Question 3.6: A parallel magnetic circuit with 2000 turns on its central l...

The flux from the central limb will get equally divided in the two outer limbs. So that flux in the outer limbs will be 0.55 mWbs.
The flux density will be the same throughout as flux density, B is

for central limb               B=\frac{Φ}{A}  =\frac{1.1  ×  10^{-3}}{10  ×  10^{-4}}= 1.1  Wb/m²,   or Tesla

for outer limbs                B=\frac{Φ}{A}  =\frac{0.55  ×  10^{-3}}{5  ×  10^{-4}}= 1.1  Wb/m²,   or Tesla

The magnetic field strength, H corresponding to B = 1.1 Wb/m² is 650 AT/m.
We have to calculate the AT required to be provided for one of the parallel paths, i.e., for the central limb, one outer limb, and one air gap.
length of central limb, l_{c} = diameter = 20 cm

length of the outer limbs, (including air gap)  l_{o}=\frac{πd}{2}=\frac{3.14  ×  20}{2}=31.4  cm

length of air gap, l_{g} = 2  mm = 0.2  cm

H=\frac{NI}{l},  H=1.1  Wb/m²

(a) NI for central limb                    = H  ×  l_{c}=650  ×  \frac{20}{100} = 130

(b) NI for outer limb                     = H  ×  (l_{o}  –  l_{g})=650   \frac{(31.4  –  0.2)}{100}

= 650  ×  \frac{31.2}{100} = 202.8

NI for air gap = H_{g} × l_{g}
we have to calculate H_{g} for air

B = μ_{o}  H  as  μ_{r} = 1

H= \frac{B}{μ_{o}}= \frac{1.1}{4π  ×  10^{-7}}= 8.758  ×  10^{5}  AT/m

(c) NI for air gap               = H_{g}  ×  l_{g} = 8.758  ×  10^{5}  ×  2  ×  10^{−3} = 1751.6
Total AT required                              = a + b + c

= 130 + 202.8 + 1751.6
= 2084.6

Total number of turns of the exciting coil placed on the central limb, N = 2000.
Current,                          I= \frac{Total  AT}{N}

= \frac{2084.6}{2000} = 1.04  A.