dF
dt
=
mol
2p
ln
æ è
a
s
ö ø
dI
dt
=
mol
2p
ln
æ è
a
s
ö ø
(-Iowsinwt)
=
-
mol
2p
ln
æ è
a
s
ö ø
sinwt
Equate the two expressions.
El
=
mol
2p
ln
æ è
a
s
ö ø
sinwt
E
=
mo
2p
ln
æ è
a
s
ö ø
sinwt
E
=
é ë
mo
2p
ln
æ è
a
s
ö ø
sinwt
ù û
^
z
A solenoid is filled with a material which results in the following
hysteresis loop.
(a) If the material in the solenoid fills a volume 15.6 cm3, how
much electrical energy does an external AC power source supply to the material
during each cycle around the loop?
SOLUTION
Estimating the areas,
Aparallelogram
=
640 J/m3
Atraingles
=
50 J/m3
The total energy density is then, 690 J/m3.
The electrical energy supplied per cycle is
E
=
(690 J/m3)(15.6×10-6 m3)
=
1.1×10-2 J
(b) If the external circuit drives the solenoid at a frequency of
60.0 Hz, what is the power dissipated as heat?
SOLUTION
The power dissipated is
P
=
Ef
=
(1.08×10-2 J)(60.0 Hz)
=
0.65 W
File translated from
TEX
by
TTH,
version 3.85. On 01 Feb 2010, 17:15.