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Chapter 2

Q. 2.10

A magnetic circuit (Fig.2.21a) consists of a core of very high permeability, an airgap length of l_{g}=0.4cm and a section of permanent magnet (made of Alnico 5) of length l_{m}=2.4 cm. Assume µ of core=∞ Calculate the flux density B_{g}in the air-gap. Given: A_{m}=4 mm^{2}


Verified Solution

\mu _{core} =\infty \Longrightarrow H_{core} =0

From Ampere’s circuital law

H_{m} l_{m}+H_{g}l_{g}=F      (2.39)
or H_{g}=-\left\lgroup\frac{l_{m}}{l_{g}} \right\rgroup H_{m}           (2.40)

where H_{g} and H_{m} are the magnetic field intensities in the air-gap and the PM respectively. Thus the existence of an air-gap is equivalent to the application of a negative field to the PM material.As the flux must be continuous around the path

\phi =B_{m} A_{m}=B_{g}A_{g}       (2.41)

Also B_{g}=\mu _{0} H_{g}

We obtain from Eqs. (2.40) and (2.41)

B_{m}=-\mu _{0}\left\lgroup\frac{A_{g}}{A_{m}} \right\rgroup \left\lgroup\frac{l_{m}}{l_{g}} \right\rgroup H_{m}      (2.42)

Substituting values we get

B_{m}=-6\mu _{0}H_{m}=-7.54\times 10^{-6}H_{m}

This is a straight line (also called load line) shown in Fig. 2.20(a), where its intersection with the demagnetization curve
at point ‘a’ gives the solution for B_{m}.    (2.43)

Thus  B_{g}=B_{m}=0.33 T

Note: If we repeat the above problem for M-5 electrical steel, it is easy to find the answer since the load line is the same
as given by Eq. (2.43). It can be shown that B_{m} =4\times 10^{-5} . This is much less than the value of B_{m} for Alnico 5.