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The graded exams will be returned next Tuesday, Nov 7. You will have until the next
class on Thursday, Oct 6 to rework the problems you got wrong and receive 50% added
credit. Make sure you are in class as you will no have another opportunity to rework the
exam. I will be going over the answers in class on Thursday. This will also be your only
opportunity to ask for corrections/clarifications on any grading mistakes.
The homework assignment will be on line this afternoon but will not be due until
Tuesday, Nov 14. This will give you the opportunity to start work on the problems so
that you will not be overloaded with homework and the exam rework next week.
Forces and Torques on the Current Loop
Net force is zero
F  IaB
F '  IbB sin(90   )  IbB cos 
The Torque is not zero
  2 F (b / 2)sin   ( IBa )(b sin  )
  IBA sin 
Magnetic Dipole Moment
In general:
τ  IA  B  IAn  B  μ  B
Magnetic (dipole) moment μ  IA
(valid for any orientation and shape)
Potential energy of the loop
U  μ  B
Torque wants to rotate the loop so
that its moment is oriented along B
Loops and Coils
For solenoid
  NIAB sin 
Example: Force and Torque on a Circular Current Loop



d l  Rd ( sin ) i  Rd (cos) j

 
d F  Id l  B
i
j
k
d l  B  Rd ( sin  ) Rd (cos ) 0
Bx
0
0
 



d   r  dF

r  R cos i  R sin  j
i
r
r
r  dF  Rcos 
0


 
  j IBA   B


j
k
Rsin 
0
0  IBx Rd  cos
Example: Force on a Current Loop in Non-Uniform Field

B0 z  B0 y 
B
j
k (only z and y components)
L
L

 
d F  Id l  B
(2)
B0 y 
Component  to dl1 is B=
k
L
(3)
(1)
(4)
 L
B0 y
IB0 L 
F1  i I 
dy 
i
L
2
0
i
F2  I dl  B d l  B  dx

 
0
IB0 L 
F3  
i
2

F4  j I
j
0
k
0
B0 z
L
B0 y
L
B0 y
dx  0
L

F2   j IB0 L
Magnetic Dipole in a Non-uniform Magnetic Field



F  (   ) B
y
B B
F   y
;
0
y y
F  0 (along y  axis)
B
0
y
F  0 (opposite to y  axis)
Magnetic Dipoles and How Magnets Work
The Direct-Current Motor
Magnetic Field of a Moving Charge
0 qv sin 
B
4 r 2

B
 
0 q v  r
4 r 2
Magnetic field of a point charge
moving with constant velocity

1

1
0
T

m
/
A
;



4

c
0

7
00 2
Example: Force between two moving protons
Find the ratio of electric and magnetic forces on the protons
1 q2
F
E
4
0 r2
0 qv
B
k
2
4
r

Magnetic field of the lower proton at
the position of the top one



F

qv
( )B
B

4
r
22

q
0 v
F
j
B
2
2
F
2 v
B


v 2
00
F
c
E
Magnetic Field of Current Element
The Biot-Savart law.
d
Q

n
q
A
d
l
f
l
o
w
w
i
t
h
v
e
l
o
c
i
t
y
v
d
For element of a (fine) wire:
 I dl  rˆ
dB  0
4 r 2
constant permeability of free space:
For the whole "circuit":
0 I dl  rˆ
B
4  r 2
For arbitrary distribution of charge flow:
 j(1)  rˆ12
B(2)  0 
dV1
2
4
r12
(rˆ12 is from point 1 to point 2)
Magnetic field around a straight wire
For the
fieldmagnitude
:
0I  sin dx
B

2

4  r
a
ad
[r 
; x  acot; dx 2 ] 
sin
sin 
0I 
0I

sin

d


4a 0
2a
(where
ais thedistance
fromthewire)
Magnetic Field of Two Wires
Field at points on the x-axis
to the right of point (3)

I
0
B

;
1
2

(
xd
)

I
0
B

;
2
2

(
xd
)

I
d
0
B

B

B

t
o
t
a
l
2 1
2 2

(
x

d)
Magnetic field outside of a conductor pair falls off more rapidly
Magnetic field of a circular arc
For the field magnitude at O :
0 I
B
4R 2
0 I
 ds  4R 2 R
0 I

4R

Magnetic Field of a Circular Current Loop
For field on the axis :
 I cos  ds
B ( x )  Bx ( x )  0  2
4 x  R 2
0 I
R
2R

2 3/ 2
2
4 ( x  R )

0 IR 2

2( x 2  R 2 )3 / 2 2 ( x 2  R 2 )3 / 2
[ x  R]

Falls off just as the electric field of
the electric dipole
Magnetic Field on the Axis of a Coil
Bx 
Bx 
0 NIR 2
2( x 2  R 2 )3/ 2
0 
2 ( x  R )
2
2 3/ 2
;
0 
  NIA
0 
2 x 3
The magnetic field of a (small) loop behaves “on the outside” like the electric field
of the electric dipole of the same orientation – that’s why “magnetic dipole”.
Magnetic force between two parallel conductors with currents
Magnetic field from conductor 2:
0 I
B2 
2 r
Magnetic force on conductor 1:
'

II
F1  I ' LB2  0 L
2 r
Absolutely the same magnitude is
for the magnetic force on conductor 2
but F1   F2
FB 0 II '

L
2 r
Currents in the same direction attract
Currents in opposite directions repel
Definition of 1 Ampere :
Identical current in two wires
separated by 1 m
is 1 Ampere
when the force per 1 meter
is 2 10 7 N/m
Example: Two straight, parallel, superconducting wires 4.5 mm apart carry
15,000 A current each in opposite directions
Should we carry about the mechanical strength of the wires?
F 0 II '

 104 N / m
L 2 r