# Physics 121 Fall 2004

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```							       Physics 121 Lecture 10
Sources of Magnetic Fields (Currents)
SJ Chapter 30, Sec. 1 - 5

•   Magnetic fields are due to currents
•   The Biot-Savart Law
•   Calculating field at the centers of current loops
•   Magnetic dipole moment
•   Field due to a long wire
•   Force between two parallel wires carrying currents
•   Ampere’s Law
•   Solenoids and toroids
•   Summary

Fall 2008
Last time: a magnetic field exerts a force on
moving charges and currents
•   Magnets only come in pairs of N
and S poles (no monopoles).
N      S    N   S   NS
•   Magnetic field exerts a force on
moving charges (i.e. on currents).

•   The force is perpendicular to both
the field and the velocity (i.e. it           
uses the use cross product). The
magnetic force can not change a
FB  q v  B
particle’s speed or KE

•   A charged particle moving in a
uniform magnetic field moves in a
circle or a spiral. The magnetic
force can not change a particle’s
speed or KE
2 qB
c     
c   m
•   Because moving charges are a
current, a wire in a magnetic field
feels a force also using cross
product. This force is responsible
for the motor effect.
      
FB  iL  B
     
m    B
•   Magnetic dipole moment,          
•   torque, and potential energy
ˆ
  Ni A n                     
Um    B Fall 2008
Magnetic fields are due to currents
Oersted - 1820: Magnetic compass deflected by current
 Magnetic fields are due to currents (free charges & in wires)

In fact, currents are the
only way to create
magnetic fields

Fall 2008
The Biot-Savart Law (1820)
•   Same basic role as Coulomb’s Law: magnetic field due to a source
•   Source strength measured by “current-length” i ds                                           
•   Fall off with distance as inverse-square of distance                                      B
•   New “permeability” constant 0                                              i s           x

•   Direction of B field depends on a cross-product ( another RH rule)                   r
use RH rule for current
segments: thumb along ids        differential addition to field at P due
curled fingers show B                 to distant source ids
P’                                                           
  0 ids  r                  unit

ˆ                 vector
dB 
dB
(out of page)                                                                           along r
4 r 2
10-7 exactly
“permeability” 0 = 4x10-7 T.m/A.
Find total field B by integrating over the
whole current region (need lots of symmetry)
The magnetic field lines are circles wrapped around a                ids sin( )
wire. Use the right hand rule to find the direction
dB  0
4    r2

Fall 2008
Magnetic field due to a long, straight wire
•   Wire is infinitely long, current i flows up                                                               
  0 ids  r
•                                                                                                                 ˆ
Find B at point P                                                                                dB 
•   dB is into page at P for ds anywhere along wire                                                       4 r 2
•   Magnitude of ids X r = ids sin()
•   Integrate dB using Biot Savart Law from –infinity to +infinity
                  0   sin( )ds
B                   dB 
4   R 2  s 2
i

• Denominator has same sign for s + and –
R
• Substitute: sin()  sin(  )         sin()  - sin(-)
1/ 2
R   2
 s2   
0           ds
B               i R
2     0 [R 2  s2 ]3 / 2

• From integral              ds                                s
0                       
1

1
1   - 0   
tables:               [R 2  s 2 ]3 / 2        R 2 [R 2  s 2 ]1 / 2   0       R2

RIGHT HAND RULE FOR A WIRE
 0i
B 
2R
R is distance
perpendicular to                                                                           FIELD LINES ARE CIRCLES
wire through P                                                                                            Fall OR
THEY DO NOT BEGIN 2008 END
Direction of Magnetic Field

Which sketch below shows the correct direction of the
magnetic field, B, at the point P?

B into                           B into               B into
page    B                        page                 page
B
i                        i       i                i                    i
P         P                      P             P       P

A                        B       C                D                    E

Fall 2008
Magnetic field at the center of a current arc

  0 ids  r
•   Circular arc, radius R is constant                                          ˆ
•   Find B at center, point C                                      dB 
•   f is included arc angle, not the cross product angle
4 r 2
•   Angle  for the cross product is always 900
•   dB at center C is up out of the paper
•   ds = Rdf’                                                               i

 0 ds  0 d'                                     ids
dB     i 2    i                  B                       
4 R     4 R                                      R
R
• integrate on arc angle f’ from 0 to f
f                                   Right hand rule
 0i           0i
 df'  4R f
for wire segments
4R 0
• For a full loop of current - f = 2 radians:

                              0i
B               B           B      (loop)
Thumb points along the
2R
current. Curled fingers
Another RH rule (for loops):                                   show direction of B
Curl fingers along current, thumb shows direction of B at center
Fall 2008
?? What would formula be for f = 45o, 180o, 100 radians ??
Examples:
FIND B FOR A POINT LINED UP WITH A SHORT STRAIGHT WIRE
P
0
dB              i ds sin()  0          
4r   2
ˆ
ds  r  0
B AT CENTER OF A HALF LOOP, RADIUS = r

 0i      0i
B          
4r      4r
into page
B AT CENTER OF TWO HALF LOOPS

 0i      0i
OPPOSITE                                      B  2x       
CURRENTS                                             4r      2r
same as closed loop

 0i  0i
PARALLEL                                       B         -      0
CURRENTS                                              4r    4r
into    out of
page     page
Fall 2008
Magnetic Field from Loops

The three loops below have the same current. Rank
them in terms of the magnitude of magnetic field at the
point shown, greatest first.

A. I, II, III.
B. III, I, II.
C. II, I, III.
D. III, II, I.
E. II, III, I.

I.              II.               III.

 0i
4R                                                            Fall 2008
Magnetic Field lines          When there are two
near a straight wire          parallel wires carrying
current, the magnetic field
from one causes a force on
the other.
i up                       
i                                 Fa,b  ibLb  Ba

. When the currents are parallel,
 0ia                 the force is attractive.
Ba                        . When the currents are anti-
2R                   parallel, the force is repulsive.
Fall 2008
Magnitude of the force between two long parallel wires
L
• Third Law says: F12 = - F21
i1                  • Use result for B due to infinitely long wire
 0i1              due to 1 at 2
x    x    x    x   x
B1 
r     x    x    x    x   x                   2r                into page via RH rule
x    x    x    x   x          • Evaluate F12 = force on 2 due to field of 1
x    x    x    x   x                                     i2L is normal to B
i2                      | F1,2 |  i2L2B1        Force is toward 1

End View                                 0 i1i2
 F1,2              L     F21= - F12
2 d
i1
• Attractive force for parallel currents
i2                              • Repulsive force for opposed currents

Example:    Two parallel wires 1 cm apart         |i1| = |12| = 100 A.
2x10 -7 x 100 x 100
F/L  force per unit length                       0.2 N / m
.01
F  0.2 N for L  1 m
Fall 2008
Forces on Parallel Currents

Which of the four situations below results in the greatest force
to the right on the central conductor?

F greatest?
A.   I.
B.   II.
C.   III.                   I.
D.   IV.
E.   Cannot                 II.
determine.
III.

IV.

Hints: Sketch the field directions. What is the net field midway
between two wires with parallel currents?
Which way is the net force?                
Ftot  i L  Btot
Fall 2008
Ampere’s Law
• Another way to find B, given the current
• Derivable from Biot-Savart Law
• But B is inside an integral  usable only for high symmetry (like Gauss’ Law)

• An “Amperian loop” is a
                          ienc= net current

 B  ds  0ienc
closed path of any shape
• Add up (integrate) components                               passing through the
of B along the loop (path)                                           loop

To find B, you have to be able to do the integral, then solve for B

Picture for applications:
• Only the tangential component of
B contributes to the dot product
• Current outside the loop (i3) doesn’t
contribute
• Another version of RH rule:
- curl fingers along loop
- thumb shows + direction for net current
Fall 2008
Example: Magnetic field outside a long
straight wire carrying current
 0i
We already used the Biot-Savart Law to show that:   B
2r
Now show it again more simply, using
Ampere’s Law:
 
 B  ds  0ienc
Draw an Amperian loop of any shape, but since
the magnetic field goes in circles around a
wire, choose a circular loop (of radius r).
B and ds are then parallel, and B is constant
everywhere on the Amperian path
 
 B  ds  B2r  0ienc
The integration is simple.
ienc is the total current.
Solve for B to get our earlier expression.
 0i
B             outside wire
2r                           Fall 2008
Magnetic field inside a long straight wire
carrying current
 
Using Ampere’s Law, we can even calculate B
inside the wire.                                           B  ds  0ienc
Assume the current density is uniform across the
wire. J = i/A. Current is evenly distributed
over the cross-section of the wire and must be
cylindrically symmetric.
Again draw a circular Amperian loop around the
axis, of radius r < R.
The enclosed current is less than the total
current i, because some is outside the Amperian
loop. The amount enclosed is
r 2
ienc  i
R 2                           B

Apply Ampere’s Law:                                        ~r
~1/r
                             r2
 B  ds  B2r   0ienc   0i R 2                            R              r

  i                            Outside wire (r>R) it looks like an
B   0 2 r   r  R inside wire      infinitely thin wire (previous expression)
 2R                            Inside: B grows linearly            Fall 2008
Counting currents within Amperian Loops

 
For the Amperian paths shown, rank the value of  B  ds
along each path, taking direction into account and putting the
more positive ahead of less positive values. All of the wires are
carrying the same current.

A. I, II, III, IV, V.
I.
B. II, III, IV, I, V.
II.
C. III, V, IV, II, I.
D. IV, V, III, II, I.               III.
E. I, II, III, V, IV.
IV.
V.
 
 B  ds  0ienc
Fall 2008
Another Ampere’s Law example
Why use COAXIAL CABLE for CATV and other applications?
Find B inside and outside of the cable
Cross section:

Amperian                            shield wire
loop 1                            current i into
sketch
Amperian
loop 2                                  center wire
current i out of
sketch

Inside – use Amperian loop 1:
                              0i   Outer shield does not
 B  ds  0i  Bx2r        B
2r
affect field inside
Reminds one of Gauss’s Law

Outside – use Amperian loop 2:
Zero field outside due to opposed
                                   currents + radial symmetry
 B  ds  0ienc  0                  Losses and interference suppressed
Fall 2008
Solenoids: strengthen field by using many loops
L
cancellation

d

n  # coils / unit length  N/L
strong uniform
Approximation: field is constant inside and                               field in center
zero outside (just like capacitor)
“Long solenoid”  d << L
IDEAL SOLENOID WITH AMPERIAN LOOP
• outside B = 0, no contribution from path c-d
• B perpendicular to ds of a-d and b-c
• inside B is uniform and parallel to ds on a-b
 
 B  ds  Bh  0ienc  0inh
B   0in
ideal solenoid

only section that has non-zero
Fall 2008
contribution
Toroids: bend a long solenoid into a circle
Find the B field inside
   Draw an Amperian loop parallel to the field, with
   The coil has a total of N turns
LINES                 The Amperian loop encloses current Ni.
OF
B ARE
 
CIRCLES                      B  ds  B2r  0ienc  0iN
 0iN
B             inside toroid     B0       outside
2 r
• N times the result for a long wire
• Depends on r
• Also same result as for long solenoid

N
n          (turns/uni t length)     B   0in
AMPERIAN LOOP IS             2 r
A CIRCLE ALONG B                                               Fall 2008
B on z-axis of a current loop (dipole)
•   Find B at a point P on the z axis of the current loop?
•   We use the Biot-Savart Law directly
 
  i ds  r
dB  0
4 r 3
 0 i ds cos()                    R
dBz  dB||  dB cos()                                cos 
4 R 2  z 2                       r
dB   cancels by symmetry
0     iR                                 r  R2  z2
dB z                   ds
4 (R  z )
2   2 3/2

•   Integrate on f around the current loop, noting
that the field is perpendicular to r.
•   By symmetry, the perpendicular part of B cancels
around the loop - only the parallel part survives.

0       iR                 0      iR 2           2
Bz   dBz 
4 (R 2  z 2 )3 / 2       4 (R 2  z 2 )3 / 2 0
ds                            df

as before               recall dipole
 0iR 2
i
B(z)                           B(z  0)  0                 moment
2(R  z )
2    2 3/2
2R              NiA  iR2   Fall 2008
B field far from the dipole (current loop)

Fall 2008
Summary: Lecture 10   Chapter 29 – Magnetic Fields from Currents

Fall 2008

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