Transcript Document

Lecture 9 Magnetic Fields due to Currents Chp.
30
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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
Physlet
Topics
– Magnetic field produced by a moving charge
– Magnetic fields produced by currents. BigBite 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
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Demos
– Torque on a current loop(galvanometer)
– Iron filings showing B fields around wires with currents.
– Compass needle near current carrying wire
– BigBite as an example of using a magnet as a research tool.
– Force between parallel wires carrying identical currents.
Magnetic Fields due to Currents
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Torque on a coil in a magnetic field demo– left over from last time
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So far we have used permanent magnets as our source of magnetic
field. Historically this is how it started.
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In early decades of the last century, it was learned that moving charges
and electric currents produced magnetic fields.
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How do you find the Magnetic field due to a moving point charge?
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How do you find the Magnetic field due to a current?
– Biot-Savart Law – direct integration
– Ampere’s Law – uses symmetry
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Examples and Demos
Crossed magnetic and electric fields
F  qE  qvxB

y
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 oriiginal position
qE  qvB
v  E /B
4 Solve for

m B 2 L2

q 2yE
Show demo of CRT

This ratio was first measured by Thompson
to be lighter than hydrogen by 1000
Hall Effect
Crossed fields to measure charge carrier density n
neA
v 
i
d
Bid
n
VeA
V=Ed
eE  evd B
E V
vd  
B Bd
Topic
A moving charge produces a magnetic field.
q

rˆ
v
r
  0 v  rˆ
B
q 2
4
r
 0  4  10 7
N
A2
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 at other points.
3.
The direction is given by the RHR rotating v into rˆ .
Example:
A point charge q = 1 mC 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?
rˆ
90o
108 m/s
0 v
B
q 2
4 r
B  10
7 N
A2
B  104 T
3
10 C 10
8 m
s

1
10 6 m 2
To find the E field of a charge distribution use:
Use:
 kdq rˆ
dE  2
r
To find the

B field of a current distribution use:
  0 ids  rˆ
dB 
4 r 2
Note that ids 
dq
ds

ds  dq
 dqv
dt
dt
Topic: Biot – Savart Law

 0 idl  rˆ
B

Use
 4 r 2 to find B field at the center of a loop of wire.
i
R
Loop of wire lying in a plane. It has radius R and total
current i flowing in it.
First find

dl
rˆ

dl
B
rˆ
 0i
2R

dl  rˆ

dl  rˆ
is a vector coming out of the paper at the same
angle anywhere on the circle. The angle is
constant.
B   dB 
 0 idl  0 i
0 i

dl

2R
2
2 
2

4 R
4 R
4 R
Magnitude of B field at center of
loop. Direction is out of paper.
i
R
kˆ
  0i
B
kˆ
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
10A
2(.05m)
B  1.2 106 102 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?
i

dl
rˆ
B
0 i
dl
2 
4 R
Total length of arc is S.
0
R
P
B
0 i
S where S is the arc length S = R0
2
4 R
0 is in radians (not degrees)
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
i
i
0
Radius R
i
Radius R/2
What is the magnitude and direction of B at the origin?
0 i
S
2
4 R
0 i
B
0
4 R
 i
i
 i
B  0 ( 0 
0 )   0 0
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
 
 B  dl   0 Ic
Current enclosed by the path
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).
i

By symmetry B

dl r
i

dl
 
 B  dl   Bdl  B  dl  B2r   0i
0 i
B
2 r
Suppose
i = 10 A
B  2 107 10101
R = 10 cm
B  2 107 T
 0  4  10 7
N
A2
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
 oi
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.
iaib 0
2d
iaib 0
Fba /L 
2d
Fba  L
 0 ia
Ba 
2d
Fba  ib L x Ba
0ia
 ibL
2d

LB

and directed towards
wire a (wires are attracted).
Suppose one of the currents is
in the opposite
 direction? What
direction is Fba ?
Example: Find magnetic field inside a long, thick wire of radius a
Cross-sectional view
a
Path of integral dl
r
2r
 
 B  dl   Bdl  B  dl  B 2r
B 2r  u 0 IC
0
IC = current enclosed by the path
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  0 ni
n is the number of turns per meter

dl
d
c

dl

dl
B

dl
a
b
First evaluate the right side, it’s easy
 
 B  d l   0 IC
IC is the total current enclosed by the path
IC = nhi
The number of loops of current; h is the length of
one side.
Right side = 0nhi

dl
d
c

dl

dl
B

dl
a
b
Evaluate left side:
c
d
a
  b
 B  dl   B  dl   B  dl   B  dl   B  dl
a
=
b
Bh
+
c
0
+
d
0
+
0
Bh = 0nhi
B = 0ni
n = the number of loops or turns per meter
 
 B  d l   0 IC
Toroid
 Bdl  
0
Ni
B2r   0 Ni
 0 Ni
B
2r
N is the total
number of turns
a<r<b
a
b

dl
 
B  dl  Bdl
cos0o  1
Tokamak Toroid at Princeton
I = 73,000 Amps for 3 secs
B = 5.2 T
Magnetic dipole inverse cube law
z
0 
B
3
2 z
Warm up Problem Set 9
Warm up set 9 Due 8:00 am Thursday
1.
HRW6 30.TB.02. [119973] A "coulomb" is:
one ampere per second
an abbreviation for a certain combination of kilogram, meter and second
the quantity of charge which will exert a force of 1 N on a similar charge at a distance of 1 m
the amount of current in each of two long parallel wires separated by 1 m,
which produces a force of 2 10-7 N per meter
the amount of charge which flows past a point in one second when the current is 1 A
2. HRW6 30.TB.03. [119974] Electrons are going around a circle in a counterclockwise direction as shown.
At the center of the circle they produce a magnetic field that is:
to the left
into the page
out of the page
zero
to the right
3. HRW6 30.TB.07. [119978] Lines of the magnetic field produced by a long straight wire carrying a current:
are circles concentric with the wire
leave the wire radially
are in the direction of the current
are lines similar to those produced by a bar magnet
are opposite to the direction of the current