Magnetic Forces, Materials
and Devices
1
Introduction
• To study the force a magnetic field exerts on
charged particles, current elements (open ended)
and (closed) loops.
• Important to problems on electrical devices such as
ammeters, voltmeters, galvanometers, cyclotrons,
plasmas, motors and magnetohydrodynamic
generators.
• Consider the magnetic field in material media.
• Discussions cover inductors, inductances, magnetic
energy and magnetic circuits.
2
Forces due to magnetic fields
There are at least three ways in which force
due to magnetic fields can be experienced.
1. Force acting on a moving charged particle due to
an external B field.
2. Force on a current element due to an external B
field.
3. Force between two current elements.
I
Q
B
B
I1
I2
3
1. Force on a charged particle
• A magnetic field can exert a force on a charge only if
the charge is moving.
• The magnetic force Fm experienced by a charge Q
moving with velocity u in a magnetic field B is
Fm = Q u  B
• Fm is perpendicular to both u and B.
• Fm does not perform work (i.e., no energy loss or gain)
because Fm is perpendicular to the direction of motion
of the charge. (recall that work done is dW = F  dl )
4
Lorentz force equation
• For a moving charge Q in the presence of both
electric and magnetic fields, the total force on
the charge is given by
F = Fe + Fm
F = Q(E + u  B)
• This is known as the Lorentz force equation.
• If the charged particle has mass m, then
du
F= m
= Q(E + u  B)
dt
5
Force on a charged particle
State of
particle
E field
only
Stationary QE
Moving
QE
B field
only
_
QuB
E and B fields
QE
Q(E+ uB)
6
2. Force on a current element
• Let I dl be the current element of a currentcarrying conductor in a magnetic field B.
• The current density is J = vu
• Relationship between current elements:
I dl = K dS = J dv (= vu dv) and
dl dQ
dQ u = dQ

dl  Idl
dt
dt
• An elementary charge dQ moving with velocity u
is equivalent to a conduction current element I dl.
7
2. Force on a current element
• The force acting on a current element I dl due to a
magnetic field B is given by dF = I dl  B
• If the path of the current I is a closed path L or a
circuit, the force on the circuit is F  Idl  B

L
For current element I dl: F 
 Idl  B
L
For surface current element K dS: F  KdS  B

s
For volume current element J dv: F 
 Jdv  B
v
8
3. Force between two
current elements
• Consider the force between two current
elements I1 dl1 and I2 dl2.
d(dF1)
R21
I1dl1
I2dl2
I1
I2
9
Force between two
current elements
• The force d(dF1) on element I1dl1 due to the
field dB2 produced by element I2dl2 is
d (dF1 )  I1 dl1  dB 2
Since dB 2 
d (dF1 ) 
o I 2 dl 2  a R
21
4 R
2
21
,
o I1 dl1  ( I 2 dl 2  a R )
o I1 I 2
F1 
4
21
4 R
2
21

L1 L2
dl1  (dl 2  a R 21 )
2
R21
10
Example
A charged particle of mass 2 kg and charge
3 C starts at point (1, -2, 0) with velocity
(4ax + 3az) m s-1 in a uniform electric field
(12ax + 10ay) V m-1. At time t = 1 s,
determine
(a) The acceleration of the particle.
(b) Its velocity.
(c) Its kinetic energy.
(d) Its position.
11
Solution
12
Solution
13
Solution
(c) Kinetic energy
1
1
2
2
2
2
m u  (2)(22  15  3 )  718 J
2
2
dl d
(d) u   ( x, y, z )  (18t  4, 15t , 3)
dt dt
dx
 u x  18t  4  x  9t 2  4t  A1
dt
dy
2
 u y  15t
 y  7.5t  B1
dt
dz
 uz  3
 z  3t  C1
dt
14
Solution
(d) At t = 0, (x, y, z) = (1, -2, 0)
x(t = 0) = 1  1 = 0 +A1 or A1 = 1
y(t = 0) = -2  -2 = 0 +B1 or B1 = -2
z(t = 0) = 0  0 = 0 +C1 or C1 = 0
(x, y, z) = (9t2 + 4t + 1, 7.5t2 - 2, 3t)
At t = 1, (x, y, z) = (14, 5.5, 3)
15
Example
A charged particle moves with a uniform
velocity 4ax m s-1 in a region where
E = 20 ay V m-1 and B = Boaz Wb m-2.
Determine Bo such that the velocity of the
particle remains constant.
16
Solution
17
Example
A rectangular loop carrying
current I2 is placed parallel
to an infinitely long
filamentary wire carrying
current I1 as shown. Show
that the force experienced
by the loop is given by
o I1 I 2b  1
1 
F
 
 aρ N
2  o o  a 
I1
I2
o
b
a

18
Solution
• The force on the loop
Fl = F1 + F2 + F3 + F4
 I 2  dl 2  B1
o I1
B1 
a
2o
z
F2 2
F1
I1
3
I2
F3
1
4
O

F4
19
Solution
 o I1
F1  I 2  dl 2  B1  I 2  dz a z 
aφ
2

o
z 0
b
 o I1 I 2 b
a
2 o
(attractive)
 o I1
F2  I 2  da  
aφ
2

 o I1I 2 o  a

ln
az
(parallel)
2
o
 a
o
o
20
Solution
 o I1
F3  I 2  dl 2  B1  I 2  dz a z 
aφ
2

(


a
)
o
z b
0
 o I1 I 2 b

a
2 (  o  a)
(repulsive)
 o I1
F4  I 2  d a  
aφ
2
   a
 o I1 I 2  o  a
ln
az
(parallel)
2
o

o
o
21
Solution
• Total force on the loop Fl = F1 + F2 + F3 + F4
 o I1 I 2 b  1
1 
F
 
a ρ N
2   o  o  a 
22
Magnetic Torque and Moment
• The concept of a current loop experiencing a
torque in a magnetic field is important in
understanding the behavior of orbiting charged
particles, d.c. motors and generators.
• If a loop is placed parallel to a magnetic field, it
experiences a force that tends to rotate it.
• The torque T (or mechanical moment of force) on
the loop is the vector product of the force F and
the moment arm r.
• Torque T = r  F
• The unit of torque is Newton-meter.
23
Rectangular planar loop in a
uniform magnetic field
Fo
z
3
l
2
4
w
Fo
Fo
w
B

B
an
1
Fo
Axis of rotation
24
Rectangular planar loop in a
uniform magnetic field
• A rectangular loop of length l and width w placed in
a uniform magnetic field B.
• dl is parallel to B along sides 1-2 and 3-4 of the
loop and no force is exerted on those sides. For
sides 2-3 and 4-1, we get
3
1
2
4
F  I  dl  B  I  dl  B
l
0
0
l
 I  dza z  B  I  dza z  B
• F = Fo – Fo = 0 where |Fo| = IBl
• No force is exerted on the loop as a whole.
25
Rectangular Planar Loop in a
uniform Magnetic Field
• However Fo and – Fo act at different points on
the loop, thereby creating a couple.
• If the normal to the plane of the loop makes an
angle  with B, the torque of the loop is
T  Fo w sin 
T  BIlw sin 
 BIS sin  where S  lw is area of the loop
26
Magnetic Dipole Moment
• Define magnetic dipole moment as m = IS an
• Unit of magnetic dipole moment is A-m2.
• The magnetic dipole moment is the product of
current and area of the loop. Its direction is
normal to the loop.
• A bar magnet or a small filament of a current
loop is usually referred to as a magnetic
dipole. N
N
S
S
27
B lines due to a magnetic dipole
m
m
N
S
Small
current
loop
Bar
Magnet
28
Magnetization In Materials
• All materials are composed of atoms.
• Each atom may be regarded as consisting of
electrons orbiting about a central positive
nucleus.
• The electrons also rotate (or spin) about their
own axes.
• Thus, an internal magnetic field is produced in
an atom by electrons orbiting around the
nucleus or electrons spinning.
29
Electron orbiting around the nucleus
Bi
nucleus
an
Ib
electron
30
Internal Magnetic Field
• Both of these electronic motions (electrons
orbiting the nucleus and electron spin) produce
internal magnetic fields Bi that is similar to the
magnetic field produced by a current loop.
• Magnetic dipole moment is m = IbS an where S
is the area of the loop and Ib is the bound
current.
• Without an external B field applied to the
material, the sum of m (over all atoms) is zero
due to random orientation.
31
Magnetization in Materials
M0
B = 0, M = 0
B
v
v
Before B applied
After B applied
32
Magnetization In Materials
• When an external B field is applied, the magnetic
moments of the electrons more or less align
themselves with B so that the net magnetic moment
is not zero.
• The magnetization M (in amperes/meter) is defined
as the magnetic dipole moment per unit volume.
• If there are N atoms in a given volume v and the kth
atom has a magnetic moment mk, then
N
m
M  lim k 1
v
v 0
k
33
Relationship between H, B, M
• For linear material M =mH, where m is the
magnetic susceptibility of the medium. It is a measure
of how susceptible (or sensitive) the material is to
magnetic field.
• B =o(H + M) =o(1 + m) H = orH = H
•  is called the permeability of the material and is
measured in Henrys per meter (H m-1).
• r = 1 + m
34
Classification of Magnetic Materials
•
•
•
•
•
For nonmagnetic materials, r = 1 (or m = 0).
For diamagnetic materials, r  1 (or m < 0).
For paramagnetic materials, r  1 (or m > 0).
For ferromagnetic materials, r >> 1 (or m >> 0).
Relative permeability r for some materials:
Diamagnetic – mercury r = 0.999968
bismuth r = 0.999833
Paramagnetic – air r = 1.00000037
manganese r = 1.001
Ferromagnetic – cobalt r = 250
nickel r = 600
35
B-H Curve for Ferromagnetics
H
S
B
N
H
H
N
B
S
H
36
B-H Curve for Ferromagnetics
• If H is reduced to zero, B is not reduced to zero but to
Br which is referred to as the permanent flux density.
• The value of Br depends on Hmax.
• If H increases negatively, B becomes zero when H
becomes Hc, which is known as the coercive field
intensity.
• Materials for which Hc is small are said to be
magnetically hard.
• The value of Hc depends on Hmax.
37
B-H Curve for Ferromagnetics
• Further increase in H in the negative direction
to reach Q.
• A reverse in its direction to reach P gives a
closed curve called a hysteresis loop.
• The shape of the hysteresis loops varies from
one material to another.
• Some ferrites have an almost rectangular
hysteresis loop and are used in digital
computer as information of storage devices.
38
B-H Curve for Ferromagnetics
• The area of the hysteresis loop gives the
energy loss per unit volume during one cycle
of the periodic magnetization of the
ferromagnetic material.
• This energy loss is in the form of heat.
• It is desirable that materials used in electric
generators, motors, and transformers should
have tall but narrow hysteresis loop so that
hysteresis losses are minimal.
39
Example
Region 0  z  2 m is occupied by an infinite
slab of permeable material (r = 2.5).
If B = 10yax - 5xay mWb m-2 within the slab,
determine
(a) J,
(b) Jb,
(c) M,
(d) Kb on z = 0.
40
Solution
41
Solution
42
Magnetic Boundary Conditions
• The conditions that H (or B) field must satisfy
at the boundary between two different media.
• Use Gauss’ law for magnetic fields
 B  dS  0
and Ampere’s circuit law
 H  dl  I
43
Boundary Condition for normal
components of B
1
B1
B1n
an
1
1
S
h
B1t
K
2
B2
2
2 B
2n
an
B2t
44
Boundary condition for normal
components of B
• Consider the boundary between two magnetic
media 1 and 2 characterized by 1 and 2,
respectively.
• Applying Gauss’ Law to the pillbox and letting
h  0:
B1nS  B2nS = 0
 B1n = B2n or 1H1n = 2H2n
• Normal component of B is continuous at the
boundary.
• Normal component of H is discontinuous at the
boundary.
45
Boundary Condition for tangential
components of H
1
1
K
H1n
H1
1
a
H1t
d
H2
H2t
a
h
an12
2
b
w
c
H2n
2
2
46
Boundary condition for tangential
components of H
• Consider the boundary between two magnetic media
1 and 2 characterized by 1 and 2, respectively.
• Apply Ampere’s circuit law to the closed path abcda.
Assume the surface current K on the boundary is
normal to the path.
K · l = H1tw + H1nh/2 – H2nh/2 - H2tw
+ H2nh/2 - H1nh/2
Letting h  0, we have H1t – H2t = K
B1t B 2t
or

K
1 2
• Tangential component of H is discontinuous at the
boundary.
47
Boundary conditions for tangential
components of H
• We can write, in general, (H1 - H2)  an12 = K
where an12 is a unit vector normal to the
interface and is directed from medium 1 to
medium 2.
• If the boundary is free of current or the media
is not a conductor, K = 0. Then,
H1t = H2t or B1t 1  B2t 2
• Tangential component of H is continuous.
• Tangential component of B is not continuous.48
Fields that make an angle
with the normal
• B1cos1= B1n = B2n = B2cos2
B1
1
sin 1  H1t  H 2t 
B2
2
sin  2
Dividing, we get
tan θ1 1

tan θ2 2
• The law of refraction for magnetic flux line at a
boundary with no surface current.
49
Inductors
• A circuit (a closed loop with any shape) carrying
current I produces a magnetic field B which causes a
flux    B  dS to pass through the loop.
• If the circuit has N turns, the flux linkage is  = N.
• If the medium surrounding the circuit is linear, the
flux linkage  is proportional to the current I
producing it:
  I   = LI
where L is the called the inductance of the circuit.
• A circuit or part of a circuit that has inductance is
called an inductor.
50
Magnetic field B produced
by a circuit
• Inductance is a measure of how much
magnetic energy is stored in an inductor.
I
I
51
Inductance
• Define inductance L of an inductor as the ratio
of the magnetic flux linkage  to the current I
through the inductor.
 N
L 
I
I
• The unit of inductance is Henry (H) which is
the same as Weber / Ampere.
52
Inductance of Wire
 ol
L
8
l
2a
53
Inductance of Hollow Cylinder
 ol  2l 
L
 ln  1
2  a

l  a
l
2a
54
Inductance of Parallel Wires
 ol d
L
ln

a
l  d , d  a
2a
l
d
55
Inductance of Coaxial Conductor
 ol b
L
ln

a
a
b
56