Transcript Magnetic Sources The Biot-Savart Law •Magnetic fields go around the wire – they are perpendicular I ˆ dB 2 ds  r to the direction of current r •Magnetic.

Magnetic Sources
The Biot-Savart Law
•Magnetic fields go around the wire – they are perpendicular
I
ˆ
dB
2 ds  r
to the direction of current
r
•Magnetic fields are perpendicular to the separation
0 I ds  rˆ
between the wire and the point where you measure it
B
 r2
4

•Sounds like a cross product!
r
I
ds
0  4 10 T  m/A
7
•Permeability of free space
•The Amp is defined to work out this way
Sample Problem
A loop of wire consists of two quarter circles of radii R
and 2R, both centered at a point P, and connected with
wires going radially from one to the other. If a current I
flows in the loop, what is the magnetic field at point P?
rˆ
rˆ
I
rˆ ds
P
R
0 I ˆ
0 I ˆ 2 R
0 I ˆ ds

k
k

k 2
2

8R
4 R
4
4
R
•Outer loop opposite direction, similar
ds
rˆ ds
•Do one side at a time
•First do one of the straight segments
•ds and r-hat are parallel ds  rˆ  0
•No contribution to the integral
•Other straight segment is the same
•Now do inner quarter loop ds  rˆ  kˆ ds
Binner
ds
B outer
0 I ˆ

k
16 R
2R
0 I ˆ
B
k
16 R
Magnetic Field from a Finite Wire
•Magnetic field from a finite straight wire:
•Let a be the distance the point is from the wire
•Let x be the horizontal separation
P
1
x1
a r
x
cos 1  
cos 2 
ds 2
I
x1
x12  a 2
x2
x22  a 2
0 I
B
4
x2
x2

r   xˆi  aˆj
r  x2  a2
ds  ˆidx
 ˆidx     xˆi  aˆj   Ia kˆ
0
4
x  a 

x2
x1


2
2
2
2 
x2  a
x1  a 
2
x1
0 Ikˆ 


4 a 

2 32
0 I
B
 cos 1  cos  2 
4 a
Warning: My 2 differs
from that of the book
x2

x1
x
dx
2
a

2 32
Magnetic Field from a Wire
•Magnitude is found from the formula
•Direction is found from the right hand rule
•Place thumb in direction of current flow
•Fingers curl in direction of B-field
•Infinite wire:
1  2  0
•Angles are simple
cos 1  cos 2  1
0 I
B
 cos 1  cos  2 
4 a
1
2
I
B
0 I
2 a
a
Sample Problem
A regular hexagon whose center is a distance a = 1 cm from the nearest
side has current I = 4.00 A flowing around it. The current flows N = 500
times around. What is the total magnetic field at the center?
•Draw in the two directions from the center to the corners of
one segment
•Top angle is one-sixth of a circle, or 60 degrees
•Total angles in circle is 180, so other two angles are 60 each
•Use formula to get magnetic field – right hand rule says up.
0 I
0 I
B
 cos 60  cos 60   4 a
4 a
•Multiply by all six side, and then by 500 cycles
Btot 
60 IN

4 a
6  4 107 T  m/A   4 A  500
4  0.01 m 
 0.120 T
I
60
60 60
a
Right Hand Rule for Loops
I
•If you curl your fingers in the direction the current
flows, thumb points in direction of B-field inside the loop
•Works for solenoids too (later)
Force Between Parallel Wires
I1
I2
F
d
L
F2  I 2L  B
•One wire – infinite – creates a magnetic field
•Other wire – finite or infinite – feels the force
•Attractive if current is parallel, repulsive if anti-parallel
B
0 I1
2 d
F 0 I1 I 2

L
2 d
Ampere’s Law (original recipe)
•Suppose we have a wire coming out of the plane
•Let’s integrate the magnetic field around a closed path B   0 I
2 a
•There’s a funky new symbol for such an integral
•Circle means “over a closed loop”
•The magnetic field is parallel to direction of integration
ds cos
0 I
 B  ds  B  ds  2 r 2 r  0 I
•What if we pick a different path?
 0 I 
 B  ds   Bds cos     2 r  rd
ds cos   rd
 B  ds   0 I
ds
d
I
•We have demonstrated this is true no matter what path you take
•Wire need not even be straight infinite wire
•All that matters is that current passes through the closed Ampere loop
r
Announcements
Day
Today
Friday
Monday
ASSIGNMENTS
Read
Quiz
Sec. 30.4-30.5 Quiz 30b
Sec. 31.1-31.3 Quiz 31a
Sec. 31.4-31.6 Quiz 31b
Homework
none
Hwk. 30a
Hwk. 30b
Test Next Wednesday
Review session here
4:30-6:00 Monday
Right
angle
5 cm

13 cm
B
Sorry About Today’s
Reading Quiz
12 cm

2/29
Total angle:
270 + ( - )
Understanding Ampere’s Law
•If multiple currents flow through, add up all that are inside the loop
•Use right-hand rule to determine if they count as + or –
 B  ds   0 I
•Curl fingers in direction of Ampere loop
•If thumb points in direction of current, plus, otherwise minus
•The wire can be bent, the loop can be any shape, even non-planar
7A
5A
2A
1A
 B  ds    3 A 
0
4A
We will later realize that
this formula is imperfect
Using Ampere’s Law
•Ampere’s Law can be used – rarely – to calculate magnetic fields
•Need lots of symmetry – usually cylindrical
A wire of radius a has total current I distributed uniformly across its cross-sectional area.
Find the magnetic field everywhere.
I
I
End-on view
•Draw an Ampere loop outside the wire – it contains all the current
•Magnetic field is parallel to the direction of this loop, and constant around it
•Use Ampere’s Law
 I  B  ds  B ds
0


 2 rB
•But we used a loop outside the wire, so we only have it for r > a
0 I
B
2 r
Using Ampere’s Law (2)
•Now do it inside the wire
•Ampere loop inside the wire does not contain all the current
•The fraction is proportional to the area
Ir  r 2
 2
2
2
Ir  Ir a
I a
a
 0 I r   B  ds  2 rB
I
 Ir
B 0 r  0 2
2 r 2 a
 0 I 2 r
B
2

Ir
2

a
 0
ra
ra
End-on view
Solenoids
•Consider a planar loop of wire – any shape – with a
current I going around it
•Now, stack many, many such loops
•Treat spacing as very closely spaced
•Assume stack is tall compared to size of loop
•Can show using symmetry that magnetic field is
only in vertical direction
•Can use Ampere’s Law to show that it is constant
inside or outside the solenoid
B1  B2
0  0 I   B  ds  B1L 0  B2 L 0
•But magnetic field at infinity must be zero
Boutside  0
Field Inside a Solenoid
•It remains only to calculate the magnetic field inside
•We use Ampere’s law
•Recall, no significant B-field outside
•Only the inside segment contributes
0 I tot   B  ds  Bin L
I tot  NI
•There may be many (N) current loops within this
Ampere loop
•Let n = N/L be loops per unit length
 NI
Bin  0
Bin  0 nI
L
•Works for any shape solenoid, not just cylindrical
•For finite length solenoids, there are “end effects”
•Real solenoids have each loop connected to the next,
like a helix, so it’s just one long wire
L
Magnetic Flux
•Magnetic flux is defined exactly the same way   B  nˆ dA
B

for magnetism as it was for electricity
A cylindrical solenoid of radius 10 cm has length 50 cm and
has 1000 turns of wire going around it. What is the
magnetic field inside it, and the magnetic flux through it,
when a current of 2.00 A is passing through the wire?
Bin 
0 NI
L
 4 10
7
1000  2 A 
T  m/A
0.5 m
Bin  0.00503 T
2
4
2
2


1.579

10
T

m

0.00503
T

0.1
m

B

R
 B  BA

 
 B
A Tesla meter2 is also
called a Weber (Wb)
Gauss’s Law for Magnetism
•Magnetic field lines always go in circles – there are no magnetic monopole sources
•For closed surfaces, any flux in must go out somewhere else
B 
 B  nˆ dA  0
Earth’s Magnetic Field
•Deep inside the Earth there are currents flowing
•These generate magnetic field lines
•Exiting near the south geographic pole
•Entering near the north geographic pole
•Current loops near the surface of the Earth try to line up with B-field
•Compass needles are really just loops of current (sort of)
•Compass needles point
approximately north on
most of the Earth
Equations for Test 2
Resistors:
V  IR
P  I V
R  R1  R2
1 1
1
 
R R1 R2
I  JA
RC Circuits:
  RC
End of material
for Test 2
Kirchoff’s Laws:
 Iin   Iout
0   V
loop
Magnetic Forces:
F  q  E  v  B
F  IL  B
With Capacitors:
I
dQ
dt
V  Q C
Gauss’s Law for
Magnetism
B 
 B  nˆ dA  0
Units:
N
T
Am
V  Nm C
A  C/s