Download Sources of Magnetic Fields (Currents)

Electricity and Magnetism - Physics 121
Lecture 10 - Sources of Magnetic Fields (Currents)
Y&F Chapter 28, Sec. 1 - 7
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Magnetic fields are due to currents
The Biot-Savart Law for wire element and moving charge
Calculating field at the centers of current loops
Field due to a long straight wire
Force between two parallel wires carrying currents
Ampere’s Law
Solenoids and toroids
Field on the axis of a current loop (dipole)
Magnetic dipole moment
Summary
Copyright R. Janow – Fall 2015
Previously: moving charges and currents
feel a force in a magnetic field
•
Magnets come only as dipole pairs
of N and S poles (no monopoles).
•
Magnetic field exerts a force on
moving charges (i.e. on currents).
•
The Lorentz force is perpendicular
to both the field and the velocity
Magnetic force can not change a
particle’s speed or KE
•
A charged particle moving in a
uniform magnetic field moves in a
circle or a spiral.
2 qB
mv
c 

R
c
m
qB
•
A wire carrying current in a
magnetic field feels a force also
using cross product.
This force is the motor effect.
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Current loops behave as Magnetic
dipoles, with dipole moment, torque,
and potential energy given by:
N

S
N
S
NS
 
FB  q v  B

 
FB  iL  B

  N i A n̂

 
dF  i ds  B

 
m    B
 
Um     B
Copyright R. Janow – Fall 2015
Magnetic fields are created by currents
Oersted - 1820: A magnetic compass is deflected by current
 Magnetic fields are due to currents (free charges & in wires)
Why do the un-magnetized
filings line up with the field?
In fact, currents are the only
way to create magnetic fields.
The magnitude of the field
created is proportional to


i s or qv (current - length)
Copyright R. Janow – Fall 2015
Biot-Savart Law (1820): Field due to a current element
• Same basic role as Coulomb’s Law: magnetic field due to a source
• Source strength measured by “current-length” i ds
• Falls off as inverse-square of distance
• New constant 0 measures “permeability”
• Direction of B field depends on a cross-product (Right Hand Rule)
Differential addition to field at P due to
distant source current element ids
  0 id s  r̂
dB 
4 r 2
10-7 exactly
Unit vector
along r from source to P

id s

dB
 x
r
P’

dB
(out of page)
“vacuum permeability”
0 = 4x10-7 T.m/A.
Find total field B by integrating over the
whole current region (need lots of symmetry)
For a straight wire the magnetic field lines are circles wrapped
around it. Another Right Hand Rule shows the direction:
 ids sin()
dB  0
4
r2
Copyright R. Janow – Fall 2015
Magnetic field due to a moving point charge
• A variant of Biot-Savart Law
• Source is qv having dimensions of current-length
• Magitude of B falls off as inverse-square of distance
• Direction of B field depends on a cross-product (Right Hand Rule)

qv

dB

r
x
Field at P due to distant source qv

  qv
 r̂
0
B
4 r 2
Unit vector
along r from source to P
10-7 exactly
r̂
The magnetic field lines are circles
wrapped around the current length qv.
r̂
Velocity
into page
B
 0 qv sin()
4
r2
Copyright R. Janow – Fall 2015
Direction of Magnetic Field
10 – 1: Which sketch below shows the correct direction of the
magnetic field, B, near the point P?
B into
page
i
P
A
B into
page
B
P
i
i
B
P
C
Use RH rule for current segments:
thumb along ids - curled fingers show B
i
B
B into
page
P
P
D
i
E
  0 id s  r̂
dB 
4 r 2
Copyright R. Janow – Fall 2015
Example: Magnetic field at the center of a current arc
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Circular current arc, radius R
Find B at center, point C
 is included arc angle, not the cross product angle
Angle  for the cross product is always 900
dB at center C is up out of the paper
ds = Rd’
i

B
 0 ds  0 d'
dB 
i 2 
i
4 R
4 R
• integrate on arc angle ’ from 0 to 

 0i
 0i
B
d' 


4R 0
4R
 in radians
 
ids  R
  0 id s  r̂
dB 
4 r 2

R
Right hand rule
for wire segments
• For a circular loop of current -  = 2 radians:

B

B
 0i
B
(loop)
2R
Thumb points along the
current. Curled fingers
show direction of B
Another Right Hand Rule (for current loops only):
Curl fingers along current, thumb shows direction of B at center
?? What would formula be for  =
45o,
180o,
Copyright R. Janow – Fall 2015
4 radians ??
Examples:
FIND B FOR A POINT LINED UP WITH A SHORT STRAIGHT WIRE

0

i ds
P
dB

i ds sin()
i ds  r̂  0
2
r̂
4r
 0
Find B AT CENTER OF A HALF LOOP, RADIUS = r
B 
 0i
 0i
 
4r
4r
into page
Find B AT CENTER OF TWO HALF LOOPS
OPPOSITE
CURRENTS
 0i
 0i
B  2x
 
4r
2r
same as closed loop
PARALLEL
CURRENTS
B 
 0i  0i
 0
4r
4r
into
page
out of
page
Copyright R. Janow – Fall 2015
Magnetic Field from Loops
10 – 2: The three loops below have the same current.
The smaller radius is half of the large one. Rank the
loops by the magnitude of magnetic field at the center,
greatest first.
A.
B.
C.
D.
E.
I, II, III.
III, I, II.
II, I, III.
III, II, I.
II, III, I.
 0i
B

4R
I.
 in radians
II.
III.
Hint: consider radius, direction, arc angle
Copyright R. Janow – Fall 2015
Magnetic field due to current in a thin, straight wire
• Current i flows to the right along x – axis
• Wire subtends angles 1 and 2 at P
• Find B at point P, a distance a from wire.

  ids
 r̂
0
dB 
4 r 2
• dB points out of page at P for ds anywhere along wire
a
Evaluate dB due to ids using Biot Savart Law

ids  i î dx
  i  dx  cos()
dB  0
k̂
4
r2
• Angles , 1, 2 measured CW about –y axis
• x negative as shown,  positive, 1 positive, 2 negative
d
x   a tan()
r  a / cos()
[tan()]  sec 2 ()  1/ cos2 ()
d
 dx  a d(tan()) / d  a  d / cos2 ()

i
| dB |  0 cos()d
4 a
Integrate on  from 1 to 2:

i
B   | dB |   0
1
4a
2

2
1
Symmetric Case:
cos()d 
 0i
[sin(1)  sin(2 )]
4a
2   1  B 
 0i
sin(1)
2a Copyright R. Janow – Fall 2015
Magnetic field due to current in thin, straight wires
B 
 0i
[sin(1 )  sin(2 )]
4 a
Special Case: Infinitely long, thin wire:
Set 1   / 2, 2   / 2 [  direction was CW in sketch ]
RIGHT HAND RULE FOR A WIRE
 0i
B 
2a
a is distance perpendicular to
wire through P
FIELD LINES ARE CIRCLES
THEY DO NOT BEGIN OR END
Example: Field at P due to Semi-Infinite wires:
A
Set 1   / 2, 2  0 (piece B)
Zero
contribution
B
 0i
|B|
4 a
Field is into slide at point P
Half the magnitude for a
fully infinite wire
Copyright R. Janow – Fall 2015
Magnetic Field lines near a
straight wire carrying current
i out of slide
i
When two parallel wires are
carrying current, the
magnetic field from one
causes a force on the other.
 0ia
Ba 
2R



Fon b  ibLb  Ba
. The force is attractive when the
currents are parallel.
. The force is repulsive when the
Copyright R. Janow – Fall 2015
currents are anti-parallel.
Magnitude of the force between two long parallel wires
L
i1
d
• Third Law says: F12 = - F21
• Use result for B due to infinitely long wire
x
x
x
x
x
x
x
x
x
x
i2x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
End View
 0i1
B1 
2d
• Evaluate F12 = force on 2 due to field of 1



Fon 2  i2L2  B1

| Fon 2 |  i2L B1
 F1,2
i1
i2
Example:
Due to 1 at wire 2
Into page via RH rule
 0 i1i2

L
2 d
i2L is normal to B
Force is toward wire1
F21= - F12
• Attractive force for parallel currents
• Repulsive force for opposed currents
Two parallel wires are 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
Copyright R. Janow – Fall 2015
Forces on parallel wires carrying currents
10 – 3: Which of the four situations below results in the greatest force
to the right on the central conductor? The currents in all the wires
have the same magnitude.
 0i
B 
2R
A.
1

 
Ftot  i L  B tot
2
greatest F ?
3
4
B.
C.
D.
Answer
B
Hints: Which pairings with center wire are attractive, repulsive, or zero?
What is the field midway between wires with parallel currents?
What is the net field direction and relative magnitude at center wire
Copyright R. Janow – Fall 2015
Ampere’s Law
• Derivable from Biot-Savart Law – A theorem
• Sometimes a way to find B, given the current that creates it
• But B is inside an integral  usable only for high symmetry (like Gauss’ Law)
• Integrate over an “Amperian
loop” - a closed path.
• Adds up components of B
along the loop path.
• Any path enclosing a set of
currents has same result.
 
 B  ds  0ienc
ienc= net current
passing through the
loop
Picture for applications:
• Only the tangential component of B
along ds contributes to the dot product
• Current outside the loop (i3) creates field
but doesn’t contribute to the path integral
• Another version of RH rule:
- curl fingers along Amperian loop
- thumb shows + direction for net current
Copyright R. Janow – Fall 2015
Ampere’s Law Example: Find magnetic field outside a long,
straight, possibly fat, cylindrical wire carrying current
The Biot-Savart Law showed that
for a thin infinitely long wire:
B
 0i
2r
Now Ampere’s Law shows it again more simply and for a long
 
fat wire.
End view of wire with radius R
 B  ds  0ienc
Amperian loop outside R can have any shape
Choose a circular loop (of radius r>R) because
field lines are circular around a wire.
B and ds are parallel and B is constant
everywhere on the Amperian path
R
 
 B  ds  Bx2r  0ienc
The integration was simple. ienc is the total current.
Solve for B to get our earlier expression:
 0i
B
2r
outside wire
R has no effect on the result.
(reminiscent of
Shell
Theorem)
Copyright
R. Janow
– Fall 2015
Magnetic field inside a long straight
wire carrying current, via Ampere’s Law
 
 B  ds  0ienc
Assume current density J = i/A is uniform across the
wire cross-section and is cylindrically symmetric.
Field lines are again concentric circles
B is axially symmetric again
Again, choose a circular Amperian loop path 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
ienc  i
r 2
R 2
Apply Ampere’s Law:
 
r2
 B  ds  B2r  0ienc  0i R 2
  i  r 
B   0  
 2R  R 
r  R inside wire
B
~r
~1/r
R
r
Outside (r > R), the wire looks like an
infinitely thin wire (previous expression)
Copyright
R. Janow
Fall
Inside: B(r) grows
linearly
up –to
R2015
Counting the current enclosed by an Amperian Loop


 B  ds
10 – 4: Rank the Amperian paths shown by the value of
along each path, taking direction into account and putting the
most positive ahead of less positive values.
All of the wires are carrying the same current..
A. I, II, III, IV, V.
B. II, III, IV, I, V.
C. III, V, IV, II, I.
D. IV, V, III, I, II.
E. I, II, III, V, IV.
 
 B  ds  0ienc
I.
II.
III.
IV.
V.
Copyright R. Janow – Fall 2015
Another Ampere’s Law example
Why use COAXIAL CABLE for CATV and other applications?
Find B inside and outside the cable
Cross section:
 
 B  ds  0ienc
shield wire
current i into
sketch
Amperian
loop 1
Amperian
loop 2
center wire
current i out of
sketch
Field Inside – use Amperian loop 1:
 
 B  ds  0i  Bx2r
 0i
B
2r
Field Outside – use Amperian loop 2:
 
 B  ds  0ienc  0
B0
Outer shield current does not
affect field inside
Reminiscent of Gauss’s Law
Zero field outside due to opposed
currents + radial symmetry
Copyright R. Janow
– Fall 2015
Losses and interference
suppressed
Solenoids strengthen fields by using many loops
“Long solenoid”  d << L
L
d
strong uniform
field in center
n  # coils / unit length  N/L
Approximation: field is constant inside and
zero outside (just like capacitor)
fields from adjacent coils cancel
FIND FIELD INSIDE IDEAL SOLENOID USING AMPERIAN LOOP abcda
 
 B  ds  0ienc
 
 B  ds  Binsideh  0ienc  0inh
only section that has nonzero contribution
• Outside B = 0, no contribution from path c-d
• B is perpendicular to ds on paths a-d and b-c
• Inside B is uniform and parallel to ds on path a-b
B   0in
inside ideal solenoid
Copyright R. Janow – Fall 2015
Toroid: A long solenoid bent into a circle
i outside
flows up
Find the magnitude of B field inside


LINES OF
CONSTANT
B ARE
CIRCLES


Draw an Amperian loop parallel to the field, with
radius r (inside the toroid)
The toroid has a total of N turns
The Amperian loop encloses current Ni.
B is constant on the Amperian path.
 
 B  ds  B  2r  0ienc  0iN
 0iN
B
2 r
inside toroid
• N times the result for a long thin wire
• Depends on r
• Also same result as for long solenoid
n
N
2 r
(turns/uni t length) 
B   0in
Find B field outside (ienc=0)
AMPERIAN LOOP IS
A CIRCLE ALONG B
Answer
B0
outside
Copyright R. Janow – Fall 2015
Find B at point P on z-axis of a current loop (dipole)
•
Use the Biot-Savart Law directly

  i ds
 r̂
dB  0
4 r 2
dB
r  R 2  z2
cos  
R
r
cancels opposite side by symmetry (normal to z - axis)
Need only component along z
 0 i ds cos()
dBz  dB||  dB cos() 
4 R 2  z2

iR
ds  R d
dBz  0 2
ds
2
3
/
2
4 (R  z )
•
•
Integrate around the current loop on  – the angle
at the center of the loop.
The field is perpendicular to r but by symmetry the
component of B normal to z-axis cancels around the
loop - only the part parallel to the z-axis survives.
Bz 
 dBz
B(z) 
0
0
iR
iR 2

ds 
4 (R 2  z 2 ) 3 / 2 
4 (R 2  z 2 )3 / 2
 0i R 2
2(R 2  z 2 )3 / 2
as before
 i
B(z  0)  0
2R
2
0
d
i is into
page
recall definition of
Dipole moment
2
  NiA
 Ni–RFall
Copyright
R. Janow
2015
B field on the axis of a dipole (current loop), continued
Far field: suppose z >> R
B(z) 
 0iR 2

2(R  z )
2
2 3/ 2
Same 1/z3 dependence as for
electrostatic dipole
 0iR 2
2z 3

is normal to loop (RH Rule).
Dipole moment vector

ˆ
  NiA
N  number of turns  1
A  area of loop  R 2
R2i  |  | above
For any current loop, along z axis with |z| >> R


0 
 B(z) 
2 z 3
For charge dipole

E(z) 

p
1
2 0 z 3
Current loops are the elementary sources of magnetic
field:


• Create dipole fields with source strength

  
• Dipole feels torque due to another  in external B field     B
Dipole-dipole
interaction:

1
r3

2


Torque
1   2
depends
on
Copyright R. Janow – Fall 2015
Try this at home
10-5: 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.
B
 0i

4R
II.
III.
Hint: consider radius, direction, arc angle
 in radians
Answer: B
Copyright R. Janow – Fall 2015