Chapter 33
Magnetic Field
Phys 133 -- Chapter 32
1
The questions
What are magnetic phenomena?
What causes magnetic fields?
Magnetic field of moving charges/currents
Magnetic force on moving charges/currents
Magnetic torque
Magnetic materials
Phys 133 -- Chapter 32
2
Demo magnets
Phys 133 -- Chapter 32
3
Magnetism: experiments 1
Orients a compass
Phys 133 -- Chapter 32
4
Magnetism: experiments 2
Phys 133 -- Chapter 32
5
Do Workbook 32.1, 3 & 4
Phys 133 -- Chapter 32
6
Magnetic field: wire
Current produces a magnetic field too - same as magnet
Right thumb rule
Phys 133 -- Chapter 32
7
Magnetic field: symbol conventions
In general
Thumb along, fingers
around (think circles!)
Phys 133 -- Chapter 32
8
Magnetic field: vectors and field lines
Arrow indicates direction compass north would point.
Longer arrow, larger field
Closer spacing, larger field
Phys 133 -- Chapter 32
9
Do Workbook 32.7 & 8
Phys 133 -- Chapter 32
10
Magnet field: cause
remember
!
D
Moving charges produce magnetic fields
!
C
C ´ D = C D sin a C , D
(
)
dir C ´ D = RHR
Bmoving
point
charge
m0 qv ´ rˆ
=
2
4p r
(cross product, 3D)
Phys 133 -- Chapter 32
r
11
Do Workbook 32.11 & 12
Phys 133 -- Chapter 32
12
Problem 32.7
What are the magnetic field strength and direction at
the dot in the figure?
Phys 133 -- Chapter 32
13
Problem 32.7 (ans)
m0 qv ´ rˆ
Bmoving =
4p r 2
point
charge
ÞB=
r
rˆ
m0 q
v rˆ sinq
2
4p r
m 0 1.26 ´ 10 -6 Tm
A
=
= 1.0 ´10 -7 Tm
A
4p
4p
q = 1.6 ´10-19 C
r = 1.41´10-2 m
m q
B = 0 2 v rˆ sinq
4p r
=1´10-7 Tm
A
v = 2.0 ´10 7 ms
-19
1.6 ´10 C
7 m
(2.0
´10
s )sin135°
-2
2
(1.4 ´10 m)
r√ = 1
q = 135° Þ sin135° = 0.707
=1.13´10-15 T
Direction-RHR = out of page
Phys 133 -- Chapter 32
14
Magnetic field: current segment
Current (many moving charges) produce magnetic fields
Bwire
segment
m0 IDs ´ rˆ
=
4p r 2
To find magnetic field produced by a wire configuration
--pick a generic current segment
--field due to segment at location of interest
--convert current segment to coordinate
--sum them up, becomes integral
Phys 133 -- Chapter 32
15
Problem 32.50
Find an expression for the magnetic field at the center (P)
of the circular arc.
Phys 133 -- Chapter 32
16
Problem 32.50 (ans)
m0 IDs ´ rˆ
Bwire =
4p r 2
segment
3
2
1
BP = Bsegment 1 + Bsegment 2 + Bsegment 3
Direction into page (RHR)
m0 I
m0 I
B
=
(2
p
)
=
For entire loop P Phys4133
p R-- Chapter 322 R
17
Magnetic fields: typical situations
m0 I
=
2p d
Bwire
Bloop =
m0
d
IR 2
2 (z + R
2
2
z = 0 Þ Bloop =
center
)
3
2
m0 I
2 R
Phys 133 -- Chapter 32
18
Demo loop
Phys 133 -- Chapter 32
19
Magnetic dipole
A current loop
Bdipole
m0 2m
=
4p z 3
m = (AI , from south to north )
Phys 133 -- Chapter 32
20
Do Workbook 32.18,19, 20
Phys 133 -- Chapter 32
21
Ampere’s law
B
×
d
s
=
m
I
0
through
ò
(Currents cause magnetic fields)
--always true
--sometimes helpful (high symmetry)
Phys 133 -- Chapter 32
22
Path integrals
B
×
d
s
ò
Take a segment of path, dot product with
field, write it down, find another segment,
repeat, add up
B ^ ds
ò
i
f
B ds, B = const
B × ds = 0
ò B × ds = BL
Phys 133 -- Chapter 32
23
Do Workbook 32.23 & 24
Phys 133 -- Chapter 32
24
Ampere’s law
B
×
d
s
=
m
I
0
through
ò
(Currents cause magnetic fields)
--always true
--sometimes helpful (high symmetry)
Phys 133 -- Chapter 32
25
Problem Ampere
A long wire is surrounded by a thin cylindrical shell of
radius R1. The inner wire carries a current I0 in one
direction and the outer shell carries a current I0 in the
other direction. Use Ampere’s law to find the magnetic
field strength everywhere.
I0
I0
R1
Phys 133 -- Chapter 32
26
Problem Ampere (ans)
ò B × ds = m0 I through
•
•
B is circular (RTR)
Choose concentric circle as path
Then B ds
Ampere’s law becomes
B2pr = m0Ithrough
m0 Ithrough
ÞB=
2pr
Phys 133 -- Chapter 32
Side view
I0 (out)
x
R1
I0 (in)
•
•
r < R1
Ithrough = I0
m0 Ithrough m 0 I0
ÞB=
=
2pr
2pr
R1 < r
Ithrough = I0 - I0 = 0
m0 Ithrough
ÞB=
=0
2pr
27
Solenoid
Phys 133 -- Chapter 32
28
Solenoid
Uniform inside, “zero”
out
Bsolenoid =
m0 NI
L
= m0 nI
Phys 133 -- Chapter 32
29
Force on moving charge
FM = qv ´ B = (qvBsina,dir = RHR)
Phys 133 -- Chapter 32
30
Do Workbook 32.27 & 28
Phys 133 -- Chapter 32
31
Demo Tesla Coil
Phys 133 -- Chapter 32
32
Problem 32.31
An electron moves in the magnetic field B=0.50 i T with a speed of
1.0x107m/s in the direction shown. What is the magnetic force (in
component form) on the electron?
Phys 133 -- Chapter 32
33
Problem 32.31 (ans)
B = ( 0.50 T ) iˆ
v = - (1.0 ´10 7 ms ) jˆ
æ 1 ˆ 1 ˆö
v = (1.0 ´10 7 ms ) ç j+
k÷
è 2
2 ø
Fm = qv ´ B
Fm = qv ´ B
Fm = q v B sinq
= e(1.0 ´ 10 7 ms )(0.5T)sin90°
= 8 ´ 10
B = ( 0.50 T ) iˆ
-13
Fm = q v B sinq
= e(1.0 ´ 10 7 ms )(0.5T)sin90°
= 8 ´ 10 -13 N
N
Direction RHR
Direction RHR
Fm = (8´10
-13
N ) kˆ
Fm = (8´10-13 N )
Phys 133 -- Chapter 32
(
1
2
jˆ +
1
2
kˆ
)
34
Do Workbook 32.29 & 31
Phys 133 -- Chapter 32
35
Problem Velocity Selector
Find the strength and direction of electric and
magnetic fields confined to the dashed region, such
that a charged particle (q>0, m) coming from the left
at speed v0 will pass through undeflected and without
speeding up.
(easy one: E=0, B=0)
Phys 133 -- Chapter 32
36
Problem Velocity Selector (ans)
+++++++++++++
Put electric field down
E
E = -Ejˆ
+
Then
Fe = qE = -qEˆj
Need force up if proton is to be undeflected
Put magnetic field into page
B = -Bkˆ
Then
Fm = qv ´ B = q v B jˆ
-------------
åF
y
= may
( Fe ) y + ( Fm ) y = may
-qE + qBv = may
We want ay = 0
Þ qE = qBv
Phys 133 -- Chapter 32
Þ E = Bv
37
Force on current-carrying wire
m0 I 2 m0 LI1I 2
Fparallel = I1LB2 = I1 L
=
2pd
2pd
wires
Phys 133 -- Chapter 32
38
Demo wire in magnetic field
Phys 133 -- Chapter 32
39
Do Workbook 32.35 & 37
Phys 133 -- Chapter 32
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Force/torque on current loop
Phys 133 -- Chapter 32
41
Do Workbook 32.40 & 41
Phys 133 -- Chapter 32
42
Magnetic material and induced dipole
Phys 133 -- Chapter 32
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