Chapter 7
Magnetism
7.1 Introduction
Magnetism has been known thousands of years dating back to the discovery
recorded by the ancient Greek.
1900 Maxwell combine the theory of electric and magnetic to predict the
electromagnetic wave theory.
7.2 Magnets, magnetic Poles and Magnetic Field Direction
Poles of a magnet are the ends where objects are most strongly attractednorth and south poles.
Like magnetic poles repel each other and unlike magnetic poles attract each
other.
Examples: bar magnet
7.21 Defining Magnetic Field Direction
The direction of magnetic field (B) at any location is the direction that the
north pole of a compass at that location would point.
7.22 Earth’s Magnetic Field
The Earth’s magnetic field resembles that achieved by burying a huge bar
magnet deep in the Earth’s interior
The Earth’s geographic north pole corresponds to a magnetic south pole
The Earth’s geographic south pole corresponds to a magnetic north pole
7.3 Current and Magnetism
There is a strong connection between electricity and magnetism
1820 – Hans Christian Oerstead – a professor of Physics at the University of
Copenhagen discovered that electric current could indeed affect magnet field
7.31 Magnetic Forces on Electric Current.
There will be a force perpendicular to a current carrying wire. The direction
of the force is according to Fleming left hand rule (motor pinciple).
I
I
F
B Bin
F= B I L sin 
 = Direction of current to the direction of magnetic field
L = Length of wire in magnetic field
Example 1. A wire 5.0 m length with 10.0 A current makes an angle of 300 with
magnetic field of 0.3 T. Calculate force on it.
Magnetic force F=LIB sin θ = 5.0x10.0x0.3 sin 300 = 7.5 N
F
B
300
I
Example 2. A thin horizontal copper rod 1.0 m long has mass of 50 g. The interaction
between electric current I with magnetic field of 2.0 T, let the rod ‘floating’ on air.
Determine the minimum current that result this phenomenon.
F magnetic
.x x x x x x x x x x x x x Bin
F gravity
Imin
F magnetic = F gravity
L Imin B sin θ = mg
Imin = mg/ L B sin θ
= mg/ L B sin 900 = 0.05x9.81/1x2 =0.245 A
7.32 Torque on a Current Loop
b/2
b/2
B
a
I
From the formula of force,
F= B I L sin 
 = Force (F) x distance perpendicular
 = F x b/2 sin 
 = B x I x a x b/2 sin 
Total torque = 1 + 2
 = 2 x B x I x a x b/2 sin 
 = B I A sin  where a x b = A (area of coil)
if the coil has N number of turns (loop)
 = N B I A sin 
we define magnetic dipole moment as  = N I A
 =  B sin 
Example: The electric motor
Example 3. A circular loop wire radius of 50 cm with current 2.0 A in a 0.4 T magnetic
field region, calculate maximum torque on it.
I clock
.
B
rotation
maximum torque τ max =μB sin 900 =NIAB
= Iπr2B = 2 x 3.14 x 0.52 x 0.4 = 0.628 mN
Example 4. Magnetohydrodynamic (MHD) propulsion is a type of vessel drive where
thrust is generated through interaction of magnetic and electric field. MHD propulsion is
preferable to the marine propellers because there are no mobile parts, no propeller noises
and no vibration.
Fthrust
Ffoward
d
E
E
XBX
The schematic design is as above. E is the electrical field between two plates 5.0x107
N/C, and B is the magnetic field supply from a superconducting magnet which can
generate 10 tesla strength (into page). Given that sea water average resistance 0.04 ohm,
and d = 15 cm.
Fthrust(water) = F forward (ship) = LIB sin 900
= d (V/ R) B = d (E d) B/R = [0.152 x 5.0x107 x 10 / 0.04 ]N
7.33 Ammeter and Voltmeter
Ammeter measures current through circuit elements and voltmeter measures
voltage across circuit elements
The basic component of an ammeter and voltmeter is a galvanometer.
The resistance of a galvanometer is very small, since the coil consists of
metal wires. It is a current-sensitive device which needle deflection (due to
torque on current coil) is proportional to the current through the coil.
(a) Ammeter
To build an ammeter, a resistor has to be added in parallel to the
galvanometer
G
I
Ig
r
Is
Rs
Vg=Vs or Igr =IsRs
Igr = (I-Ig)Rs
Ig =
IRs
r  Rs
(b) Voltmeter
To build a voltmeter, a resistor have to be added in series with the
galvanometer.
Ig
Rm
r
G
V = Vg +Vm =Igr + IgRm = Ig(r+Rm)
Ig =
V
r  Rm
7.34 Magnetic Forces on Moving Charged Particle
A current in a wire consists of moving charges.
I=
q
t
Since a current carrying wire may experience a force when placed in a
magnetic field, it is not surprising that a moving charge that is not confine
within a wire may also experience a force due to magnetic field.
F= B I L sin 
F=B
q
L sin 
t
Since v (velocity) = Length (L) / time
F = q v B sin 
Motion of free charge in a magnetic field
v
Bin
F
q
When the charge particle enter the magnetic field in perpendicular direction,
it will move in a circular motion.
The particle will experience a centripetal force Fc.
FE = q v B sin 
since the angle is 90o, sin 90o = 1
Fc = FE
mv2
=
r
r =
Bqv
mv
qB
Period of Rotation T
Since, Velocity = Distance / Time
T=
2r
v
2r
2m
T=
Br = qB (period for 1 rotation)
q
m
Example 5. A positive ion mass of 2.5 x 10-26 kg been accelerated through p.d of 250 V.
It enters perpendicular into region of 0.5 T field. Determine the path radius.
+p
v
x x x x x Bin
250 V
PE = qV
KE = 0.5mv2
r
v a-clock circular motion
.r = mv/qB = m (qV/0.5m)0.5 /qB = 1.77 cm
Cyclotron
An accelerator which used a magnetic field to bend the path of particles into
nearly circular orbits.
Oscillating Voltage Source
X X X X X X
BIN X X s X X X X
S-ion source within the `dees’ experiences changes electric potential difference between
the dees passes with velocity v. This will result the ion with circular path of radius r
within of the dees.
7.4 Ampere’s Law
Ampere’s Law states that the summation of the contributions around the
entire loop is proportional to the net current I through the loop.
 B  L cos  = o I
7.41 Magnetic field of a long straight wire
For the magnetic field surrounding a wire;
I
r
l
B
B=
o I
2r
o = 4 x 10-7 Tm/A
o = 4 x 10-7 N/A2
Example 6. Two wires with 3.0 A and 5.0 A separated 20 cm each other (both currents
out of page). Calculate the field at P which just above the 5.0 A wire distance of 20 cm.
B1
B2
R1
450
B1 =μ0 I1 /2π R1 = 2.12 x 10-6 T
P
B2 =μ0 I2 /2π R2 = 5.00 x 10-6 T
R2
BP = B1 + B2 = (-6.5 x + 1.5 y )10-6 T
Bp =6.67 x 10-6 T (at 770 left of the vertical axis)
wire1
wire2
3A
5A
Bp
from Pythagoras R1 =0.283 m
770
7.42 Magnetic field at the center of the coil
I(a-clockwise)
o I
Bout =
2r
7.43 Magnetic Field at the center of a solenoid
N turns
I
B axis
L
B = o n I
where n = N/L (no of turns per unit length)
7.44 Magnetic field of a toroid
Toroid- think as solenoid bent into a circle of radius r (doughnut-like shape)
r
B=
 o NI
2r
7.5 Forces between two current carrying wires
I1
I2
F1
F2
 o I1
B1 =
2d
F2 = B1 I2 L
F2=
Force per unit length on wire 2:
 o I1
I L
2d 2
 o I1 I 2
F2
=
2d
L
Example 7. Two parallel wires 10.0 cm apart each carries 10 A current in same direction.
Force per unit length F/L = μ0 I2 / 2πd = 2.0 x 10-4 N/m –attracted each other
7.6 Magnetic materials
The direction of a magnetic material is determined by its resultant magnetic
domain.
The magnetic domain arises from the electron spin in an atom.
In atoms with two or more electron, the electron usually is arranged in pairs
with their spin oppositely aligned. The magnetic field then cancels each
other, and the material is not magnetic. Aluminum (Diamagnetism) is an
example.
In unmagnetized ferromagnetic materials, the domains are randomly
oriented and there is no magnetization. But when it is placed in an external
magnetic field
a. Domain boundaries change and the domain with magnetic orientations
in the direction of the external field grow at the expense of the other
domain.
b. The magnetic orientation of some domain may change slightly so as
to be more aligned with the field.