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## Q. 2.12

Consider the magnetic circuit of Fig.2.21(a). The permanent magnet material Alinco-5 is in demagnetized state. It is required to be magnetized to a reduced flux density $B_{r} = 1.25$ T.Magnetic circuit dimensions are: $A_{m} =A_{g} =2.5cm^{2}$,$l_{m} = 4$ cm, $l_{g} = 0.2$ cm. Excitation coil turns,N = 200.

## Verified Solution

To find the first quadrant tip of the hysteresis loop, we assume

$H_{max} =3.5H_{c}$

From Fig. 2.20(a)
$H_{c}=50$ (magnitude)
So, $H_{max} =170$KA/ m
Extrapolating B-H curve into first quadrant, we get
$B_{max}\approx 2$T
Substituting values in Eq. (2.50)
$B_{m}=-\mu _{0} \left\lgroup\frac{A_{g} }{A_{m} } \right\rgroup \left(\frac{l_{m} }{l_{g} } \right)H_{m} +\mu _{0} \left\lgroup\frac{A_{g} } {A_{m} } \right\rgroup \frac{1}{l_{g} } \left(Ni\right)$
$B_{max} =\mu _{0} \left[-\left\lgroup\frac{2}{2} \right\rgroup \left\lgroup\frac{4}{0.2} \right\rgroup H_{max}+\left\lgroup\frac{2}{2} \right\rgroup \frac{1\times 200i}{0.2\times 10^{-2} } \right]$
$\mu _{0}=4\pi \times 10^{-7}$
$B_{max} =-2.51\times 10^{-5} H_{max}+ 12.57\times 10^{-2}i$
Substituting for$B_{max}$ and $H_{max}$
$2=-2.51\times 10^{-5} \times 170\times 10^{3} +12.57\times 10^{2} i$      (2.51)
Solving Eq.(2.51), we get
i=49.86 A

Note: Equation (2.51) represents a straight line in B-H plane for a given i. Its intersection with B-H curve gives the state
of the permanent magnet at that value of exciting current