View
1
Download
0
Embed Size (px)
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
• Warmup problem
• Topics
– Magnetic field produced by a moving charge
– Magnetic fields produced by currents. Big Bite as an example.
– Using BiotSavart 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 of segment 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?
– BiotSavart 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
i a
b F
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 it about 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.phyastr.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
qEL y =
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
E v
qvBqE
=
=
4 Solve for yE2
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
rv qB
! =
! !
"
µ
2
7 0 104
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 (1x103C) 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/s 90o
r̂ 2
0
4 r
v qB
!
µ =
26
83
2
7
10
1 101010
m CB
s
m
A
N
!
!! """=
TB 4
10=
Recall the E field of a charge distribution
“Coulombs Law”
r̂ r
kdq Ed
2 =
!
To find the field of a current distribution use:B !
2
0
r
r̂sid
4 Bd
!
"
µ =
! !
BiotSavart 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
r 2
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" k̂ idl
R 2! =
µ0
4" k̂ i
R 2 dl! =
µ0
4" k̂ i
R 2 2"R
k̂ R2
i B
0µ = !
Magnitude of B field at center of
loop. Direction is out of paper.
R k̂ i
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" k̂ idl
R 2!
= µ0
4"
i
R 2 k̂ dl li
l f
! = µ0
4"
i
R 2 k̂ l
li
l f
= µ0
4"
i
R 2 k̂(l f # li )
= µ0
4"
i
R 2 k̂(2"R # 0)
= µ 0 i
2R k̂
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
i B
2
0µ =
)05(.2
10 104
2
7
m
A B
A
N! "= #
TB 26
10102.1 !"= #
GaussTB 2.1102.1 4
=!= " 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
i ld !
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
R 2 k̂ dl!
= µ0
4"
i
R 2
k̂ S
! B =
µ0
4!
i
R "
0 k̂
where S is the arc length S = R"0
"0 is in radians (not degrees)
B = µ0
4!" i d ! l # r̂
r 2
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
i B =
Suppose you had the following loop.
What is the magnitude and direction of B at the origin?
S R
i B
2
0
4!
µ =
0 0
00 0
2 )
2/ (
4 !
"
µ !!
"
µ
R
i
R
i
R
i B #=#=
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
i B
!
µ
2
0
= Suppose i = 10 A
R = 10 cm T 102
10
10102
5
1
7
!
!
!
"=
"" =
B
B
Show Fe fillings around a straight wire with current,
current loop, and solenoid.
2
7 0 104
A
N! "= #µ
Rules for finding direction of B field from a current
flowing in a wire
r2