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]]>The post Hysteresis Loss | Eddy Current and Core Losses appeared first on Electrical Academia.

]]>With appropriate constants, the hysteresis loss can be given in watts per unit volume. An empirical relationship developed by Charles P. Steinmetz gives the hysteresis loss as;

${{P}_{h}}={{k}_{h}}fB_{m}^{n}$

Where P_{h} is the watts per unit volume, k_{h} a constant term, f the number of magnetization cycles per second, B_{m} the maximum flux density, and n the Steinmetz constant normally taken as 1.6. It follows that the greater the energy required to magnetize a sample, the greater the energy needed to demagnetize it. Large hysteresis loops are, therefore, required for permanent magnets, because the large hysteresis loop represents a large storage of energy.

A changing magnetic field induces an emf in a conducting material in that field. Such emf, within a magnetic core, create circulating or eddy currents. The eddy currents encounter the electrical resistance of the core producing power loss proportional to I^{2}R losses. Although the eddy current values cannot be determined directly, the power loss has been found to be given by empirically,

${{P}_{e}}={{k}_{e}}{{f}^{2}}B_{m}^{2}$

Where P_{e }is the eddy current loss in watts per unit volume and k_{e} a constant; f and B_{m} are as previously defined. In order to reduce the magnitude of eddy currents and hence reduce the power loss in a core, magnetic cores are constructed by stacking thin laminations as shown in the following figure.

The laminations are insulated from each other by a thin coat of varnish.

In conclusion, the combined hysteresis and eddy current loss are known as the core losses.

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]]>The post Hysteresis Loop | Magnetization Curve appeared first on Electrical Academia.

]]>**Hysteresis Definition**

Hysteresis is the lagging of the magnetization of a ferromagnetic material behind the magnetizing force H.

By using a graph having B-H coordinates, we can plot the hysteresis characteristics of a given ferromagnetic material. Such a curve is plotted in the following figure and called a hysteresis loop. By periodically reversing a magnetizing force, we can plot the changing values of B within the material.

Actually, a hysteresis loop is a B-H curve under the influence of an AC magnetizing force. Values of flux density B are shown on the vertical axis and are in Tesla. Magnetizing force H is plotted on the horizontal axis.

In above figure, the specimen is assumed to be unmagnified, and the current is starting from zero in the center of the graph. As H increases positively, B follows the red dotted curve from origin to saturation point a, indicated by **B _{max}**.

As H decreases to zero, the flux follows the curve ab and drops to **B _{r}** which indicates the retentively or residual induction. This point represents the amount of flux remaining in the core after the magnetizing force is removed.

When H starts in the negative direction, the core will lose its magnetism, as shown by following the curve from point b to c. The amount of magnetizing force required to completely demagnetize the core is called the coercive force and is designated as **–H _{c} **in the figure.

As the peak of the negative cycle is approached, the flux follows the portion of the curve labeled cd. Point **–B _{max}** represents saturation in the opposite direction from

A coercive force of **+H _{c} **is required to reduce the core magnetization to zero. As the magnetic force continues to increase in the positive direction, the portion of the loop from point f to a is completed. The periodic reversal of the magnetizing force causes the core flux to repeatedly trace out the hysteresis loop.

The process by which the magnetization within the ferromagnetic materials is reduced to zero by exposing it to a strong alternating magnetic field that is gradually reduced to zero.

To demagnetize any magnetic material, we must reduce its residual magnetism **B _{r} **to zero. This can be done by connecting a suitable coil to a source of alternating current and placing it close to the object to be degaussed. Slowly moving the coil and object away from each other, causes the hysteresis loop to become progressively small. Finally, a point is reached where the loop is reduced to zero and no residual magnetism remains.

From the hysteresis loop, we can conclude different magnetic properties of a material such as:

**Reluctance**– The opposition that a magnetic circuit presents to the passage of magnetic lines through it.

**Retentivity**– The ability of a ferromagnetic material to retain residual magnetism is termed its retentivity.

**Residual Magnetism**– The magnetism remaining after the external magnetizing force is removed.

**Coercive Force** – The magnetic field strength required to reduce the residual magnetism to zero is termed the coercive force.

**Permeability**– Permeability is the measure of the ease, with which magnetic lines of force pass through a given material.

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]]>The post Absolute and Relative Magnetic Permeability appeared first on Electrical Academia.

]]>The ability of a material to concentrate magnetic flux is called permeability and its symbol is the Greek lower case letter μ. Any material that is easily magnetized tends to concentrate magnetic flux.

Because soft iron is easily magnetized, so it has a high permeability. The permeability of a material is a measure of how easy it is for flux lines to pass through it. Numerical values of μ for different materials are assigned by comparing their permeability with the permeability of air or vacuum.

Since soft iron has a high permeability (several hundred times that of air) it is much easier for magnetic flux to be conducted through it. The typical value of μ for iron vary from as low as 100 to as high as 5000, depending on the grade (quantity) used. The permeability of magnetic materials also varies according to the degree of magnetization.

Mathematically μ can be defined as the ratio of flux density to magnetizing force

$\mu =\frac{B~\left( Tesla \right)}{H~\left( {}^{A-t}/{}_{m} \right)}~~~~~\text{ }\cdots \text{ }~~~~~\left( 1 \right)$

The permeability of free space, μ_{o,} is

${{\mu }_{o}}=4\pi *~{{10}^{-7}}~{}^{H}/{}_{m}$

and is constant. The absolute permeability of another material can be expressed relative to the permeability of free space. Then,

$\mu ={{\mu }_{o}}{{\mu }_{r}}$

Where _{ }is the dimensionless quantity called relative permeability.

The relative permeability of a magnetic material, designated μ_{r} , is the ratio of its absolute permeability μ to that of air μ_{o} .

The μ_{r} of a nonmagnetic material such as air, copper, wood glass and plastic is, for all practical purposes, equal to unity. On the other hand, the μ_{r} of magnetic materials such as cobalt, nickel, iron, steel and their alloys are far greater than unity and are not constant.

**Example**

Calculate the absolute value of μ for a magnetic material whose μ_{r} is 800.

**Solution**

${{\mu }_{r}}=\frac{\mu }{{{\mu }_{0}}}$

From above equation, we have

$\mu ={{\mu }_{0}}{{\mu }_{r}}=800*4\pi *~{{10}^{-7}}$

$\mu =3200\pi *~{{10}^{-7}}~{}^{H}/{}_{m}$

If the field intensity and permeability of a circuit are known, we can calculate the flux density B using equation 1.

$\mu =\frac{B}{H}$

$B=\mu H$

From above equation, flux density can be calculated and expressed in Tesla.

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]]>The post Magnetic Field Intensity | Definition Formula appeared first on Electrical Academia.

]]>Magnetic field intensity is also known as the magnetizing force which is measured is ampere-turns per meter (A-t/m).

Of primary concern, however, is the magnetomotive force needed to establish a certain flux density, B in a unit length of the magnetic circuit.

The letter symbol for magnetizing force (magnetic field intensity) is H. The following relationship defines H as;

$H=\frac{\mathfrak{F}}{l}$

Where

ℑ =applied MMF in ampere-turns

*l* =average length of the magnetic path in meters

**Example**

Find the magnetic field intensity in the magnetic circuit shown below:

Solution:

We can calculate the intensity using following formula:

$H=\frac{F}{l}=\frac{(2.5*{{10}^{2}})*(1.5*{{10}^{-1}})}{1.2*{{10}^{-1}}}$

$H=\frac{3.75*{{10}^{1}}}{1.2*{{10}^{-1}}}=3.125*{{10}^{2}}A-t/m$

If the dimensions of the magnetic path were changed, the value of H would also change. Fox example, if the total length of the magnetic path doubled, we should expect the value of H to decrease to one-half its previous amount.

If physical dimensions are double for the above circuit, then magnetizing force will be;

$H=\frac{3.75*{{10}^{1}}}{2.4*{{10}^{-1}}}=1.5625*{{10}^{2}}A-t/m$

So, we can observe that, for a given number of ampere-turns, the magnetizing force varies inversely per unit length of the magnetic path.

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]]>The post Magnetic Flux Density | Definition and Formula appeared first on Electrical Academia.

]]>Flux density is the measure of the number of magnetic lines of force per unit of cross-sectional area.

While the total amount of the flux produced by a magnet is important, we are more interested in how dense or concentrated, the flux is per unit of cross-sectional area. Flux per unit of cross-sectional area is called flux density.

**Magnetic Flux Density Formula**

Its letter symbol is B. The relationship between total flux and flux density is given by the following equation:

$B=\frac{\varphi }{A}$

Where

B=flux density in Tesla

φ=total magnetic flux in weber

A= Cross-sectional area in square meter

**Magnetic Flux Density Unit**

The SI Unit for flux density is the Tesla (T) which is defined as;

$B=\frac{\varphi }{A}$

$B=\frac{Wb}{{{m}^{2}}}=Tesla$

“If one line of magnetic field passes normally through m^{2} area, the magnetic flux density, B, will be one Tesla,

Calculate the flux density in a ferromagnetic material with a cross-sectional area of 0.01 m^{2} containing 100 lines.

**Solution**

We know that 100 lines equal to 1 μWb. By using following formula, we can calculate the flux density B

$B=\frac{\varphi }{A}$

$B=\frac{1~\mu ~Wb}{1*{{10}^{-2}}{{m}^{2}}}=\frac{1*{{10}^{-6}}~Wb}{1*{{10}^{-2}}{{m}^{2}}}=1*~{{10}^{-4}}~T$

Determine the cross- sectional area of a toroid that has a flux of 0.5 Wb and a flux density of 25 T.

**Solution**

We have the following formula to calculate cross-sectional area:

__$A=\frac{\varphi }{B}=\frac{0.5~Wb}{25~{}^{Wb}/{}_{{{m}^{2}}}}=2*~{{10}^{-2~}}{{m}^{2}}$__

An air core coil has 0.65 μ Wb of flux in its core. Calculate the flux density if the core diameter is 4 cm.

**Solution**

First calculate the area of the core:

$A=\pi ~{{r}^{2}}=3.14*{{\left( 0.02m \right)}^{2}}=1.256*~{{10}^{-3}}~{{m}^{2}}$

Now, calculate the flux density using following formula:

$B=\frac{\varphi }{A}=~\frac{0.65*~{{10}^{-6}}~Wb}{1.256*~{{10}^{-3}}~{{m}^{2}}}=5.175*~{{10}^{-4}}~T$

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]]>The post Magnetic Flux Definition and Unit appeared first on Electrical Academia.

]]>**“a total number of lines of magnetic force passing through any surface placed perpendicular to the magnetic field. It is denoted by φ (Greek Letters Phi).**

**Magnetic Flux Formula**

The magnetic flux is also defined as the dot product of magnetic field B and vector Area A, as shown in the following figure.

$\text{Magnetic }\!\!~\!\!\text{ Flux}=\text{ }\!\!\varphi\!\!\text{ }=B.A~~~\text{ }\cdots \text{ }~~~~~\left( 1 \right)$

Where θ is the angle between B and vector area A or the outward normal drawn to the surface area as shown in above figure. A is a vector area whose magnitude is the area of the element and its direction is along the normal to the surface of the element.

**Maximum Flux**

If the magnetic field is directed along the normal to the area, so θ is 0 as shown in the following figure.

Now by equation (1), we get

$\varphi =BAcos~\left( {{0}^{}} \right)$

Hence

$\varphi =BA~~~~\cdots \text{ }~~~~\left( 2 \right)$

Equation (2) shows that the magnetic flux is maximum when the angle between a magnetic field and normal to an area is zero.

**Minimum Flux**

If the magnetic field is parallel to the plane of area and angle between the field and normal to an area is 90^{° }as shown in the figure.

Then

$\varphi =BAcos~\left( {{90}^{o}} \right)$

$\varphi =0$

It means that flux through an area in this position is zero or minimum.

SI unit of magnetic flux is Weber, as explained below,

$\varphi =BAcos\theta $

$\varphi =BA~~~~\text{ }~~\therefore ~~surface~area~perpendicular~to~B,~so~Cos\theta =1~~$

$\varphi =\frac{Newton}{Ampere-meter}*~mete{{r}^{2}}$

$\varphi =\frac{N}{A-m}*~{{m}^{2}}$

$\varphi =\frac{Nm}{A}$

$\varphi =Weber$

Hence magnetic flux unit is NmA^{-1}^{ }or weber. Whereas

\[1\text{ }Weber={{10}^{8}}lines\text{ }of\text{ }force\]

**Example**

A certain magnet has a flux φ = 1.5 Weber. How many lines of force does flux represent?

Solution:

By definition,

\[1\text{ }Weber={{10}^{8}}lines\text{ }of\text{ }force\]

So,

\[1.5\text{ }Weber\text{ }=\text{ }1.5\text{ }*\text{ }{{10}^{8}}lines\]

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]]>The post Magnetomotive Force (MMF) Definition appeared first on Electrical Academia.

]]>**“The force produced by current through a coil of wire is called magnetomotive force (mmf), as shown in the following figure.**

It is the force by which a magnetic field is produced. Just as an increase in electromotive force produces more current in a circuit, so an increase in the mmf increases the strength of the magnetic field.

Magnetomotive force (mmf), represented by ℑ, is measured in ampere- turs (A-t) of the coil shown above.

“One Ampere-turn is the amount of magnetic force or flux produced when one ampere flows through a single turn of an electrical conductor.”

The magnetomotive force (mmf) formula is expressed by the following relationship.

$\mathfrak{F}=A-t$

A coil having 10 turns and 1A of current produces the same magnetic flux or force as one having 5 turns and 2A current because the ampere-turns are the same.

**Example**

What is the magnetizing force of a coil having 4000 turns and 10 mA current flowing through it?

**Solution**

We have following formula to calculate ampere – turns (magnetizing force):

$\mathfrak{F}=A-t$

So, putting values of current and turns into formula,

$\mathfrak{F}=4000*0.01=40A-t$

Thus, if a large amount of magnetizing force is needed and only a small amount of current is available, it is necessary to use a great number of turns on the coil.

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]]>The post Magnetic Force on a Current Carrying Conductor appeared first on Electrical Academia.

]]>Consider a copper rod which can move on a pair of copper rails. The whole arrangement is placed in a uniform magnetic field. The magnetic field on the rod is directed vertically upwards as shown in fig.

When a current is passed through a copper rod from a batter, a rod moves on the rails. The relative direction of current, magnetic field and motion of conductor (force) are shown in the figure.

It can be observed that the force on a conductor is always at right angle to the plane containing the rod and the direction of the magnetic field B and is predicted by Fleming’s Left-Hand rule. The magnitude of the force acting on the rod depends on the following factors.

- The force F is directly proportional to Sin𝛂 where 𝛂 is the angle between conductor L and magnetic field B.

$F\propto Sin\alpha $

It means that the force is zero if the rod is placed parallel to the magnetic field and force is maximum when the conductor is placed at right angle to a field.

- The force F is directly proportional to the current I flowing through the conductor.

$F\propto ~I$

- The force F is directly proportional to the length of the conductor L inside the magnetic field B.

$F\propto ~L$

- The force F is directly proportional to the strength of magnetic field applied.

$F\propto B$

Combining all four factors, we can write a simple following equation

$F\propto BIL~Sin\alpha $

Or

$F=KBIL~Sin\alpha $

Where K is the constant of proportionality. The SI units of force, current, conductor length and magnetic field are given below:

Force is measured in Newton (N)

Current is measured in ampere (A)

Conductor length is in Meter (m)

Magnetic field strength is measured in Tesla Whereas;

$Tesla={}^{N}/{}_{Am}$

By putting all units is above formula, we can observe that K has no units.

**Example**

A 20 cm wire carrying a current of 10 A is placed in a uniform magnetic field of 0.3 Tesla. If the wire makes an angle of 40^{o} with the direction of magnetic field, find the magnitude of the force acting on the wire?

**Solution**

Length of the wire = L= 20 cm = 0.2 m

Current in the wire = I= 10 A

Strength of magnetic field = B = 0.30 T

Angle along magnetic field =α= 40^{o}

In order to calculate Force, we have following formula:

$F=BIL~Sin\alpha $

So,

$F=0.30*10*0.2*Sin~({{40}^{o}})$

$F=0.386~N$

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]]>The post Difference Between Diamagnetism, Paramagnetism, and Ferromagnetism appeared first on Electrical Academia.

]]>A material that turns at a right angle to the field by producing a magnetic response opposite to the applied field is called diamagnetic material such as silver, copper, and carbon have permeability’s slightly less than free space (for copper, μ_{r}= 0.9999980).

A material aligning itself with the applied field is called paramagnetic material. Paramagnetic materials such as aluminum and air have permeability’s slightly greater than that of free space (for air μ_{r} =1.0000004). The effects of diamagnetic and paramagnetic are negligibly small so that materials possessing these weak phenomena are said to be non-magnetic.

Within the paramagnetic class of materials, is a special classification of materials called ferromagnetic material. These materials are strongly attracted to magnets and exhibit Paramagnetism to a phenomenal degree. Ferromagnetic materials such as iron, steel, cobalt and their alloys have relative permeability’s extending into the hundreds and thousands, are said to be magnetic.

The magnetic properties of matter are associated with the spinning motion of electrons in the third shell of the atomic structure.

An electron revolving in an orbit around the nucleus of an atom is equivalent to a tiny current loop, which gives rise to a magnetic field. Also, the magnetic field is associated with the angular momentum of the electrons’ spin on its own axis.

In most atoms, there is a tendency for both the orbital and spin angular momentums to cancel each other by pair formation. For example, an electron spinning clockwise can pair with an electron spinning counter clockwise. Their total momentum and magnetism are then zero. Variations in this electron pairing account for the weak magnetism of the nonmagnetic materials. Diamagnetism results from an unbalance of the orbital pairing of electrons, whereas Paramagnetism results from an unbalance of the spin pairing of electrons.

Properties | Ferromagnetic Materials | Paramagnetic Materials | Diamagnetic |
---|---|---|---|

State | They are solid. | They can be solid, liquid or gas. | They can be solid, liquid or gas. |

Effect of Magnet | Strongly attracted by a magnet. | Weakly attracted by a magnet. | Weakly repelled by a magnet. |

Behavior under non-uniform field | tend to move from low to high field region. | tend to move from low to high field region. | tend to move from high to low region. |

Behavior under external field | They preserve the magnetic properties after the external field is removed. | They do not preserve the magnetic properties once the external field is removed. | They do not preserve the magnetic properties once the external field is removed. |

Effect of Temperature | Above curie point, it becomes a paramagnetic. | With the rise of temperature, it becomes a diamagnetic. | No effect. |

Permeability | Very high | Little greater than unity | Little less than unity |

Susceptibility | Very high and positive | Little greater than unity and positive | Little less than unity and negative |

Examples | Iron, Nickel, Cobalt | Lithium, Tantalum, Magnesium | Copper, Silver, Gold |

written by* Ahmed Faizan M.Sc. (USA), Research Fellow (USA)*

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