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.
Chapter 2
Q. 2.11
Step-by-Step
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}=Fl_{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}
