Sources of the Magnetic
Field
March 23, 2009
Note – These slides will be updated for
the actual presentation.
Remember the wire?
Try to remember…
1
rdq
 r  dq
dE 
  2 
3


40  r  r
40 r
1
r
 UNIT VECTOR
r
The “Coulomb’s Law” of Magnetism
A Vector Equation … duck
For the Magnetic Field,
current “elements” create the field.
This is the Law of
Biot-Savart
In a similar fashion to E field :
0 ids  runit 0 ids  r
B

2

4
r
4  r 3
permeabili ty
0  4 10 7 Tm / A  1.26 10 7 Tm
Magnetic Field of a Straight Wire
• We intimated via magnets that the
Magnetic field associated with a
straight wire seemed to vary with 1/d.
• We can now PROVE this!
From the Past
Using a Compass
Right-hand rule: Grasp the
element in your right hand with
your extended thumb pointing
in the direction of the current.
Your fingers will then naturally
curl around in the direction of
the magnetic field lines due to
that element.
Let’s Calculate the FIELD
Note:
For ALL current elements
in the wire:
ds X r
is into the page
The Details
 0 ids sin(  )
dB 
4
r2
Negative portion of the wire
contribute s an equal amount so we
integrate from 0 to  and DOUBLE it.

 0i sin(  )ds
B
2 0
r2
Moving right along
r  s R
2
2
sin   sin(    ) 
R
s2  R2
So

 0i
 0i
Rds
B

3
/
2

2 0 s 2  R 2 
2R
1/d
A bit more complicated
A finite wire
P1
NOTE : sin(  )  sin(    )
ds  r  ds r sin(  )
r


ds
R
sin(  ) 
r
 0i ds sin(  )
dB 
2
4
r

r s R
2

2 1/ 2
More P1
L/2
 0i
ds
B
3/ 2

2
2
4  L / 2 s  R 
and
 0i
L
B
2R L2  4 R 2
when L  ,
 0i
B
2R
P2
 0iR
ds
B
4 L s 2  R 2 3 / 2
0
or
 0i
L
B
4R s 2  R 2
APPLICATION:
Find the magnetic field B at point P
Center of a Circular Arc of
a Wire carrying current
More arc…
ds
ds  Rd 
 0 ids  0 iRd 
dB 

2
4 R
4 R 2


 0 iRd   0i
B   dB  

d
2

4 R
4R 0
0
 0 i
B
at point C
4R
The overall field from a
circular current loop
Sorta looks like a magnet!
Iron
Howya Do Dat??
ds r  0
No Field at C
Force Between Two Current
Carrying Straight Parallel
Conductors
Wire “a” creates
a field at wire “b”
Current in wire “b” sees a
force because it is moving
in the magnetic field of “a”.
The Calculation
The FIELD at wire " b" due to
wire " a" is what we just calculated :
 0ia
Bat "b" 
2d
Fon "b"  ib L  B
Since L and B are at right angles...
 0 Lia ib
F
2d
Definition of the Ampere
The force acting between currents in parallel
wires is the basis for the definition of the
ampere, which is one of the seven SI base
units. The definition, adopted in 1946, is
this: The ampere is that constant current
which, if maintained in two straight, parallel
conductors of infinite length, of negligible
circular cross section, and placed 1 m apart
in vacuum, would produce on each of these
conductors a force of magnitude 2 x 10-7
newton per meter of length.
Ampere’s Law
The return of Gauss
Remember GAUSS’S LAW??
E

d
A


Surface
Integral
qenclosed
0
Gauss’s Law
• Made calculations easier than
integration over a charge distribution.
• Applied to situations of HIGH
SYMMETRY.
• Gaussian SURFACE had to be defined
which was consistent with the
geometry.
• AMPERE’S Law is the Gauss’ Law of
Magnetism! (Sorry)
The next few slides have been
lifted from Seb Oliver
on the internet
Whoever he is!
Biot-Savart
• The “Coulombs Law of Magnetism”
  0  ids  rˆ
dB   
2
 4  r
Invisible Summary

Biot-Savart Law
  0  ids  rˆ
dB   
2
 4  r
0 I
(Field produced by wires)
B
2R
 Centre of a wire loop radius R
 0 NI
B

 Centre of a tight Wire Coil with N turns
2R
 Distance a from long straight wire

Force between two wires
 Definition of Ampere

0 I
B
2a
F  0 I1 I 2

l
2a
Magnetic Field from a long wire
0I
B 
Using Biot-Savart Law
r
I
Take a short vector
on a circle, ds
B
ds
B  ds  B  ds cos
  0  cos  1
Thus the dot product of B &
the short vector ds is:
2r
B  ds  B ds
0 I
B  ds 
ds
2r
Sum B.ds around a circular path
0 I
B  ds 
ds
2r
r
I
B
Sum this around the whole ring
ds
Circumference
of circle
 B  ds
 ds  2r
0 I

ds
2r
0 I

ds
2r
0 I
  B  ds 
2r   0 Ι
2r
Consider a different path
B ds  0
i
• Field goes as 1/r
• Path goes as r.
• Integral
independent of r
SO, AMPERE’S LAW
by SUPERPOSITION:
We will do a LINE INTEGRATION
Around a closed path or LOOP.
Ampere’s Law
B

d
s


i
0
enclosed

USE THE RIGHT HAND RULE IN THESE CALCULATIONS
The Right Hand Rule .. AGAIN
Another Right Hand Rule
COMPARE
B

d
s


i
0
enclosed

Line Integral
E

d
A


Surface Integral
qenclosed
0
Simple Example
Field Around a Long Straight Wire
 B  ds   i
0 enclosed
B  2r   0i
 0i
B
2r
Field INSIDE a Wire
Carrying UNIFORM Current
The Calculation
 B  ds  B  ds  2rB   i
0 enclosed
ienclosed
r 2
i 2
R
and
  0i 
B
r
2 
 2R 
B
 0i
2R
R
r
Procedure
• Apply Ampere’s law only to highly symmetrical
situations.
• Superposition works.
▫ Two wires can be treated separately and the
results added (VECTORIALLY!)
• The individual parts of the calculation can be
handled (usually) without the use of vector
calculations because of the symmetry.
• THIS IS SORT OF LIKE GAUSS’s LAW WITH
AN ATTITUDE!
The figure below shows a cross section of an infinite conducting sheet
carrying a current per unit x-length of l; the current emerges
perpendicularly out of the page. (a) Use the Biot–Savart law and symmetry
to show that for all points P above the sheet, and all points P´ below it,
the magnetic field B is parallel to the sheet and directed as shown. (b)
Use Ampere's law to find B at all points P and P´.
FIRST PART
Vertical Components
Cancel
Apply Ampere to Circuit
L
B
Infinite Extent
B
  current per unit length
Current inside the loop is therefore :
i  L
The “Math”
B
Infinite Extent
B
 B  ds   i
0 enclosed
BL  BL   0L
B
 0
2
A Physical Solenoid
Inside the Solenoid
For an “INFINITE” (long) solenoid the previous
problem and SUPERPOSITION suggests that the
field OUTSIDE this solenoid is ZERO!
More on Long Solenoid
Field is ZERO!
Field looks UNIFORM
Field is ZERO
The real thing…..
Finite Length
Weak Field
Stronger - Leakage
Another Way
Ampere :
 B  ds   i
0 enclosed
0h  Bh   0 nih
B   0 ni
Application
• Creation of Uniform Magnetic Field Region
• Minimal field outside
▫ except at the ends!
Two Coils
“Real” Helmholtz Coils
Used for experiments.
Can be aligned to cancel
out the Earth’s magnetic
field for critical measurements.
The Toroid
Slightly less
dense than
inner portion
The Toroid
Ampere again. We need only worry
about the INNER coil contained in
the path of integratio n :
 B  ds  B  2r   Ni (N  total # turns)
0
so
 0 Ni
B
2r