Advanced Topic - Electromagnetism (Optional)

This topic is about magnetism and the magnetic effect of a current. It has a very close relation with electromagnetic induction. However the content covered has very little applications in A.S.L. Electronics syllabus. Students can skip this part if they do not have time.
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Magnetic field and magnetic field lines

A magnet has a north pole and a south pole. It exerts magnetic forces and the space experiences these forces is called a magnetic field. The direction of the magnetic field at a point is defined as the direction of the force experienced by a magnetic north pole. A magnetic field can be represented by magnetic field lines, coming out from the north pole and entering into the south pole of a magnet. The magnetic field lines of a bar magnet is shown in Fig 2.8.


Fig. 2.8 field lines of a bar magnet


Force on a current carrying conductor

When a current flows through a wire, a magnetic field is produced around the wire (Fig. 2.9).



Fig. 2.9

A wire carrying a current can be denoted by a circle. A dot in the circle indicates the current is flowing out of the screen and a cross in the circle indicates the current is flowing into the screen.

(a) (b) (c)
Fig. 2.10

Consider a wire carrying a current as shown in Fig. 2.10(a) is placed in the magnetic field in Fig. 2.10(b). The resultant magnetic field is shown in Fig. 2.10(c). It can be seen that the magnetic field immediately above the wire is strengthened, whilst the field below is weakened. Therefore, a net downward force F is produced to move the wire from the stronger field to the weaker field. The magnetic flux density B (or the magnetic induction) is defined as the force acting per unit length on a wire which carries unit current and is perpendicular to the direction of the magnetic field.

B = F / I l

where I is the current flowing through the wire, and
l is the length of the wire.

The SI unit for B is tesla (T). 1 T = 1 N A-1 m-1. It is a vector quantity whose direction at a point is the direction of the field line at that point.

Example 2.3

Calculate the force on a conductor of length m which carries a current of A at right angle to a magnetic flux density of T.

B = T,      I = A,     l = m
F = B I l = = N.

The direction of the force on the wire can be determined by using the Fleming's left hand rule shown below :



All three directions are mutually perpendicular.


Magnetic flux

The magnetic flux through a small surface is defined as the product of the magnetic flux density B and the area of the surface A. The SI unit of magnetic flux is Weber (Wb).

= B A

Example 2.4

Calculate the magnetic flux density when a flux of Wb passes through an area of m2.

= 10-6 Wb,      A = m2
B = / A = 10-6 / = T.


Faraday's law of electromagnetic induction

If a conductor is moving in a magnetic field, an electromotive force (e.m.f.) is induced in the conductor. In fact, an e.m.f. is induced whenever the conductor cuts across the lines of magnetic flux.

Faraday's Law states that the magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linking with the conductor or the rate of flux cutting by the conductor.

It is not difficult to show, by using the Faraday's Law, when a conductor of length l cuts a magnetic field of flux density B at a uniform velocity v, the e.m.f. E induced in the conductor is given by

E = B l v

If the conductor is moving at some angle to the magnetic field, the equation is modified to

E = B l v sin

Example 2.5

A straight conductor of active length m is moving at a velocity of m s-1 perpendicular to the magnetic field of flux density T. Calculate the e.m.f. induced in the conductor.

B = T,     l = m,     v = m s-1
E = B l v = = V.



The direction of the induced e.m.f. can be determined by the Fleming's right hand rule shown below :


All three directions are mutually perpendicular.

Magnetomotive force and magnetic field strength


Fig. 2.11

In Fig 2.11, lines of magnetic flux form closed paths. The completed closed path followed by magnetic flux lines is referred as a magnetic circuit.

The source which produces the magnetic field in a magnetic circuit is known as the magnetomotive force (m.m.f.) which is given by

m.m.f. = I N

where I is the current flowing through the coil and N is the number of turns of the coil. The SI unit of m.m.f. is Ampere (A).

Example 2.6

Calculate the m.m.f. produced by a coil of turns when it carries a current of A.

I = A,     N =
m.m.f. = IN = = A.

The magnetic field strength H is defined as the m.m.f. per unit length of the magnetic circuit.

H = m.m.f. / l
 or
H = IN / l

The field strength is also known as the magnetizing force or magnetic field intensity. The SI unit for H is Ampere per metre (Am-1).

Example 2.7

A coil of length m is wound with turns of wire and carries a current of A. Calculate the magnetic field strength at the centre of the coil.

I = A,      N = ,      l = m
H = IN / l = / = Am-1

Permeability

The absolute permeability of a material is defined as

= B / H

The SI unit of is henry per metre (Hm-1).

The absolute permeability of a vacuum is constant, and is called permeability of free space 0 ,

0 = 4 10-7 Hm-1

The permeabilities of air and all non-magnetic materials have similar values as 0. For practical purposes, we can take 0 as the value of the permeabilities of all non-magnetic materials.

For the same field strength, a core made of magnetic material can produce a much stronger magnetic flux than a core made of non-magnetic material. In other words, permeability of a magnetic material is much greater than permeability of free space. The ratio between the two permeabilities is called the relative permeability r.

r = / 0
     or    
= 0 r

The relative permeability of a magnetic material is not constant, it depends on the field strength.

Example 2.8

A coil of 200 turns is wound uniformly over a wooden ring having a mean circumference of 600 mm and a uniform cross-sectional area of 500 mm2. If the current through the coil is 4 A, calculate

(a) the magnetic field strength,
(b) the magnetic flux density, and
(c) the total flux linkage.

N = 200,     l = 600 mm,    A = 500 mm2,     I = 4 A

(a) H = I N / l = 4 200 / (600 10-3) = 1333 Am-1.
(b) B = 0 H = 4 10-7 1333 = 1.675 mT.
(c) = B A = 1.675 10-3 500 10-6 = 0.8375 Wb.


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