MAGNETIC FIELD AROUND CURRENT-CARRYING WIRES
How will we tackle this? Plan:
r
1 : Will look at contribution dB to the total magnetic field atr some
point in space due to the current in a small segment ds of wire
dB
st
ds
I
• Biot-Savart Law
r
2 : Will use Biot-Savart Law to find B along the axis of a current loop.
nd
3rd: Will use Biot-Savart Law to get result for field around a “long” wire
µ0 I
−7
B
=
µ
=
4
π
×
10
T ⋅ m/A
a
•
dir.
by
RH-rule,
0
2π a
4th: Will look at force between parallel current-carrying wires
µ0 I
th
B
=
5 : Ampere’s Law: another way to get
2π a for field around long wire
I
B
I
P
BIOT-SAVART LAW (22.7)
r
field at
Expression for contribution dB to magnetic
r
P due to current I in small segment ds of wire
r
r
I
θ
r
• r̂ is UNIT VECTOR pointing ds → P
ds
r
d
s
• r is DISTANCE from
→P
r
• θ is the ANGLE between ds and r̂
r
r
Then contribution to B at P from ds is
r µ 0 I dsr × rˆ
dB =
4π r 2
−7
• µ 0 = 4π × 10 T ⋅ m/A is the permittivity of free space
r
r
d
s
× rˆ (can also use RH-rule with thumb → current)
d
B
• Direction of
is from
r µ 0 I dsr × rˆ
BIOT-SAVART LAW: understanding dB = 4π r 2
r
• For drawing, direction of d s × rˆ is out of screen/page
P
r
r
ds → r̂ ; thumb shows direction
o Fingers
sweep
r
d
s
× rˆ
of
r
r
d
B
• So
at P due to ds points out of screen/page
r
θ
ds
r
d
s
× rˆ = ds sin θ
• Magnitude
r
r
dB from ds are maximum for points on
o For a given r, contributions
r
d
s
plane perpendicular to
r
r
r
o Current in ds makes NO contribution to dB at points along direction ds
• For TOTAL FIELD at P, must sum (integrate) contributions from all
segments of wire (don’t panic. We will only do special cases)
I
EXAMPLE: 22.6 in text – Magnetic field on axis of circular current loop
Look at a loop of radius r located in the yz plane carrying a current I. What is the
magnetic field a distance x from the centre of the loop along its axis?
y
ds
dB
θ
r
R
R
θ
dB
x
x
z
r
ds
P
I
MAGNETIC FIELD AROUND A LONG (INFINITE) WIRE
r
r
Biot-Savart Law gives contribution dB at P due to current I in segment ds
r µ 0 I dsr × rˆ
• dB = 4π r 2
dB
P
r
r
r
o dB points out of screen/page at P by RH-rule
ds
r
To get B at P due to WHOLE wire, must sum contributions from all angles
r
• Angle between ds and r̂ changes with position along wire!
o Will state result and then sketch derivation
I
MAGNETIC FIELD AROUND A LONG (INFINITE) WIRE
• Result of using Biot-Savart Law:
r
o Magnetic field lines circle wire → no component of B parallel to wire
B
I
Iout
r
B
• Magnitude of
inversely proportional to perpendicular a distance from wire
r µ0 I
B=
2π a
(IMPORTANT RESULT)
• Direction of magnetic field lines:
r
B
o Another Right-Hand Rule: thumb along I ; fingers curl in direction of
r µ0 I
“Sketch” of how we get B = 2π a from the Biot-Savart Law
r
Looking for B at a point P located perpendicular distance a from wire
• Start with contribution
to field at P due to
r
segment ds located distance s along the wire
from the point closest to P :
r µ 0 I dsr × rˆ
dB =
4π r 2
r
r
d
s
× rˆ = ds sin θ
o Note that ds × rˆ points out of the screen/page and that
r
r
To get total B at P from infinite wire, could sum dB from s = −∞ to s = +∞ .
r
• Helps to express r and ds in terms of angle θ between ds and r̂ and then
sum from θ = 0 to θ = π
r µ0 I
“Sketch” of how we get B = 2π a from the Biot-Savart Law (continued)
Changing Variables:
a
a
r
=
=
sin
θ
• From
r get
sin θ
a
=
−
tan
θ
s = −a cot θ
• Tricky point: for θ as shown, s is negative so that
s or
• Use derivative to relate ds to dθ
ds
a
2
=
a
csc
θ
=
o From dθ
sin 2 θ
we get
ds =
a dθ
sin 2 θ
r µ0 I
“Sketch” of how we get B = 2π a from the Biot-Savart Law (continued)
Doing the integral. So far, we have:
a
r
=
•
sin θ
a dθ
ds
=
•
sin 2 θ so
r
a dθ
ds × rˆ = ds sin θ =
sin θ
r
So contribution to field at P from ds is
r
µ I ds × rˆ µ 0 I
dB = 0
=
sin θ dθ
4π r 2
4π a
µ0 I π
µ0 I
=
=
−
B
sin
θ
d
θ
cos θ
Magnitude of total field is:
4π a ∫0
4π a
RESULT: field at distance a from wire with current I:
π
=
0
µ0 I
2π a
B=
µ0 I
2π a
µ0 I
=
B
USE
2π a TO FIND MAGNETIC FORCE BETWEEN PARALLEL WIRES
I1
a
B2
F1
B2
a
I2
µ0 I 2
B
=
• Field at I1 due to I2 is 2 2π a
o Exerts a force on length l of I1:
I1
r r
r
F1 = I 1l × B2
µ 0 l I1 I 2
F
I
lB
=
=
o Magnitude of force on length l of I1 is 1 1 2
2π a
F1
I2
MAGNETIC FORCE BETWEEN PARALLEL WIRES
F1 µ0 I1 I 2
• Force on I1 per unit length is l = 2π a toward I2 (for currents in same dir.)
By Newton’s 3rd law,
F2 µ 0 I1 I 2
• Force on I2 per unit length is l = 2π a toward I1 (for currents in same dir.)
In general:
• For parallel conductors, current in same direction:
F µ 0 I1 I 2
o Wires ATTRACT with force per unit length l = 2π a
• For parallel conductors, current in opposite direction:
F µ 0 I1 I 2
o Wires REPEL with force per unit length l = 2π a
Provides definition of AMPERE:
−7
• For 2 wires, 1 m apart, I 1 = I 2 = 1 A → force per unit length is 2 × 10 N/m