example: calculate the magnetic field at point o due to the wire segment shown. the wire carries...
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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 .
I
A´
C
A
R
Thanks to Dr. Waddill for the use of the diagram.
C´
O
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 .
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 .
I see three “parts” to the wire.
As usual, break the problem up into simpler parts.
A’ to A
A to C
C to C’
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 .
I
A´
C
A
R
C´
O
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 .
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 .
For segment A’ to A: dsr
ˆ ˆ 0
ds r =ds r sin =0
A'AdB =0
A'AB =0
ˆμ
4π
02
I ds rdB=
r
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´
C
A
R
C´
O
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 .
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 .
For segment C to C’: dsr
ˆ ˆ
ds r =ds r sin 180 =0
CC'dB =0
CC'B =0
ˆμ
4π
02
I ds rdB=
r
I
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´
C
A
R
C´
Important technique, handy for homework and exams:
The magnetic field due to wire segments A’A and CC’ is zero because ds is either parallel or antiparallel to along those paths.
rO
dsr
ds
r
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 .
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 .
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 .
For segment A to C:
ˆ ˆ
ds r =ds r sin 90 =0
ˆμ
4π
02
I ds rdB=
r
A´
C
A
R
C´
O
Ir
ds
=ds 1 1
=ds
μ
4π
0
AC AC 2
I dsdB =dB =
R
A´
C
A
R
C´
O
Ir
ds
μ
4π0
AC 2
I dsdB =
R
μ
4π 0AC AC 2
arc arc
I dsB = dB =
R
μθ
4π0
AC
IB =
R
μ
4π 0AC 2
arc
IB = ds
R
The integral of ds is just the arc length; just use that if you already know it.
μθ
4π 0AC 2
arc
IB = R d
R
μθ
4π 0AC
arc
IB = d
R
We still need to provide the direction of the magnetic field.
A´
C
A
R
C´
O
Ir
ds
μ θ ˆ4π
0
A'A AC CC'
IB = B + B + B =- k
R
If we use the standard xyz axes, the direction is ˆ-k.
Cross into . The direction is “into” the page, or .
ds r
x
y
z
A´
C
A
R
C´
O
Ir
ds
Important technique, handy for exams:
Along path AC, ds is perpendicular to .
r
ˆ ˆ
ds r = ds r sin 90
ˆ
ds r =ds