advanced electromagnetism

Advanced electromagnetism

E.m.f. in a conductor

•Moving a conductor through a magnetic field can induce an emf.

• The faster the conductor moves through the field the greater the emf and hence the greater the current

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E.m.f. in a conductor

• The bigger the length of the conductor moving through the field the greater the emf and hence the greater the current

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E.m.f. in a conductor

• The greater the flux density (B) of the field, the greater the emf and hence the greater the current

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•Magnitude of induced e.m.f. in a conductor •E = Blv volts

(B = flux density in Teslas, l = length in metres and v = velocity in m/s)

E.m.f. in a conductor

•It is the rate at which the conductor “cuts” through the magnetic field

= dФ/dt volts

E.m.f. in a conductor

Right hand rule

The direction of the current can be found using Fleming’s right hand rule

Example

A conductor, 800 mm long, is moved at a uniform speed at right-angles to a magnetic field

of density 0.8Tesla. Calculate the velocity required to generate an e.m.f. in the conductor of 10 V.

Example

A conductor, 800 mm long, is moved at a uniform speed at right-angles to a magnetic field

of density 0.8Tesla. Calculate the velocity required to generate an e.m.f. in the conductor of 10 V.

Example

From: E = Bℓv V = e / Bℓ

= 10 / (0.8 x 800 x 10-3) = 15.65 m/s

E.m.f induced in a rotating coil

E = - d/dt(NΦ)

example

A coil of 3000 turns, when energised, produces a magnetic flux of 3.5 mWb. If the energising

current is reversed in 0.3 seconds, determine the direction and average value of e.m.f. induced in

the coil.

example

From: E = -N x (dΦ) / dtE = -3000 x (-2 x 3.5 x 10-3) / 0.3 = 70V

Flux reversal hence +ve

Flux linkage

Flux relates to the number of field linesΦ = BA

Flux linkage takes into consideration the number of turns in a coil

Flux linkage = NΦ

Induction in terms of Flux linkage (NΦ)

L = NΦ/I

= N.dΦ/dI

Rate of change in flux linkage

Example

•A coil of 250 turns is wound on a non-magnetic ring. If a current of 5 A produces a magnetic flux

of 0.4mWb, calculate: •a) Inductance of coil

•b) Average e.m.f. induced in the coil when switching on if the current takes 3 ms to rise to

its final value.

Example

•From: L = NΦ / IL = 250 x (0.4 x 10-3) / 5= 20 mH

b) From: e.m.f.av = - NΦ ./ t•e.m.f.av = - 250 x (0.4 x 10-3) / (3 x 10-3) = - 33.3V.

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Pushing a magnet into a coil induces a current in the coil wire

Pulling the magnet out of the coil induces a current in the opposite direction

Current growth in an inductive circuit

I = Io(1-eRt/L)

I

t

Current decay in an inductive circuit

I = Io(e-Rt/L)

I

t

120V

120 Ω

40 Ω

Example

•A non-reactive resistor of 120 Ω is connected in parallel with a coil of inductance 4 H and

resistance 40 Ω. Calculate the current flowing in the coil 0.06 seconds after the circuit is disconnected from a 120 V dc supply.

Example

• Initial current through coil = V/RI= 120 / 40 = 3 A

•From i = I x e -Rt/L

I = 3 x e-(120 x 0.06) / 4= 3 x e- 1.8

e-1.8 = 0.1653I= 3 x 0.1653 = 495.9 mA

Energy stored in an inductor

W(energy) =1/2LI2

Example

A current of 5 A flows through a coil of 3 H. Calculate the amount of energy stored in the

coil.

Example

From: energy stored = 1/2 x LI2

Energy stored = 0.5 x 3 x (52) = 37.5 Joules

Mutual Inductance

E = - MdI/dt = -N2dΦ/dt

M = N2Φ2/I1

Example

•Two coils have a mutual inductance of 300 μH. Calculate the e.m.f. produced in one coil when the current in the other coil changes at the rate

of 20 x103 A/s.

Example

From: E = -M x dI / dt•E = -300 x 10-6 x 20000 = -6 V

Magnetism

The relationship between magnetic field strength and magnetic flux density is:

B = H × µ

where µ is the magnetic permeability of the substance

Magnetic field strength equation in a coil H = (NI) / l

where: H = magnetic field strength (ampere per metre)

I = current flowing through coil (amperes) N = number of turns in coil

l = length of magnetic circuit

Magnetism

Magnetism

The magnetomotive force in an inductor or electromagnet consisting of a coil of wire is given

F = NI where N is the number of turns of wire in the coil

and I is the current in the wire.

Magnetism

Permeability

Is a measure of how easily a magnetic field can set up in a material

It is the ratio of the flux density of the magnetic field within the material to its field strength

µ =B/HPermeabilty of free space µo is 4π x10-7 H/m

Magnetism

Relative Permeablity µr

• This is how much more permeable the material is compared to free space (a vacuum). The permeability

of the material can be calculated by multiplying its relative permeability by the permeability of free

space.•µ = µo x µr

Magnetism

Magnetic Flux

The rate of flow of magnetic energy across or through a (real or imaginary) surface. The unit of

flux is the Weber (Wb)

Magnetic Flux Density

A measure of the amount of magnetic flux in a unit area perpendicular to the direction of magnetic flow, or the amount of magnetism

induced in a substance placed in the magnetic field.

The SI unit of magnetic flux density is the Tesla, (T).

One Tesla, (1T), is equivalent to one weber per square metre (1 Wb/ m2).

Magnetism

• To summarise• The magnetic flux density , B, multiplied by the area

swept out by a conductor, A, is called the magnetic flux, Φ.•Φ = BA

• . Unit of flux: weber, Wb• Unit of flux density: Tesla, T

Example

•A coil of400 turns is wound uniformly over a wooden ring of mean circumference 200 mm and cross-

sectional area 150 mm2. If the coil carries a current of 2 A, calculate:

• a) Magneto-motive-force •b) Magnetic Field Strength

• c) Flux Density •d) Total Flux

Example

• a) From: m.m.f = NI = 400 x 2 = 800 AT•b) From: H = m.m.f. / ℓ = 800 / 200 x 10-3 = 4 x 103 A/m• c) From: B / H = μ0 . B = μ0 x H = 4π x 10-7 x 4 x 103 =

5.027 x 10-3 Tesla•d) From: Φ = B x area = 5.027 x 10-3 x 150 x 10-6 =

0.754 μWeber

‘Hard’ magnetic materials

Hard magnets, such as steel, are magnetised, but afterwards take a lot of work to de-magnetise. They're

good for making permanent magnets.

.

‘Soft’ magnetic materials

•Soft magnets are the opposite. With an example being iron, they are magnetised, but easily lost their magnetism, be it through vibration or any other means. These are best for things that only

need to be magnetised at certain points, eg magnetic fuse/trip switch

Magnetism

Solenoid

Magnetism

The magnetomotive force in an inductor or electromagnet consisting of a coil of wire is given by:

F = NI where N is the number of turns of wire in the coil

and I is the current in the wire.The unit is amp.turns

(AT)

Magnetism

Magnetic field strength in a coil = mmf/ length of the coil

Magnetic field strength equation in a coi

H = (NI) / lwhere:

H = magnetic field strength (ampere per metre) I = current flowing through coil (amperes)

N = number of turns in coil l = length of magnetic circuit

Retentivity –

A measure of the residual flux density corresponding to the

saturation induction of a magnetic material. In other words, it is a

material's ability to retain a certain amount of residual magnetic field

when the magnetizing force is removed after achieving saturation

• Residual Magnetism or Residual Flux - the magnetic flux density that remains in a material

when the magnetizing force is zero.

• Coercive Force - The amount of reverse magnetic field which must be applied to a magnetic material

to make the magnetic flux return to zero. (The value of H at point c on the hysteresis curve

Magnetism

• Example

• Starting with the concept of molecular magnets in a magnetic material, explain• a) Relative permeability of a material

• b) Loss of magnetisation in a ‘soft’ material

• c) Magnetic saturation

Magnetism

• a) Relative permeability of a material, how easily molecular magnets align with applied field

• b) Loss of magnetisation in a ‘soft’ material, how easily molecular magnets take up random alignment

• c) Magnetic saturation, molecular magnets all aligned in field direction

Example

• A coil, uniformly wound over a mild steel ring, produces a magnetic field strength of 3000 A/m when energised.

Using an appropriate curve :•

• i) Flux density

• ii) Relative permeability of mild steel under the stated conditions •

Example

• From Curve - Flux Density v Magnetic field strength

• Flux density = 1.5Tesla

• ii) From B / H = μ0 x μr

. Then μr = B / H μ0

= 1.5 / (3000 x 4π x 10-7 )

• = 398

Example

m(relative) = m(material) / (m air)

Example

Reluctance of a magnetic circuit (S)S = mmf/Ф

S = ℓ / μ0 μr

a

Example

A mild steel ring has a cross- sectional area of 400 mm2

and mean circumference of 300 mm. If a coil of 250 turns is wound uniformly over the ring, calculate:

i) Flux density in the ring assuming a total flux of 600 μWb

ii) Reluctance of the ring assuming μr

iii) Current to produce the required flux density

Example

Example

• i) From Φ = B x area. Flux density B = 600 x 10-6 / 400 x 10-6 = 1.5 Tesla

ii) From curve – μr = 960

From: S = ℓ / μ0 μra

= 300 x 10-3 / (4π x 10-7 x 960 x 400 x 10-6) = 6.2 x 105A/Wb

Example

•From: m.m.f. = ΦS = 600 x 10-6 x 6.2 x 105 •= 372 Ampere-turns.

•Current m.m.f/ turns =372/ 250 •= 1.48 Amperes