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Chapter 2

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.

Step-by-Step

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} =170KA/ m
Extrapolating B-H curve into first quadrant, we get
B_{max}\approx 2T
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