roofner (bar784) – Homework 13 – Weathers – (22202)
1
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001
(part 1 of 3) 10.0 points
An infinite wire is bent as shown in the figure.
The current is
I
.
It consists of 3 segments
AB
,
BDE
, and
EF
.
AB
is vertical with
A
extending to infinity.
EF
is horizontal with
F
extending to infinity.
BDE
is onequarter
of a circular arc. We use the Roman numerals
I, II, III and IV to denote the direction of a
vector in the
xy
plane, which begins at the
origin and is pointing into the quadrants I, II,
III and IV respectively.
B
O
D
C
E
F
A
O
y
x
I
II
III
IV
The direction of the magnetic vector at
O
due to the current segment
BD
, that due
to
DE
and that due to
EF
are given by
respectively (“in, out” mean “into, or out
of”the
xy
plane)
1.
B
BD
= in,
B
DE
= out,
B
ED
= out
2.
I, II,
B
ED
= in
3.
IV, IV,
B
ED
= out
4.
B
BD
= in,
B
DE
= out,
B
ED
= in
5.
B
BD
= out,
B
DE
= out,
B
ED
= out
correct
6.
B
BD
= out,
B
DE
= out,
B
ED
= in
7.
B
BD
= in,
B
DE
= in,
B
ED
= in
8.
B
BD
= out,
B
DE
= in,
B
ED
= in
9.
B
BD
= in,
B
DE
= in,
B
ED
= out
10.
B
BD
= out,
B
DE
= in,
B
ED
= out
Explanation:
The BiotSavart Law is given by
d
vector
B
=
μ
0
4
π
I d
vector
l
×
ˆ
r
r
2
.
Since both
d
vector
l
and ˆ
r
lie in the
xy
plane, their
cross product must be perpendicular to this
plane.
Apply the BiotSavart Law to each
line element of the wire,
d
vector
B
is found always
to direct out the
xy
plane at
O
.
002
(part 2 of 3) 10.0 points
What is the magnitude of
vector
B
at the center of
the arc due to the arc
BDE
alone?
1.
B
=
μ
0
I
8
π r
2.
B
=
μ
0
2
π I
r
3.
B
=
μ
0
I
8
r
correct
4.
B
=
μ
0
I
4
r
5.
B
=
μ
0
I
2
r
6.
B
=
μ
0
I
2
π r
7.
B
=
μ
0
I
r
8.
B
=
μ
0
I
4
π r
9.
B
=
μ
0
π I
r
10.
B
=
μ
0
I
π r
Explanation:
Note:
The distance
r
from a current el
ement of
BDE
to
O
is a constant and the
current element
I d
vector
l
is always perpendicular
to ˆ
r
.
Hence, the magnitude of the
vector
B
at the center
of the arc due to the arc BDC alone is
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roofner (bar784) – Homework 13 – Weathers – (22202)
2
B
BDE
=
μ
0
4
π
I
r
2
integraldisplay
dl
=
μ
0
4
π
I
r
2
π
2
r
=
μ
0
I
8
r
.
003
(part 3 of 3) 10.0 points
Let
r
= 0
.
063 m and
I
= 1
.
1 A.
The
permeability
of
free
space
is
1
.
25664
×
10
−
6
N
/
A
2
.
Find the magnitude of
vector
B
at
O
due to the
entire wire
ABDEF
?
Correct answer: 6
.
23472
×
10
−
6
T.
Explanation:
Let :
μ
0
= 1
.
25664
×
10
−
6
N
/
A
2
,
I
= 1
.
1 A
,
and
r
= 0
.
063 m
.
First
calculate
the
magnetic
field
of
a
straight wire carrying a current(See figure be
low).
I
r
s
∆
y
y
O
θ
B
From the BiotSavart Law, the contribution
to the magnetic field at
O
due to a current
element
I
∆
y
at
y
is given by
∆
B
=
μ
0
4
π
I
∆
y
s
2
sin
θ
=
μ
0
I
4
π
∆
θ
r
sin
θ .
Upon integration,
the magnetic field con
tributed by a current segment from
θ
1
to
θ
2
is
given by
B
=
μ
0
I
4
π r
(cos
θ
1
−
cos
θ
2
)
For the wire segment
AB
, we set
θ
1
=
π
2
and
θ
2
=
π .
Hence
B
AB
=
μ
0
I
4
π r
parenleftBig
cos
π
2
−
cos
π
parenrightBig
=
μ
0
I
4
π r
.
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 Fall '00
 Littler
 Magnetism, Work, Magnetic Field, Wire, BiotSavart law

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