MAGNETISM
APPLICATIONS OF MAGNET
 Motors,
 generators,
 telephones,
 relays,
 loudspeakers,
 computer hard drives and floppy disks,
 anti-lock brakes,
 cameras,
 fishing reels,
 electronic ignition systems,
 keyboards,
 t.v.
 radio components,
 Transmission equipment.
WHAT IS A MAGNET?
A magnet is a material or object that produces a magnetic field.
A permanent magnet is a piece of ferromagnetic material (such as iron, nickel or cobalt) which has properties
of attracting other pieces of these materials.
It has two poles, North and South.
MAGNETIC FIELD
 area around the magnet
 a force that exists around the magnet
 also called lines of force
OTHER IMPORTANT TERMS
MAGNETIC FLUX
 the amount of magnetic field (or number of lines of force) produced by the
magnetic source
 symbol is phi (ф)
 Unit is weber (Wb)
NOTE:
1 x 10^8 lines of force or magnetic field lines = 1 Weber
MAGNETIC FLUX DENSITY
 the amount of flux passing through a defined area that is perpendicular to the
direction of the flux.
 Symbol is B
 Unit is Tesla (T)
 Formula B = ф / A
 NOTE:
1 Tesla = 1Weber/m^2
SAMPLE PROBLEMS
1.
If a magnetic flux has 3000 lines, find the number of microwebers.
3000 lines * 1 wb/100000000 lines =
3000/100000000 = 0.00003 weber = 30 microweber =
30x10^-6 weber
Milliweber
0.00003 weber = 0.03milliweber = 0.03 x 10^-3 weber
SAMPLE PROBLEMS
2. What is the flux density in teslas when there exists a flux of 600 microweber
through an area of 0.0003 sq. m.?
B = ф/A
600 microweber = 0.0006weber = 600x10^-6 weber
B = 0.0006/0.0003 = 2T
SAMPLE PROBLEMS
3. A magnetic pole face has a rectangular section having dimensions 200mm by
100mm. If the total flux emerging from the pole is 150microweber, calculate the
flux density.
B = ф/A
Ф = 150microweber = 0.000150 = 150x10^-6 weber
A=
L= 200mm* 1cm/10mm*1m/100cm = 0.2m
W=100mm = 0.1m
= 0.02m^2
B = 0.00015/0.02 = 7.5x10^-3T = 7.5mT = 0.0075T
SAMPLE PROBLEMS
4. The maximum working flux density of a lifting electromagnet is 1.8T and the
effective area of a pole face is circular in cross section. If the total magnetic flux
produced is 353 mWb, determine the radius of the pole faces.
R=?
Ф = 353mwb = 0.353Wb
0.196/pi = r^2
Ac = pir^2
B = 1.8T
0.06 = r^2
B= ф/A
Ac = pir^2
0.25m = r
AB = ф
0.196 = pi(r)^2
250mm
A= ф/B
A = 0.353 Wb/1.8Wb/m^2 = 0.196m^2
ACTIVITY 1
1.
What is the density in a magnetic field of cross sectional area 20sq.cm having
a flux of 3mWb?
2.
Determine the total flux emerging from a magnetic pole face having
dimensions 5cm by 6cm, if the flux density is 0.9T.
3.
The maximum working flux density of a lifting electromagnet is 1.9T and the
effective area of a pole face is circular section. If the total magnetic flux
produced is 611 mWb, determine the diameter of the pole face.
Given:
 Ф = 611mWb = 0.611Wb; B = 1.9T; d=?
 A= Ф/B = 0.611/1.9 = 0.32m^2
 Ac=pi*r^2


0.32 = pi*r^2
0.32/pi = r^2
0.1 = r^2
r = 0.32m
d = 2r = 0.64m = 640mm
GOOD EVENING, ECE1B!
PERMEABILITY
 refers to the ability of a magnetic material to concentrate magnetic flux
 Symbol is mu (μ)
 Unit is Henries/m or Newton/A^2
 NOTES
1. the permeability of free space (or the magnetic space constant) and is equal to 4π × 10^−7
H/m = μo
2. Permeability of air or any non-magnetic medium
μo = B/H
3. Permeability of magnetic medium
μo μr = B/H
RELATIVE PERMEABILITY
 measure of permeability for different material in comparison with air or vacuum.
 Symbol is μr
 Unitless
μr = flux density in material / flux density in vacuum
NOTE:
Absolute permeability (μ) = μr μo
MAGNETOMOTIVE FORCE
 Magnetomotive force is the cause of the existence of a magnetic flux in a
magnetic circuit.
 m.m.f or Fm = NI
 Where N is the number of conductors (or turns)
 I is the current in amperes
 Unit is ampere-turns (sometimes)
 SI unit for Fm is ampere
MAGNETIC FIELD STRENGTH
 Or magnetizing force
 H = NI/l
 Where l is the mean length of the flux path in meters
 Thus Fm = Hl
SAMPLE PROBLEMS
 1. A magnetizing force of 8000 A/m is applied to a circular magnetic circuit of
mean diameter 30cm by passing a current through a coil wound on the circuit. If
the coil is uniformly wound around the circuit and has 750 turns, find the current in
the coil.
Given:
H = 8000A/m
Fm = NI
H = NI/l
I = 8000(0.942)/750 =10.048A
d = 30cm = 0.3m = l = 2pir =pid =pi(0.3) = 0.942m
N = 750 turns
I=?
Hl = NI Hl/N = I
SAMPLE PROBLEMS
2. A current of 5 A is passed through a 1000-turn coil wound on a circular magnetic
circuit of radius 120mm, calculate (a) magnetomotive force, and (b) magnetic field
strength.
Given
I = 5A
Fm = NI = (1000)(5)
N = 1000
R = 120mm =0.12m = l = 2pir= 2pi(0.12)
0.75m
= 5000 A
H = 5000/0.75 = 6666.67 A/m
SAMPLE PROBLEMS
3. An electromagnet of square cross-section produces a flux density of 0.45 T. If the
magnetic flux is 720 micro weber, find the dimension of the electromagnet crosssection.
B = 0.45T
Ф = 720microweber = 720x10^-6 =0.00072
A = Ф/B = 0.00072weber / 0.45weber/m^2 = 0.0016m^2
Weber/weber/m^2
Dimension = sqrt 0.0016m^2 = 0.04m
SAMPLE PROBLEMS
4. Find the magnetic field strength applied to a magnetic circuit of mean length
50cm when a coil of 400 turns is applied to it carrying a current of 1.2A
Given:
I = 1.2A
N = 400
L = 50cm = 0.5m
H = 400*1.2/0.5 = 960A/m
SAMPLE PROBLEMS
5. A solenoid 20cm long is wound with 500 turns of wire. Find the current required
to establish a magnetising force of 2500A/m inside the solenoid.
RELUCTANCE
It is a magnetic resistance of magnetic circuit to the presence of magnetic flux.
Symbol is S
Unit is 1/H or A/Wb
For a series magnetic circuit having n parts, the total reluctance S is given by: S = S1
+ S2 + ··· + Sn
S = Fm/ф = NI/ф = Hl/BA = l / (B/H)A
l / (μo μ r)A
H = NI/l
SAMPLE PROBLEMS
 Determine the reluctance of a piece of mumetal of length 150 mm and crosssectional area 1800 mm^2 when the relative permeability is 4000. Find also the
absolute permeability of the mumetal.
 Given:
 L = 150mm=0.15m
1000mm = 1m
 A = 1800mm^2 = 0.0018m^2
1800 mm *1/1000 = 1.8m
 μr = 4000
 S = l / μoμrA
 = 0.15 / (4pix10^-7) (4000*0.0018)
 = 16534.391 /H
μ=
μoμr = 4pi x 10^-7 (4000)
5.04x 10^-3 H/m
SAMPLE PROBLEMS
 A mild steel ring has a radius of 50 mm and a cross-sectional area of 400 mm^2. A
current of 0.5A flows in a coil wound uniformly around the ring and the flux
produced is 0.1 mWb. If the relative permeability at this value of current is 200
find (a) the reluctance of the mild steel and (b) the number of turns on the coil.
 Given:
 A = 400mm^2 = 0.0004m^2
S = 0.31 / 4pix10^-7(200)(0.0004)
 I = 0.5A
S = 3085191.08/H
 Ф = 0.1mWb = 0.1x10^-3 = 0.0001Wb
S = Fm/ Ф S = NI/ Ф
 μr = 200
N = S Ф/I
 r = 50mm = 0.05m= l = 2pi(0.05) = 0.31m
= 3085191.08(0.0001)/.5 = 617turns
SAMPLE PROBLEMS
 A closed magnetic circuit of cast steel contains a 6 cm long path of cross-sectional
area 1 cm^2 and a 2 cm path of cross-sectional area 0.5 cm^2. A coil of 200 turns is
wound around the 6 cm length of the circuit and a current of 0.4A flows.
Determine the flux density in the 2 cm path, if the relative permeability of the cast
steel is 750.
 For 6cm
S(total) = 1,061,032
 S = 0.06m/(4pix10^-7)(750)(0.0001m^2)
ф = NI/S = 200(.4)/1061032
 = 636,619/H
ф = 7.54x10^-5 Wb
For 2cm
B = ф/A = 7.54x10^-5 / 0.00005
S = 0.02/(4pix10^-7)(750)(0.00005)
B = 1.51T
= 424,413/H