Question

A cast steel frame is to be used as the core of an inductor.

an 800 turn coil is wound on the Centre limb of the frame which has two similar outer in parallel with each other

the frame has the following parameters

outer limb c. S. A. Of 2000 mm and length of 700mm

centre limb C. S. A. Of 4000 mm and length of 250 mm

Answer 1

Gap length: $L_g=0.1\text{ cm}.$

Outer limb (core) length: $L_c=70\text{ cm}.$

Outer limb (core) area: $A_c=20\text{ cm}^2.$

Gap area: $A_g=40\text{ cm}^2.$

Central limb length: $L_l=25\text{ cm}.$

Calculate the magnetic reluctance:

$$R_c=\frac{L_c}{\mu_0\mu A_c},\\ \space\\ R_g=\frac{L_g}{\mu_0 A_g},\\ \space\\ R_l=\frac{L_l}{\mu_0\mu A_g}.$$

The magnetic circuit can be calculated like an electric circuit, but with flux $\Phi$ instead of current and with magneto-motive force $F_M=NI$ instead of voltage (see the equivalent circuit above):

$$\Phi_g=\frac{F_M}{R}=\frac{NI}{R_c/2+R_l+R_g},\\$$ $$I= \frac{Φ_g}{N}(R_C/2+R_l+R _g)=\\ \space\\ =\frac{Φ_g}{\mu_0\mu N}\bigg(\frac{L_c}{2A_c}+\frac{\mu L_g}{A_g}+\frac{L_l}{A_g}\bigg)=0.48\text{ A}.$$