Question Question 2 Air gap I 12 Figure 1 A 1000-turn coil is wound on the centre limb of a steel frame as shown on figure1 above. At operating flux densities, the magnetic field intensity in the outer limbs is 500A/m while at the operating flux density the field intensity in the centre (inner) limb is 900A/m. The cross-sectional area of each limb is 15cmLength of the centre limb is 25cm, air-gap is 1mm while total outer length(two equal branches) is 50cm. Neglect the effects of magnetic leakage and for an operating flux density of 1T in the central limb: a. Calculate the relative permeability of the inner and outer limbs [10 marks] b. Determine the current flowing in the coil [10 marks]

a. Magnetic field intensity, central limb, \quad H_{1}=900 \mathrm{~A} / \mathrm{m} outer \operatorname{limb}, \quad H_{2}=H_{3}=500 \mathrm{~A} / \mathrm{m}Flux density in central limb,B_{1}=1 TAs we know,B=\mu_{o} \mu_{r} HThen relative permeability of central limb,\begin{aligned}\mu_{r 1}=\frac{B_{1}}{H_{1} \mu_{o}} & =\frac{1}{900 \times 4 \pi \times 10^{-7}} \\\boldsymbol{\mu}_{r \mathbf{1}} & =\mathbf{8 8 4 . 2}\end{aligned}As the length and cross-sectional area of two outer limb is same, thus reluctance offered to flux will be same by two outer paths. Flux will be divided in two path equally and will be half of flux in central limb.flux _{1}=B_{1} A_{1} where A_{1} \rightarrow cross sectional area of central limb \operatorname{flux}_{2}=\operatorname{flux}_{3}=\frac{\operatorname{flux}_{1}}{2}=\frac{B_{1} A_{1}}{2}flux density in outer limbs, \quad B_{2}=B_{3}=\frac{f_{l u x_{2}}}{A_{2}}=\frac{B_{1} A_{1}}{2 A_{2}} given that A_{1}=A_{2}=A_{3}=15 \mathrm{~cm}^{2}Then flux density in outer limbs,B_{2}=B_{3}=\frac{B_{1}}{2}=0.5 TThen relative permeability of outer limbs,\mu_{r 2}=\frac{B_{2}}{H_{2} \mu_{o}}=\frac{0.5}{500 \times 4 \pi \times 10^{-7}}=795.77Thus, relative permeability of central limb is \boldsymbol{\mu}_{r \mathbf{1}}=\mathbf{8 8 4 . 2} while relative permeability of outer limbs is \boldsymbol{\mu}_{r 2}=\mathbf{7 9 5 . 7 7}.b. As we know,M M F_{\text {central }}=H_{1} l_{\text {central limb }}Where, length of central limb,l_{\text {central limb }}=25-\text { airgap length }=25-0.1=24.9 \mathrm{~cm}=0.249 \mathrm{~m}Then,M M F_{\text {central }}=900 \times 0.249=224.1 \text { ATs }Also we know,M M F=\text { Current } \times \text { number of turns, }For number of turns on central limb \mathrm{N}_{\text {central }}=1000,\begin{array}{c}\text { Current in coil }=\frac{M M F_{\text {central }}}{N_{\text {central }}}=\frac{224.1}{1000}=0.2241 \mathrm{~A} \\\boldsymbol{I}_{\text {coil }}=\mathbf{2 2 4 . 1} \mathbf{m A}\end{array} If you have any doubt please let me know I will surely help you I really appreciate if you like the answer Please don't dislike the answer Thank you Have a nice day Please like the answer ...