Download Lecture

Quiz 1
100
90
80
70
60
Borderline
Series1
50
Trouble
40
30
Deep Trouble
20
10
0
10
20
30
40
From Last time
Temperature effects on resistance
Permanent magnets and gaussmeter.
Electromagnets - BigBite for Research
Force on a wire in a magnetic field - Demo
Force on a coil in a magnetic field - Demo
Crossed E/B fields. Discovery of the electron
Lecture 9 starts here
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.
F
i
.
P
a
i
F
B
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.
Equal and opposite forces F = iaB are exerted on the sides a.
No forces exerted on b since i B
Since net force is zero, we can evaluate t (torque) about any point. Evaluate it
about P (cm).
For N loops we mult by N
t = Fbsinq
F = iaB
t = iabBsinq
t = NiabBsin q = NiABsin q
A=ab=area of loop
Torque on a current loop
t = NiABsin q
n̂ q
 = NiA = magnetic dipole moment
B
t =  Bsinq
t =B
Torque tends to rotate loop until plane is^ to B
n̂
( n̂ parallel to B).
n̂
q
Must reverse current using
commutator to keep loop
turning.
Loop will rotate and stop when
the plane is perpendicular to B
Magnetic dipole moment is in the
same direction as the normal.
http: //hyperphysics .phy-astr .gsu.edu/hbase/magnetic/motdc .html#c4
Split ring
commutator
+
-
Show Galvanometer Demo
Electron moving with speed v in a crossed electric
and magnetic field in a cathode ray tube.
F = qE  qvxB

e
y
Electric field bends particle upwards
Magnetic field bends it downwards
Discovery of the electron by J.J. Thompson in 1897
1.
E=0, B=0 Observe spot on screen
2. Set E to some value and measure y the deflection
qEL2
y=
2mv 2
3. Now turn on B until spot returns to the original position
qE = qvB
E
v=
B
4 Solve for
m B2L2
=
q 2yE
Show demo of CRT
This ratio was first measured by Thompson
to be lighter than hydrogen by 1000
How do you calculate the
magnetic field from a moving charge
or from a current flowing in a wire?
What do the flux lines look like?
Lecture 9 Magnetic Fields due to Currents Ch. 30
•
•
•
•
Cartoon - Shows magnetic field around a long current carrying wire and a loop of
wire
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 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.
Polling
Magnetic Fields due to Currents
•
•
•
•
•
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
Topic: Moving charges produce magnetic fields
q
q
r̂
v
r
  0 v  rˆ
B=
q 2
4 r
Determined from
Experiment
1.
Magnitude of B is proportional to q, v, and 1/r2.
2.
B is zero along the line of motion and proportional to sin qat
other points.
3.
The direction is given by the RHR rotating
4.
Magnetic permeability
 0 = 4 10 7
v into
N
A2
Demo Sprinkle iron filings around a wire carrying current
r̂
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?
r̂
90o
108 m/s
0 v
B=
q 2
4 r
B = 10
7 N
A2
B = 10 4 T
3
10 C 10
8 m
s

1
10 6 m 2
Topic: A current produces a magnetic field
Recall the E field of a charge distribution
“Coulombs Law”
 kdq
dE = 2 r̂
r

To find the B field of a current distribution use:
r̂
Biot-Savart Law
 0 ids  r̂
dB =
4 r 2
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.
i
 0 i dl  r̂
B= 
4 r 2

d l  r̂

dl
q
P
R
r̂
is a vector coming out of the paper The angle
between dl and r is constant and equal to 90
degrees.
dl  r̂ = dl sin q k̂
sin q = sin 90 = 1
dl  r̂ = dl k̂
B =  dB =
 0 idl  0 i
0 i
k̂  2 =
k̂ 2  dl =
k̂ 2 2 R
4
R
4 R
4 R
Magnitude of B field at center of
loop. Direction is out of paper.
i
R
k̂
 0i
B=
k̂
2R
Magnitude is
Direction is k̂
0i
2R
Example
Loop of wire of radius R = 5 cm and current i = 10 A. What is B at the
center? Magnitude and direction
i
B=
 0i
2R
B = 4 107
10 A
2(.05m)
B = 1.2 10 6 10 2 T
N
A2
B = 1.2 104 T = 1.2 Gauss
Direction is out of
the page.
What is the B field at the center of a segment or circular
arc of wire?
v
 0 i dl  r̂
B= 
4 r 2
i
B =  dB =

dl
r̂
=
0 i
k̂  dl
2
4 R
0 i
k̂ S
2
4 R
Total length of arc is S.
dl r̂
dl r̂
q0
R
P
dl  r = rdl sinq k̂
sin q = sin 0 = 0
where S is the arc length S = Rq0
q0 is in radians (not degrees)
B=
0 i
q 0 k̂
4 R
q 0 = 100 deg=100 deg 
6.28 rad
= 100  0.0174 = 1.74 rad
360 deg
Why is the contribution to the B field at P equal to zero from
the straight section of wire?
Suppose you had the following loop.
Find magnetic field at center of arc length
What is the magnitude and direction of B at the origin?
B=
0 i
S
4 R 2
0 i
q
4 R 0
 i
 i
i
B = 0 ( q0 
q0 ) =  0 q0
4 R
R/2
2 R
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
Examples
 
 B  dl =  0 Ic
Current enclosed by the path
This dl is along the Amperian loop not
along the wire. dl is the same as ds below
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).

dl r
i
By symmetry

B
i

dl
 
 B  dl =  Bdl = B  dl = B2r =  0i
0 i
B=
2 r
Suppose
i = 10 A
R = 10 cm
 0 = 4 10 7
N
A2
2  10 7  10
B=
10 1
B = 2  10 5 T
Show Fe fillings around a straight wire with current,
current loop, and solenoid.
Rules for finding direction of B field from a current
flowing in a wire
 0i
B=
2r
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.
Fba ia ib  0
=
L
2 d
 0i a
Ba =
2d
(Given by Ampere’s Law)
 

Fba = i bL x Ba
 0i a
= i bL
2d
The force is directed towards the
wire (wires are attracted).
Show Demo
Suppose one of the currents is
in the opposite
 direction? What
direction is Fba ?
Example: What is the magnetic field inside the wire? Find the
magnetic field inside a long, thick wire of radius a
2r
 
 B  dl =  Bdl = B  dl = B2r
B2r = u0Ic
0
IC = current enclosed by the circle whose radius is r
r 2
r2
IC = 2 i = 2 i
a
a
r2 1
B = 0 2 i
a 2r
 0ir
B=
2a 2
Example: Find field inside a
solenoid. See next slide.
Solenoid
B = 0ni
n is the number of turns per meter

dl
d
c

dl

dl
B
a

dl
b
First evaluate the right side of the integral , it’s easy
 
 B  dl =  0 IC
IC is the total current enclosed by the path
Ic = nhi
n is the the number of loops of current per meter
and h is the length of one side.
Right side =
0 nhi
d

dl

dl
c

dl
a
B

dl
b
Evaluate left side of the integral :
c
d
a
  b
 B  dl =  B  dl   B  dl   B  dl  B  dl
a
b
c
d
= Bh  0  0  0
Bh = 0 nhi
B = 0 ni
n = the number of loops or turns per meter
 
 B  dl =  0 IC
Toroid
 Bdl = 
0
Ni
B 2r =  0 Ni
 0 Ni
B=
2r
N is the total
number of turns
a<r<b
a
b

dl
 
B  dl = Bdl
cos 0 o = 1
Tokamak Toroid at Princeton
I = 73,000 Amps for 3 secs
B = 5.2 T
Magnetic dipole inverse cube law
z
P
.

0 
B=
2 z 3
Magnetic field lines for a bar magnet
and a current loop are similar
Chapter 29 Problem 5
In Figure 29-36, two circular arcs have radii a =
13.1 cm and b = 10.6 cm, subtend angle θ = 74.0°,
carry current i = 0.388 A, and share the same
center of curvature P. What is the magnitude and
direction of the net magnetic field at P? (Take out
of the page to be positive.)
Figure 29-36
Chapter 29 Problem 41
A 180 turn solenoid having a length of 22 cm and a diameter of 10
cm carries a current of 0.39 A. Calculate the magnitude of the
magnetic field B inside the solenoid.
Chapter 29 Problem 49
A student makes a short electromagnet by winding 150 turns
of wire around a wooden cylinder of diameter d = 7.0 cm.
The coil is connected to a battery producing a current of
4.0 A in the wire.
(a) What is the magnetic moment of this device?
(b) At what axial distance z d will the magnetic field of this
dipole have the magnitude 5.0 µT (approximately onetenth that of Earth's magnetic field)?