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Lecture 9 Magnetic Fields due to Currents Ch. 30• Cartoon - Shows magnetic field around a long current carrying wire and a loop of wire
• Opening Demo - Iron filings showing B fields around wires with currents
• Warm-up problem
• Topics
– Magnetic field produced by a moving charge
– Magnetic fields produced by currents. Big Bite as an example.
– Using Biot-Savart Law to calculate magnetic fields produced by currents.
– Examples: Field at center of loop of wire, at center of circular arc of wire, at center ofsegment of wire.
– Amperes’ Law : Analogous to Gauss’ L:aw in electrostatics, Useful in symmetric cases.
– Infinitely long straight wire of radius a. Find B outside and inside wire.
– Solenoid and Toroid Find B field.
– Forces between current carrying wires or parallel moving charges
• Demos
– Torque on a current loop(galvanometer)
– Iron filings showing B fields around wires with currents.
– Compass needle near current carrying wire
– Big Bite as an example of using a magnet as a research tool.
– Force between parallel wires carrying identical currents.
Reminders
Ungraded problem solutions are on website?
UVa email activated
Quiz tomorrow
Class in afternoon 1:00 PM
Grades are on WebAssign
Magnetic Fields due to Currents
Torque on a coil in a magnetic field demo– left over from last time
• So far we have used permanent magnets as our source of magnetic
field. Historically this is how it started.
• In early decades of the last century, it was learned that moving charges
and electric currents produced magnetic fields.
• How do you find the Magnetic field due to a moving point charge?
• How do you find the Magnetic field due to a current?
– Biot-Savart Law – direct integration
– Ampere’s Law – uses symmetry
• Examples and Demos
Torques on current loops
Electric motors operate by connecting a coil in a magnetic field to a current supply,which produces a torque on the coil causing it to rotate.
P
F
ia
bF
B
B
Above is a rectangular loop of wire of sides a and b carrying current i.
B is in the plane of the loop and ̂ to a.
Equal and opposite forces are exerted on the sides a.No forces exerted on b since
Since net force is zero, we can evaluate t (torque) about any point. Evaluate itabout P (cm).
iaBF =
Bi
! = NiabBsin" = NiABsin"
i
.
A=ab=area of loop
! = Fbsin"
F = iaB
! = iabBsin"
For N loops we mult by N
Torque on a current loop
! = NiABsin"
µ = NiA = magnetic dipole moment
! = µBsin"
! = µ # B
n̂
!
B
!n̂
Must reverse current using
commutator to keep loop
turning.
Torque !!tends to rotate loop until plane is " to B
( parallel to B).n̂
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.html#c4
Split ring
commutator +
-
Electron moving with speed v in a crossed electric
and magnetic field in a cathode ray tube.
�
! F = qE + q
! v x! B
y
Discovery of the electron by J.J. Thompson in 1897
2
2
mv2
qELy =
1. E=0, B=0 Observe spot on screen
2. Set E to some value and measure y the deflection
3. Now turn on B until spot returns to the original position
B
Ev
qvBqE
=
=
4 Solve foryE2
LB
q
m 22
=This ratio was first measured by Thompson
to be lighter than hydrogen by 1000
Show demo of CRT
Topic: Moving charges produce magnetic fields
q
r
"r̂
2
0 ˆ
4 r
rvqB
!=
!!
"
µ
2
70 104
A
N!"= #µ
Determined fom
Experiment
!v
1. Magnitude of B is proportional to q, , and 1/r2.
2. B is zero along the line of motion and proportional to sin at other
points.
3. The direction is given by the RHR rotating into
4. Magnetic permeability
r̂
!v
!v
Example:
A point charge q = 1 mC (1x10-3C) moves in the x direction with
v = 108 m/s. It misses a mosquito by 1 mm. What is the B field
experienced by the mosquito?
108 m/s90o
r̂2
0
4 r
vqB
!
µ=
26
83
2
7
10
1101010
mCB
s
m
A
N
!
!!"""=
TB4
10=
Recall the E field of a charge distribution
“Coulombs Law”
r̂r
kdqEd
2=
!
To find the field of a current distribution use:B
!
2
0
r
r̂sid
4Bd
!
"
µ=
!!
Biot-Savart Law
Topic: A current produces a magnetic field
r̂
This Law is found from experiment
Find B field at center of loop of wire lying in a plane with radius R
and total current i flowing in it.
B =µ0
4!"i d!l # r̂
r2
R
i
r̂ld !
!is a vector coming out of the paper The angle
between dl and r is constant and equal to 90
degrees.
!B = d
!B! =
µ0
4"k̂idl
R2! =
µ0
4"k̂i
R2dl! =
µ0
4"k̂i
R22"R
k̂R2
iB
0µ=!
Magnitude of B field at center of
loop. Direction is out of paper.
R k̂i
d!l ! r̂ = dl sin" k̂
sin" = sin90 = 1
d!l ! r̂ = dl k̂
r̂
ld!
P
"
Magnitude is µ0i
2R
Direction is k̂
!B = d
!B! =
µ0
4"k̂idl
R2!
=µ0
4"
i
R2k̂ dlli
l f
! =µ0
4"
i
R2k̂ l
li
l f
=µ0
4"
i
R2k̂(l f # li )
=µ0
4"
i
R2k̂(2"R # 0)
=µ0i
2Rk̂
r̂
ld!
P
"
Integration Details
Example
Loop of wire of radius R = 5 cm and current i = 10 A. What is B at the
center? Magnitude and direction
R
iB
2
0µ=
)05(.2
10104
2
7
m
AB
A
N!"= #
TB26
10102.1 !"=#
GaussTB 2.1102.14
=!=" Direction is out of
the page.
i
What is the B field at the center of a segment or circular
arc of wire?
"0 R
ild!
r̂
Total length of arc is S.
Why is the contribution to the B field at P equal to zero from
the straight section of wire?
P
!0 = 100 deg=100 deg "6.28 rad
360 deg= 100 " 0.0174 = 1.74 rad
!B = d
!B! =
µ0
4"
i
R2k̂ dl!
=µ0
4"
i
R2
k̂ S
!B =
µ0
4!
i
R "
0k̂
where S is the arc length S = R"0
"0 is in radians (not degrees)
B =µ0
4!"i d!l # r̂
r2
d!l !!r = rdl sin" k̂
sin" = sin0 = 0
d!l " r̂
d!l " r̂
Find magnetic field at center of arc length
0
0
4!
"
µ
R
iB =
Suppose you had the following loop.
What is the magnitude and direction of B at the origin?
SR
iB
2
0
4!
µ=
00
000
2)
2/(
4!
"
µ!!
"
µ
R
i
R
i
R
iB #=#=
Next topic: Ampere’s Law
Allows us to solve certain highly symmetric current problems for the
magnetic field as Gauss’ Law did in electrostatics.
Ampere’s Law is cIldB 0µ=!"!! Current enclosed by the path
!E"! "d!A =
qenc
#0
Gauss’s Law Charge enclosed by surface
Example: Use Ampere’s Law to find B near a very long, straight
wire. B is independent of position along the wire and only depends
on the distance from the wire (symmetry).
rld!
i i
By symmetry ldB!!
irBdlBBdlldB 02 µ! ====" # ##!!
r
iB
!
µ
2
0
=Suppose i = 10 A
R = 10 cmT 102
10
10102
5
1
7
!
!
!
"=
""=
B
B
Show Fe fillings around a straight wire with current,
current loop, and solenoid.
2
70 104
A
N!"= #µ
Rules for finding direction of B field from a current
flowing in a wire
r2
iB
0
!
µ=
Force between two current carrying wires
Find the force due to the current element of the first wire and the magnetic
field of the second wire. Integrate over the length of both wires. This will give
the force between the two wires.
abbaBLiF!"!
x =
Force between two current carrying wires
Find the force due to the current element of the first wire and the magnetic
field of the second wire. Integrate over the length of both wires. This will give
the force between the two wires.
d
iB
a
a!
µ
2
0
=
abbaBLiF!"!
x =
d
iLi
a
b!
µ
2
0
=
d
ii
L
F
d
iiLF
baba
ba
ba
!
µ!
µ
2
2
0
0
=
=
�
! L !" B and directed towards
wire a (wires are attracted).
Suppose one of the currents is
in the opposite direction? What
direction is ?baF!
From experiment we find
(Given by Ampere’s Law)
Example: Find magnetic field inside a long, thick wire of radius a
rBdlBBdlldB
r
!!
2
2
0
===" # ##!!
cIur2B0
=!
IC = current enclosed by the circle whose radius is r
ia
ri
a
rIC
2
2
2
2
==!
!
ri
a
rB
!µ
2
1
2
2
0= 2
0
2 a
irB
!
µ= Example: Find field inside a
solenoid. See next slide.
Solenoid
�
B = µ0ni
n is the number of turns per meter
ld!
ld!
ld!
ld!
Ba
d
b
c
CIldB 0µ=!"!!
First evaluate the right side, it’s easy
IC is the total current enclosed by the path
The number of loops of current; h is the length of
one side.
Right side =
nhiIc=
nhi0
µ
Evaluate left side:
!! ! !! "+"+"+"="a
d
b
a
c
b
d
c
dlBdlBdlBdlBldB!!
ld!
ld!
ld!
ld!
B
a
d
b
c
n = the number of loops or turns per meter
000 +++= Bh
Bh = µ0nhi
B = µ0ni
Toroid! =" CIldB 0µ
!!
! = NiBdl 0µ
NirB 02 µ! =
r
NiB
!
µ
2
0
=
a < r < b
a
b ld!
BdlldB =!
!!
10cos =o
Tokamak Toroid at Princeton
I = 73,000 Amps for 3 secs
B = 5.2 T
N is the total
number of turns
Magnetic dipole inverse cube law
�
! B =
µ0
2!
" µ
z3
z
.P
Magnetic field lines for a bar magnet
and a current loop are similar
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