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.
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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 .
μ0 I ds  rˆ
dB =
4π r 2
A´
I
ds
r̂
A
For segment A’ to A:
C´

O
ds  rˆ = ds rˆ sin 0 = 0
C
R
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 .
μ0 I ds  rˆ
dB =
4π r 2
A´
A
r̂
ds

O
C
R
I
C´
For segment C to 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´
r̂
A
Important technique, handy for
homework and exams:
ds
r̂

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 r̂ 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 .
μ0 I ds  rˆ
dB =
4π r 2
A´
A
For segment A to C:
ds
r̂

O
C
I
C´
ds  rˆ = ds rˆ sin 90 = 0
= ds 1 1
R
= ds
dB AC
μ 0 I ds
= dB AC =
4π R 2
A´
A
μ 0 I ds
dB AC =
4π R 2
r̂

O
μ0I ds
B AC =  dB AC = 
2
4π
R
arc
arc
ds
C
I
C´
μ 0I
B AC =
4πR 2
R
μ 0I
B AC =
4πR 2
 ds
arc
The integral of ds is just
the arc length; just use
that if you already know it.
 R dθ
arc
μ 0I
B AC =
dθ

4πR arc
μ 0I
B AC =
θ
4πR
We still need to provide the
direction of the magnetic field.
A´
A
Cross ds into r̂ . The direction is
“into” the page, or .
ds
r̂

O
C
I
C´
If we use the standard xyz axes,
ˆ
the direction is -k.
y
R
z
B = B A'A + B AC + B CC' = -
μ 0 Iθ ˆ
k
4πR
x
Important technique, handy for
exams:
A´
A
Along path AC, ds is
perpendicular to r̂ .
ds
r̂

O
C
R
I
C´
ds  rˆ = ds rˆ sin 90
ds  rˆ = ds