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## Q. 2.10

A magnetic circuit (Fig.2.21a) consists of a core of very high permeability, an airgap length of $l_{g}=0.4$cm 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.