Version 098/ABCAC – midterm 03 – Turner – (59070)
1
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001
10.0 points
A coil formed by wrapping 55
.
9 turns of wire
in the shape of a square is positioned in a
magnetic field so that the normal to the plane
of the coil makes an angle of 24
.
8
◦
with the
direction of the field.
When the magnitude
of the magnetic field is increased uniformly
from 270
μ
T to 600
μ
T in 0
.
215 s, an emf of
39
.
3 mV is induced in the coil.
What is the total length of the wire?
1. 308.719
2. 242.655
3. 239.128
4. 272.502
5. 171.491
6. 158.831
7. 230.688
8. 314.146
9. 221.746
10. 339.031
Correct answer: 158
.
831 m.
Explanation:
From Faraday’s Law,
E
=
−
N
d
Φ
B
dt
E
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
−
N
d
Φ
B
dt
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
N
Δ
B
Δ
t
A
cos
θ
So the area of one loop in the coil is
A
=
E
N
cos
θ
·
Δ
t
Δ
B
= 0
.
504575 m
2
and the side length of the loop should be
a
=
√
A
= 0
.
710335 m
Finally, the length of the wire is
l
= (4
·
a
)
N
= 158
.
831 m
002
10.0 points
Three very long wires are strung parallel to
each other as shown in the figure below. Each
wire is at the perpendicular distance 56 cm
from the other two, and each wire carries a
current of magnitude 5
.
8 A in the directions
shown in the figure.
I
I
I
3
2
1
z
x
y
56 cm
56 cm
56 cm
×
3
2
1
y
x
z
Crosssectional View
The permeability of free space is 1
.
2566
×
10
−
6
T
·
m
/
A.
Find the magnitude of the net force per
unit length exerted on the upper wire (wire 3)
by the other two wires.
1. 2.8145e05
2. 2.08087e05
3. 3.14326e05
4. 2.56557e05
5. 6.33417e06
6. 4.09208e05
7. 1.46596e05
8. 3.21781e05
9. 6.62384e06
10. 2.77175e05
Correct answer: 2
.
08087
×
10
−
5
N
/
m.
Explanation:
Let :
r
= 56 cm = 0
.
56 m
and
I
= 5
.
8 A
.
Magnetic field due to a long straight wire is
B
=
μ
0
I
2
π r
,
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Version 098/ABCAC – midterm 03 – Turner – (59070)
2
and the force per unit length between two
parallel wires is
F
ℓ
=
μ
0
I
1
I
2
2
π r
.
There are two ways to solve this problem
which are essentially the same. The first way
is to find the net magnetic field at the upper
wire from the two wires below (
vector
B
net
=
vector
B
1
+
vector
B
2
) and then find the force from
vector
F
=
I
vector
L
×
vector
B .
The crucial step here will be to add the
magnetic fields as
vectors
.
The second way
would be to use
vector
F
=
I
vector
L
×
vector
B
to find the net
force on the upper wire from the two lower
wires
vector
F
net
=
vector
F
1
+
vector
F
2
,
where we must be
sure to add the forces as vectors. You should
recognize that the two methods are formally
identical.
Let’s do it the first way.
The
magnitude magnetic field from wire 1 is found
from Ampere’s law to be
B
1
=
μ
0
I
2
π r
.
Using the right hand rule the direction points
up and to the left of wire as shown in figure 2.
30
◦
30
◦
60
◦
60
◦
60
◦
B
1
B
B
2
×
3
2
1
ˆ
ˆ
ı
ˆ
k
Adding Magnetic Fields
30
◦
30
◦
60
◦
60
◦
60
◦
F
13
F
F
23
×
3
2
1
ˆ
ˆ
ı
ˆ
k
Adding Forces
Its components will then be
vector
B
1
=
B
[sin (30
◦
)ˆ
−
cos (30
◦
)ˆ
ı
]
.
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 Spring '08
 Turner
 Magnetic Force, Magnetic Field

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