B fields from a single current
z
Use Ampere’s law to
calculate B field from
long straight wire
r
r
B
•
d
s
=
μ
i
0
enc
∫
z
Draw Amperian loop
as a circle surrounding
the wire (like the
magnetic field lines)
z
At every point of the
loop
z
z
Magnitude of B is
constant
B and ds are || (both
tangent to the circle)
B fields from a single current
z
B and ds are || so
cosθ = cos0 = 1
r r
B • ds = Bds
z
B constant on loop so
r r
B
•
d
s
=
B
ds
∫
∫
∫ ds = 2π r
r
r
B
•
d
s
=
B
(
2
π
r
)
∫
B fields from a single current
z
Ampere’s law becomes
B ( 2π r ) = μ 0ienc
z
Current enclosed is just i so
μ 0i
B=
2π r
z
Same result as with Biot-Savart law (we used r=R before).
B fields from solenoids
z
z
z
z
What happens if there
are several loops of
wire put together?
A long, tightly wound
helical coil of wire is
called a solenoid
Bend solenoid so ends
meet to make a hollow
donut gives a toroid
Use Ampere’s law to calculate B field for
a solenoid and a toroid
B fields from solenoids
Solenoid’s B field is vector sum of fields
produced by each turn (loop) in solenoid
B fields from solenoids
z
z
z
Near each loop it
looks like an infinite
straight wire
Between the loops
fields tend to cancel
Inside the solenoid,
far from the wire, B
field is parallel to axis
z
An ideal solenoid
z
z
is infinity long with closely
packed turns of wire
has uniform B field which
is parallel to solenoid axis
B fields from solenoids
z
z
z
For points outside the
solenoid B fields from
the upper parts of the
turns tend to cancel
the lower
Ideal solenoid Boutside=0
For a real solenoid can
assume Boutside=0 if
z
z
length >> diameter
Only consider points not
near ends of solenoid
z
Use right-hand rule to
find direction of B field
z
Grasp solenoid so fingers
follow direction of i in
loops, thumb points in B
B fields from solenoids
z
Use Ampere’s law to calculate B field of ideal
solenoid
z
Draw Amperian loop a-b-c-d-a intersecting solenoid
B fields from solenoids
r
r
∫ B • d s = μ 0ienc
z
Integral can be written
as sum of four integrals,
one for each side
r r br r cr r
∫ B • ds = ∫ B • ds + ∫ B • ds
a
b
r r ar r
+ ∫ B • ds + ∫ B • ds
d
c
d
B fields from solenoids
z
b
∫
First integral B field
is || to ds
r
r
B • d s = B [s ] ba = Bh
a
z
z
For sides bc and da
B is ⊥ to ds so
For the length
outside the solenoid
B=0 d r
∫
c
r
B • ds = 0
c
∫
r
r
B • ds =
b
a
∫
r
r
B • ds = 0
d
∫
r
r
B • d s = Bh
B fields from solenoids
z
Now need to find amount
of current enclosed
∫
z
z
z
r
r
B • d s = μ 0 i enc
Single coil has current i
But Amperian loop
encloses several coils so
total current is
where n is the number of
turns per unit length
i enc = inh
N
n=
L
z
z
N = total #
of turns
L = length
B fields from solenoids
z
Substituting into
Ampere’s law
∫
r
r
B • d s = μ 0 i enc
Bh = inh
z
For ideal solenoid:
B = μ0in
z
n is # turns/length
z
B field of solenoid
z
z
does not depend on
diameter or length of
solenoid
is uniform over its cross
section
B fields from toroids
z
Calculate B field for a
toroid using Ampere’s law
∫
z
z
r
r
B • d s = μ 0 i enc
Choose Amperian loop to
be a concentric circle
inside toroid
B and ds are parallel
along entire loop so
r r
∫ B • ds = B ∫ ds = B ( 2πr )
B fields from toroids
z
Current enclosed by loop is
i enc = iN
∫
r
r
B • d s = μ 0 i enc
B( 2πr ) = μ0iN
z
B field for toroid is
μ0iN
B=
2πr
B fields from toroids
z
Toroid – B field is not
constant over its cross
section
μ0iN 1
B=
2π r
z
z
N = total # of turns
Use right-hand rule to find
direction of B field
z
z
Grasp toroid with fingers in
direction of current in
windings, thumb points in B
B = 0 outside toroid
B fields from finite wires
z
Ampere’s law becomes
B ( 2π r ) = μ 0ienc
z
Current enclosed is just i so
μ 0i
B=
2π r
z
Same result as with Biot-Savart law (we used r=R before).
B fields from finite wires
z
Calculate B field inside
a long straight wire
r r
∫ B • ds = μ0ienc
z
Again B and ds are ||
and B is a constant so
r r
∫ B • ds = B ∫ ds = B ( 2πr )
B ( 2π r ) = μ 0ienc
B fields from finite wires
z
z
Need to find ienc
Current is uniformly
distributed so i
enclosed by loop is α
to area enclosed
πr
=i 2
πR
2
ienc
πr
B ( 2π r ) = μ 0i
2
πR
2
⎛ μ 0i ⎞
B=⎜
r
⎟
2
⎝ 2π R ⎠