The Biot-Savart Law with Applications.

wire
I
dB
r
θ
ds
r
P
The objective is to find the magnetic field at
a point P created by the current in the blue
wire. We must first find the magnetic field
dB produced by the current in a differential
length ds.
ds - a differential length vector tangent to wire in the direction of the current flow
r a unit vector pointing from ds towards “P”
r - distance from ds to point “P”
θ - angle between r and ds
dB – differential field at P
1

wire
I
dB
r
r
θ
ds
P
The differential field and its magnitude are:
r
d B =
r
µ
4π
r
0 Id s × rˆ
2
µ 0 I sin θ ds
dB =
4π
2
r
dB is perpendicular to both r and ds
The field created by the complete wire would be obtained by
integration of dB over the whole wire.
µ 0 = 4π x 10 -7 T.m A
Y
wire
ds
Step 1: Field dB due to current in ds:
r
r
µ 0 Id s × rˆ
d B =
4π
2
r
dθ
θ
r
r
I
X
The magnitude of ds is rdθ. The direction of
the vector cross product is found from the
RHR.
dB
^
The direction is out of page in the +k direction:
)
ds
× rˆ = krdθ ˆ
r Id
d B kˆ µ 0 θ
=
4π
r
2

Y
θ = π/2
ds
Step 2: Integrate to find the total field. The
integration limits are zero to π/2.
dθ
r
r
I
θ
X
B θ = 0
B is out of the page
r
B
π
r kˆ
2
µ I kˆ
0 µ I
π
0 2
= ∫ dB = dθ
= [ θ ]
0
4πr
∫
4πr
0
I
kˆ
µ 0
=
8r
constants
P
wire
r^
ds
I
The field at P created by the current in the small length ds of
the current carrying wire is zero since:
ds
r × rˆ = dssin 180 =
( ) 0
The total field created by the complete straight line segment is zero.
3

Field of a Straight Wire at an Off-Axis Point
P
Y
a
wire
L 1
I L 2
X
We want to find the magnetic field at the location P (0,a)
created by the current in a straight wire located along the X
axis from L 1 to L 2 .
P
a
Y
r = (x 2 + a 2 ) 1/2
r^
180-θ θ ds X
L
x
1
L 2
Step 1: We find the field caused by the
current in ds. Using the RHR:
r
d s × rˆ
= kˆ
( dssinθ
)
We replace ds by dx and sinθ by:
sin
( θ ) = sin( 180−θ)
=
a
r
r
dB =
µ 0
4π
v
Ids × rˆ
r
2
ˆ µ 0Idx
⎛ a
= k
2
⎜
4πr
⎝ r
⎞
⎟
⎠
ˆ µ 0Iadx
= k
3
4πr
= kˆ
4π
µ
Iadx
0
3
2 2
( x + a )2
4

B
We integrate dB over the limits x = L 1 to x = L 2 :
L
r
2
ˆ µ 0Ia
= ∫ dB = k
4π
∫
L
r ⎡
= ˆ µ 0I
B k ⎢
4πa
⎢
⎣
1
dx
ˆ µ 0Ia
⎡
= k ⎢
( ) 3 4π
2 2
⎢ 2 2 2
x + a 2 ⎣a
( x + a )
L
−
2
( + ) 1
( + ) 1
⎥⎥ 2 2 2 2
L a L1
a ⎦
2
2
2
Usually it is more convenient to obtain the magnitude of this field from the
following equation and obtain the correct direction from a new right hand rule
shown on the next slide.
L
1
⎤
x
1
2
⎤
⎥
⎥
⎦
L
L
2
1
µ 0I
B =
4 π a
L
1
2
( ) 1
2 2 2
L + a ( L + a ) 2
2
2
2
−
1
L
1
B
B
These diagrams shown end views of two straight wires.
The top wire has a current directed out of the page while the
bottom wire has a current into the page. The magnetic field
is circular about each wire. The direction is easily obtained
from a new RHR of convenience.
X
Imagine grasping the wire with your right hand
with your thumb extended in the current direction.
Your fingers will wrap around the wire in the
direction of the circular magnetic field.
5

L 1
= -L
P
a
L>>a
L 2
= L
µ 0 I
B =
4 π a
1
2 2
( ) 1
2 2 2
L + a ( L + a ) 2
2
L
2
−
1
L
1
µ
0I
B =
4πa
L
−
− L
µ
≈
L
L
− L
−
2 2
( L + a ) 1
2 2 2
( L + a ) 1
2
4πa
L
0
I
µ 0I
B =
2πa
As long as we are not too close to either end.
6