The post Absolute and Relative Magnetic Permeability (µ) appeared first on Electrical A2Z.

]]>*The permeability of a material is defined as the ease with which a flux can be created in that material. If there is no material, such as in a vacuum, the value can be shown to be 4E – 7π (or 4πE – 7; if you prefer). That is the permeability of free space (a vacuum), which has the abbreviation µ _{o}.*

The permeability of free space is the reference against which the permeability of other materials is compared.

To compare the permeability of any given material with the permeability of free space, it is necessary to use a ratio µ_{r} which is known as the relative permeability of the material. For air and other non-magnetic materials, µ_{r} has the value of unity (µ_{r} = 1).

\[{{\mu }_{r}}=\frac{\mu }{{{\mu }_{o}}}\]

If the non-magnetic core of a solenoid is replaced with a magnetic material, the flux produced by the same number of ampere-turns may be greatly increased. The ratio of the flux produced by the magnetic core to that produced by the non-magnetic core is a direct result of the relative permeability of the magnetic material. For some magnetic materials µ_{r} can have a value in the thousands.

For any one magnetic material, the relative permeability value can vary considerably, being dependent on the flux density in the material. Relative permeability is higher at low values of **flux density**.

To find the absolute permeability of a material, the permeability of free space is multiplied by the relative permeability of the material:

\[\mu ={{\mu }_{o}}{{\mu }_{r}}\]

**Where:**

**µ =**absolute permeability**µ**permeability of free space_{o}=**µ**relative permeability_{r}=

**Example**

Calculate the absolute permeability for a magnetic material whose μ_{r} is 800.

**Solution**

\[{{\mu }_{r}}=\frac{\mu }{{{\mu }_{0}}}\]

From above equation, we have

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

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

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]]>The post Magnetic Flux Density (B) appeared first on Electrical A2Z.

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

The general symbol for magnetic flux density is B and the unit is the weber per square meter (Wb/m^{2}). One weber per square meter is called a tesla (T).

If both the total flux and the area of the magnetic path are known, the flux density is found from:

**Where:**

**B**= flux density in tesla (Wb/m^{2})**Φ**= total flux in webers**A**= area in m^{2}

A magnetic circuit has a cross-sectional area of 100*mm*^{2} and a flux density of 0.01T. Calculate the total flux in the circuit.

**Note**: The answer is expressed in webers and not in lines of force.

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, we’ll calculate the core area:

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

Now, we can calculate the magnetic flux density using the 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 Field Intensity appeared first on Electrical A2Z.

]]>*The m.m.f. required to magnetize a unit length of a magnetic path is termed the magnetizing force or magnetic field intensity (H) for that portion of the magnetic circuit.*

The magnetic field intensity is applicable only to that section of the magnetic circuit, made of the one material and with a constant cross-section. The unit is expressed in ampere-turns per meter and the symbol is H.

In a similar fashion to that for magnetomotive force, the ‘turns’ part of the expression is dimensionless and in pure SI units should be omitted, leaving it as amperes per meter. For all practical purposes, however, the more descriptive term is ampere-turns per meter. That is:

**Where:**

**l =**length of magnetic circuit in meters

**Note: **Be sure to convert length into meters! Magnetizing force must not be confused with magnetomotive force.

**Example**

Find the magnetic field intensity (H) in the magnetic circuit given below:

**Solution:**

Here, we will calculate the magnetic field intensity (H) using the following formula:

\[H=\frac{N\times I}{l}=\frac{(2.5*{{10}^{2}})*(1.5*{{10}^{-1}})}{1.2*{{10}^{-1}}}\]

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

If we change the dimensions of the magnetic path, the value of H would also change. For example, if we double the length of the magnetic path, we should anticipate the value of magnetic field intently (H) to decrease to one-half its original value calculated above.

If, for the above magnetic circuit, physical dimensions are, then magnetizing force (H) will be;

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

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

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]]>The post Magnetomotive Force (MMF) appeared first on Electrical A2Z.

]]>*The force required to create a magnetic field is called the magnetomotive force, abbreviated to m.m.f. with a general symbol Fm. *

The magnetomotive force is the force that creates the magnetic field. The m.m.f. acts like the electromotive force in an electric circuit and is the force necessary to set up a magnetic flux in a magnetic circuit.

The m.m.f. is **dependent** on the current flowing in a conductor, and the magnetic field of a solenoid is dependent on the number of turns of the solenoid. The field strength of a coil is therefore proportional to the product of the current and the number of turns in the coil.

In pure SI units, the m.m.f. unit is the ampere, because the number of turns in a coil or solenoid is considered dimensionless. In calculations, however, the number of turns has to be included.

Generally within the electrical trades, an m.m.f. quantity will be given directly as ampere-turns (abbreviated IN), and the units will be specified as ampere-turns (At).

Where:

**Fm =**m.m.f. or magnetomotive force, in ampere turns**I = c**urrent flowing in amperes**N =**number of turns in coil

**Example**

If a current of 5 A is flowing in a coil of 120 turns, find the value of m.m.f. creating a magnetic flux.

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