Transcript PHY 231 Lecture 29 (Fall 2006)

Physics 213
General Physics
Lecture 10
Last Meeting: Lorentz Force and
Torque
Today: Review Torque,
Magnetic Field of
Conductors and Currents,
Ampere’s Law, Solenoid,
and Magnetic Materials
1
2
Determining Magnetic Moments
and Torques for a Current Loop

yˆ

xˆ
zˆ


B
I



Directions are
always determined
with right-hand
rule.
Magnitudes are
given by
  IAN
  B sin 
3
θ
┴

θ

4
Review
μ
   B sin 
5
yˆ
I

B
xˆ
zˆ


(A)
yˆ

(B)
yˆ

(C)

B

B
xˆ
xˆ
zˆ
It will not rotate.
zˆ
6
yˆ
I

B
xˆ
zˆ


(A)
yˆ

(B)
yˆ

(C)

B

B
xˆ
xˆ
zˆ
It will not rotate.
zˆ
7
Magnetic Fields –
Long Straight Wire


A current-carrying wire
produces a magnetic
field
The compass needle
deflects in directions
tangent to the circle

The compass needle
points in the direction of
the magnetic field
produced by the current
Direction of the Field of a Long
Straight Wire

Right Hand Rule #2
 Grasp
the wire in
your right hand
 Point your thumb in
the direction of the
current
 Your fingers will curl
in the direction of the
field
Magnitude of the Field of a Long
Straight Wire
The magnitude of the field at a distance r
from a wire carrying a current of I is
o I
B
2 r
 µo = 4  x 10-7 T.m / A

 µo is
called the permeability of free space
Ampère’s Law, and Field from
Currents


Choose an arbitrary
closed path around
the current
Sum all the products
of B|| Δℓ around the
closed path
 B|| Δℓ = µo I
Ampère’s Law to Find B for a
Long Straight Wire


Use a closed circular
path
The circumference of
the circle is 2  r


o I
B
2 r
This is identical to the
result previously obtained
Magnetic Force Between Two
Parallel Conductors


The force on wire 1 is
due to the current in
wire 1 and the
magnetic field
produced by wire 2
The force per unit
length is:
F
o I1 I2

2 d
Magnetic Force Between Two
Parallel Conductors, Derived

The force on wire 1
F = B I1 ℓ sin θ

B produced by wire 2
0I2
B 
2d

sin=1, so
F
o I1 I2

2 d
Force Between Two Conductors,
cont
Parallel conductors carrying currents in the
same direction attract each other
 Parallel conductors carrying currents in the
opposite directions repel each other

16
Circular current loops act like
Magnetics, with a N and S pole.
17
18
19
20
N
Bl =μ 0 NI  B=0 I
l
21
22
23
Magnetic Materials
Types of Magnetic Materials



Ferromagnetic
Permanent magnetic
moments. Repel or attract
depending on orientation
Paramagnetic
Induced magnetism in same
direction as applied field.
Materials are attracted to
magnets.
Diamagnetic
Induced magnetism in
opposite direction as applied
field. Materials are repelled.
S N
S
S N
N
S
N
S N
S
N
S
N
N
N
S
S N
S
Magnetic Effects of Electrons –
Spins

Electrons also have
spin
 The
classical model
is to consider the
electrons to spin like
tops
 It is actually a
quantum effect
Ferromagnetic Domains



Random alignment (left) shows an unmagnetized material
When an external field is applied, the domains aligned with B
grow (right)
Each domain acts like a little magnet with a N and S pole.
Arrows below represent the magnetic moment.
S
N
A loop of wire with a weight of 1.47 N is oriented vertically
and carries a current I = 1.75 A. A segment of the wire
passes through a magnetic field directed into the plane of
the page as shown. The net force on the wire is measured
using a balance and found to be zero. What is the
magnitude of the magnetic field?
(a) zero tesla
(d) 1.5 T
(b) 0.51 T
(e) 4.2 T X
(c) 0.84 T
The magnetic force is pointing upwards and
cancels the gravitational force:
FB  BIl  Fg  1.47 N
1.47 N
B
 4.20T
(1.75 A)(0.20m)
Two long, straight, parallel wires separated by a distance d carry currents
in opposite directions as shown in the figure. The bottom wire carries a
current of 6.0 A. Point C is at the midpoint between the wires and point O
is a distance 0.50d below the 6-A wire as suggested in the figure. The
total magnetic field at point O is zero tesla.
1. Determine the value of the current, I,
in the top wire.
(a) 2 A
(c) 6 A
determined since
(e) This cannot be
(b) 3 A
(d) 18 A X
not specified.
the value of d is
2. Determine the magnitude of the
magnetic field at point C if d = 0.10 m.
(a) 2.4x10–5 T
(e) 1.4x10–4 T
(b)
4.8x10–5
T
(c) 9.6x10–5 T X
(d)
1.1x10–4
T
0 I
0 (6.0 A)
B1 
 B2 
,
2 (1.5d )
2 (0.5d )
I  3  (6.0 A)  18.0 A
0 18.0 A 6.0 A
(

)
2 0.5d 0.5d
(4 107 Tm / A)(24.0 A)

2  0.05m
B (out ) 
 9.60 10 5 T
Quiz

θ
30

θ
31