Note:
Contribution to
B
0
from the current
in the segment
CA
equals the contribution
from the current in the segment
AC
0
.
Reason:
This follows by noticing (from the
BiotSavart law) that the contribution to
~
B
0
from a current through the line element
d ‘
centered about
P
is the same as the contribu
tion from an equal incremental line element
d ‘
centered about
P
0
with
P
and
P
0
equally
distant from point
A
.
Thus each half of the extended wire gives
the desired magnetic field
B ,
and
B
+
B
=
B
0
=
μ
0
I
2
π a
. Therefore
B
=
μ
0
I
4
π a
.
Direct Way:
Set up a coordinate system
with the origin at
A ,
such that the segment
CA
goes along the negative
x
direction as in
the figure. The distance from the field point
O
to the wire is
a
. Consider a certain point
P
along
CA
with
x
coordinate
x
. The distance
from
O
to this particular point is called
r
.
Denote the (counterclockwise) angle between
Platt, David – Oldquiz 3 – Due: Nov 13 2005, 4:00 am – Inst: Ken Shih
2
the line
OP
and the horizontal by
θ .
We find
sin
θ
=
a
r
,
cos
θ
=

x
r
.
Dividing gives
x
=

a
cot
θ ,
then
dx
=
a
sin
2
θ
dθ
=
r
2
a
dθ .
The contribution to the magnetic field at
O
from an incremental line segment
dx
at point
P
is
dB
CA
=
μ
0
4
π
i dx
r
2
sin
θ
=
μ
0
i
4
π
dθ
a
sin
θ.
Every segment part of
CA
gives a contribu
tion to the field which points out of the paper.
Thus the total field due to
CA
is
B
CA
=
Z
dB
CA
=
μ
0
i
4
π a
Z
sin
θ dθ .
The integration starts at infinity, where the
angle
θ
is 0
◦
.
The integration ends at
A ,
where the angle is 90
◦
=
π/
2 radians.
Thus
the integral is
Z
π/
2
0
sin
θ dθ
=

cos
θ
fl
fl
fl
fl
π/
2
0
= 1
,
and so the magnetic field is
B
CA
=
μ
0
i
4
π a
.
002
(part 2 of 4) 10 points
Find the magnitude of the field
B
AD
at
O
due
to the circular arc
AD
alone.
1.
B
AD
=
13
96
μ
0
i
a
correct
2.
B
AD
=
5
32
μ
0
i
a
3.
B
AD
=
7
32
μ
0
i
a
4.
B
AD
=
1
6
μ
0
i
a
5.
B
AD
=
3
16
μ
0
i
a
6.
B
AD
=
23
96
μ
0
i
a
7.
B
AD
=
17
96
μ
0
i
a
8.
B
AD
=
5
24
μ
0
i
a
9.
B
AD
=
11
48
μ
0
i
a
10.
B
AD
=
7
48
μ
0
i
a
Explanation:
We sum up the incremental current contri
butions along the arc:
Δ
B
AD
=
μ
0
4
π
i
Δ
x
r
2
sin
θ.
In this case,
Δ
x
=
a
Δ
θ
which is the formula for the arc length of an
arc segment with angle Δ
θ
and radius
a
.
Note:
The
r
in the previous formula is the
distance to the field point, but for the circular
arc this is constantly
a
.
Finally, we note that the angle
θ
is always
90
◦
along the arc. Thus
Δ
B
AD
=
μ
0
4
π
i a
Δ
θ
a
2
sin 90
◦
=
μ
0
i
4
π a
Δ
θ .
The arc has an angular span (given in the
question) of
Δ
θ
=
13
48
2
π .
Thus, the magnitude of the field
B
AD
at
O
due to the circular arc
AD
alone is
B
AD
=
μ
0
i
4
π a
13
48
2
π
=
13
96
μ
0
i
a
.
003
(part 3 of 4) 10 points
Platt, David – Oldquiz 3 – Due: Nov 13 2005, 4:00 am – Inst: Ken Shih
3
Select the choice which represents the magni
tude and the direction of the resultant mag
netic field
O
due to the vector sum of all
contributions:
~
B
=
~
B
CA
+
~
B
AD
+
~
B
DF
.