A conduction electron in the pendulum will
experience a magnetic force opposite the di
rection of
v
×
B
, so using the right hand rule,
we can determine that the motion of the elec
trons will be in the clockwise direction, which
produces a counterclockwise current.
Because the magnetic field is pointing into
the paper, the magnetic flux through the pen
dulum is increasing, so by Lenz’s law we know
that the induced magnetic field is in the op
posite direction, or out of the paper.
B
B
B
B
O
e
n
t
e
ri
n
g
fi
e
l
d
i
F
w
Alternative Solution:
Use the result of
Part 1, and the right hand rule on the current
flow to determine that the induced magnetic
field must be directed out of the paper.
021
10.0 points
A solenoid with circular cross section pro
duces
a
steadily
increasing
magnetic
flux
through its cross section. There is an octago
nally shaped circuit surrounding the solenoid
as shown.
B
B
B
B
X
Y
i
i
Figure 1:
The increasing magnetic flux gives rise to a
counterclockwise induced
emf
E
. The circuit
consists of two identical light bulbs of equal
resistance,
R
, connected in series, leading to
a loop equation
E

2
i R
= 0.
Now connect the points
C
and
D
with a
wire
CAD
(as in figure 2). Label the currents
i
1
,
i
2
, and
i
3
as indicated.
B
B
B
B
X
Y
D
C
A
i
2
i
1
i
3
i
3
Figure 2:
To what do the corresponding loop and
node equations lead?
1.
E

2
i
3
R
= 0 :
i
2
=
i
3
3
,
i
1
=
2
i
3
3
2.
E

2
i
3
R
= 0 :
i
2
=
i
3
4
,
i
1
=
3
i
3
4
3.
E

i
3
R
= 0 :
i
2
=
i
3
2
,
i
1
=
i
3
2
4.
E

i
3
R
= 0 :
i
2
=
i
3
3
,
i
1
=
2
i
3
3
5.
E

i
3
R
= 0;
i
2
=
i
3
,
i
1
= 0
correct
6.
E

i
3
R
= 0 :
i
2
=
i
3
4
,
i
1
=
3
i
3
4
toupal (rgt374) – Homework 09 – Chiu – (58295)
13
7.
E

i
3
R
= 0 :
i
2
= 0
,
i
1
=
i
3
8.
E

2
i
3
R
= 0 :
i
2
=
i
3
2
,
i
1
=
i
3
2
9.
E

2
i
3
R
= 0 :
i
2
=
i
3
,
i
1
= 0
10.
E

2
i
3
R
= 0 :
i
2
= 0
,
i
1
=
i
3
Explanation:
Consider the loop
XCADX
(where
Y
indi
cates the position of bulb
Y
and
X
indicates
the position of bulb
X
), with loop equation
E

i
3
R
= 0
.
The loop
XDY CX
leads to
E

i
3
R

i
1
R
= 0
.
Subtracting the former from the latter equa
tion we find
i
1
= 0
.
We also have one node equation;
e.g.
, at the
junction
C
:
i
3
=
i
1
+
i
2
i
3
=
i
2
.
022
10.0 points
A solenoid (with magnetic field
B
) pro
duces a steadily increasing uniform magnetic
flux through its circular cross section.
A
octagonal circuit surrounds the solenoid as
shown in the figure.
The wires connecting
in the circuit are ideal, having no resistance.
Two identical resistors with resistance
R
(la
beled
X
and
Y
) are in the circuit.
A wire
connects points
C
and
D
.
The ratio of the
solenoid’s area
A
L
left of the wire
CD
and
the solenoid’s area
A
R
right of the wire
CD
is
5.
4
E
5

i
1
R
= 0
and
E
5

i
2
R
= 0
.
6.
E
5

i
1
R
= 0
and
4
E
5
+
i
2
R
= 0
.
7.
E
5
+
i
1
R
= 0
and
4
E
5

i
2
R
= 0
.
8.
4
E
5
+
i
1
R
= 0
and
E
5
+
i
2
R
= 0
.
B
B
B
B
R
Y
R
X
D
C
i
3
i
1
i
2
A
L
A
R
E
=

d
Φ
dt
Figure:
Let
i
1
,
i
2
, and
i
3
be defined
as positive if the currents flow in the
same direction as shown by the arrows in
the figure (otherwise the currents
i
1
,
i
2
,
and
i
3
have negative values).
Also,
let
the induced emf be defined as positive,
E
>
0
.
What are the equations for the (right) loop
CXDC
and the (left) loop
CDY C
, respec
tively?
Assume the induced
emf
for the closed loop
octagonal
CXDY C
is
E
.
1.
E
5

i
1
R
= 0
and
4
E
5

i
2
R
= 0
.
2.
4
E
5

i
1
R
= 0
and
E
5
+
i
2
R
= 0
.
3.
E
5
+
i
1
R
= 0
and
4
E
5
+
i
2
R
= 0
.
cor
rect
4.
4
E
5
+
i
1
R
= 0
and
E
5

i
2
R
= 0
.