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## Q. 2.11

For the magnetic circuit of Fig.2.21(a) if $A_{g}=3.0cm^{2}$ find the minimum magnet volume required to produce an air-gap flux density of 0.7 T.

## Verified Solution

The smallest magnet volume will be obtained with the magnet operating at point as shown in Fig. 2.20(a) which corresponds to $B_{m}$ = 1.0 T and $H_{m}$ = – 40 kA/m.

From Eq. (2.41)

$\phi =B_{m} A_{m}=B_{g}A_{g}$

$A_{m}=B_{g}A_{g}/ B_{m}=\left(0.7\times 3\right) / 1=2.1cm^{2}$

From Eq. (2.39)

$H_{m} l_{m}+H_{g}l_{g}=F$

$l_{m}=-\frac{H_{g}l_{g}}{ H_{m}} =-\frac{B_{g}l_{g}}{\mu _{0} H_{m}} =-\frac{0.7\times 0.4}{4\pi \times 10^{-7}\times \left(-40\times 10^{3} \right) } =5.57$ cm

Minimum magnet volume= $2.1\times 5.57 = 11.7 cm^{3}$ 