Chapter 10 – OddNumbered
10.1. In Fig. 10.4, let
B
= 0
.
2cos120
πt
T, and assume that the conductor joining the two ends
of the resistor is perfect. It may be assumed that the magnetic field produced by
I
(
t
) is
negligible. Find:
a)
V
ab
(
t
): Since
B
is constant over the loop area, the ﬂux is Φ =
π
(0
.
15)
2
B
= 1
.
41
×
10
−
2
cos120
πt
Wb.
Now,
emf
=
V
ba
(
t
) =
−
d
Φ
/dt
= (120
π
)(1
.
41
×
10
−
2
)sin120
πt
.
Then
V
ab
(
t
) =
−
V
ba
(
t
) =
−
5
.
33sin120
πt
V
.
b)
I
(
t
) =
V
ba
(
t
)
/R
= 5
.
33sin(120
πt
)
/
250 = 21
.
3sin(120
πt
) mA
10.3. Given
H
= 300
a
z
cos(3
×
10
8
t
−
y
) A/m in free space, find the emf developed in the general
a
φ
direction about the closed path having corners at
a) (0,0,0), (1,0,0), (1,1,0), and (0,1,0): The magnetic ﬂux will be:
Φ =
1
0
1
0
300
µ
0
cos(3
×
10
8
t
−
y
)
dx dy
= 300
µ
0
sin(3
×
10
8
t
−
y
)

1
0
= 300
µ
0
sin(3
×
10
8
t
−
1)
−
sin(3
×
10
8
t
)
Wb
Then
emf =
−
d
Φ
dt
=
−
300(3
×
10
8
)(4
π
×
10
−
7
) cos(3
×
10
8
t
−
1)
−
cos(3
×
10
8
t
)
=
−
1
.
13
×
10
5
cos(3
×
10
8
t
−
1)
−
cos(3
×
10
8
t
)
V
b) corners at (0,0,0), (2
π
,0,0), (2
π
,2
π
,0), (0,2
π
,0): In this case, the ﬂux is
Φ = 2
π
×
300
µ
0
sin(3
×
10
8
t
−
y
)

2
π
0
= 0
The emf is therefore 0
.
10.5. The location of the sliding bar in Fig. 10.5 is given by
x
= 5
t
+2
t
3
, and the separation of the
two rails is 20 cm. Let
B
= 0
.
8
x
2
a
z
T. Find the voltmeter reading at:
a)
t
= 0
.
4 s: The ﬂux through the loop will be
Φ =
0
.
2
0
x
0
0
.
8(
x
)
2
dx dy
=
0
.
16
3
x
3
=
0
.
16
3
(5
t
+ 2
t
3
)
3
Wb
Then
emf =
−
d
Φ
dt
=
0
.
16
3
(3)(5
t
+2
t
3
)
2
(5+6
t
2
) =
−
(0
.
16)[5(
.
4)+2(
.
4)
3
]
2
[5+6(
.
4)
2
] =
−
4
.
32 V
b)
x
= 0
.
6 m: Have 0
.
6 = 5
t
+ 2
t
3
, from which we find
t
= 0
.
1193. Thus
emf =
−
(0
.
16)[5(
.
1193) + 2(
.
1193)
3
]
2
[5 + 6(
.
1193)
2
] =
−
2
.
93 V
110
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10.7. The rails in Fig. 10.7 each have a resistance of 2.2 Ω
/
m. The bar moves to the right at a
constant speed of 9 m/s in a uniform magnetic field of 0.8 T. Find
I
(
t
), 0
< t <
1 s, if the bar
is at
x
= 2 m at
t
= 0 and
a) a 0.3 Ω resistor is present across the left end with the right end opencircuited: The ﬂux
in the lefthand closed loop is
Φ
l
=
B
×
area = (0
.
8)(0
.
2)(2 + 9
t
)
Then, emf
l
=
−
d
Φ
l
/dt
=
−
(0
.
16)(9) =
−
1
.
44 V. With the bar in motion, the loop
resistance is increasing with time, and is given by
R
l
(
t
) = 0
.
3+2[2
.
2(2+9
t
)]. The current
is now
I
l
(
t
) =
emf
l
R
l
(
t
)
=
−
1
.
44
9
.
1 + 39
.
6
t
A
Note that the sign of the current indicates that it is ﬂowing in the direction opposite that
shown in the figure.
b) Repeat part
a
, but with a resistor of 0.3 Ω across each end: In this case, there will be
a contribution to the current from the right loop, which is now closed. The ﬂux in the
right loop, whose area decreases with time, is
Φ
r
= (0
.
8)(0
.
2)[(16
−
2)
−
9
t
]
and emf
r
=
−
d
Φ
r
/dt
= (0
.
16)(9) = 1
.
44 V. The resistance of the right loop is
R
r
(
t
) =
0
.
3 + 2[2
.
2(14
−
9
t
)], and so the contribution to the current from the right loop will be
I
r
(
t
) =
−
1
.
44
61
.
9
−
39
.
6
t
A
The minus sign has been inserted because again the current must ﬂow in the opposite
direction as that indicated in the figure, with the ﬂux decreasing with time. The total
current is found by adding the part
a
result, or
I
T
(
t
) =
−
1
.
44
1
61
.
9
−
39
.
6
t
+
1
9
.
1 + 39
.
6
t
A
10.9. A square filamentary loop of wire is 25 cm on a side and has a resistance of 125 Ω per meter
length. The loop lies in the
z
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 Fall '08
 Joshi
 Electrical Engineering, Sin, Cos, Trigraph, DVD region code

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