Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two radial straight segments and a circular arc of
radius R that subtends angle .
A´
I see three “parts” to the wire.
A’ to A
A
A to C

O
C
I
C´
R
Thanks to Dr. Waddill for the use of the diagram.
C to C’
As usual, break the problem up
into simpler parts.
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two radial straight segments and a circular arc of
radius R that subtends angle .
dB =
A´
I
ds
rˆ
A
O
4π
2
ds  rˆ = ds rˆ sin 0 = 0
C
R
r
For segment A’ to A:
C´

μ 0 I d s  rˆ
dB A'A = 0
B A 'A = 0
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two radial straight segments and a circular arc of
radius R that subtends angle .
dB =
A´
4π
r
2
For segment C to C’:
A
ds
rˆ

O
μ 0 I d s  rˆ
C
R
I
C´
ds  rˆ = ds rˆ sin 180  = 0
dB CC' = 0
B CC' = 0
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two straight segments and a circular arc of radius R
that subtends angle .
A´
Important technique, handy for
homework and exams:
ds
rˆ
A
rˆ

O
C
R
ds
C´
The magnetic field due to wire
segments A’A and CC’ is zero
because ds is either parallel or
antiparallel to rˆ along those
paths.
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two radial straight segments and a circular arc of
radius R that subtends angle .
dB =
A´
A
rˆ
O
C
R
4π
r
2
For segment A to C:
ds

μ 0 I d s  rˆ
I
C´
ds  rˆ = ds rˆ sin 90  = 0
= d s 1  1 
= ds
dB AC = dB AC =
μ 0 I ds
4π R
2
A´
μ 0 I ds
dB AC =
A
ds

O

B AC =
rˆ
C
I
R
B AC =
B AC =
B AC =
2
d B AC =
a rc
C´
B AC =
4π R
μ 0I
4π R
μ 0I
4πR
μ 0I
4π R
R
2
a rc
μ 0I
4π R

2

a rc
ds
μ 0I ds
4π R
The integral of ds is just
the arc length; just use
that if you already know it.
dθ
a rc

dθ
a rc
θ
2
We still need to provide the
direction of the magnetic field.
A´
A
Cross ds into rˆ . The direction is
“into” the page, or .
ds
rˆ

O
C
I
C´
If we use the standard xyz axes,
the direction is -kˆ .
y
R
z
B = B A'A + B AC + B CC'
μ 0Iθ ˆ
=k
4π R
x
Important technique, handy for
exams:
A´
A
Along path AC, ds is
perpendicular to rˆ .
ds
rˆ

O
C
R
I
C´
ds  rˆ = ds
ds  rˆ = ds
rˆ sin 90 