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 limbs in parallel with each other.

The frame has the following parameters:

Outer limb: c.s.a. of 2000 mm2 and length of 700mm.

Centre limb: c.s.a of 4000 mm2 and length of 250 mm.



Determine the coil current to set up a flux of 1mwb if a saw-tooth (an air-gap) of 1mm is cut in the centre limb.



The relative permeability of steel is 1000.

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 magnetomotive force $F_M=NI$ instead of voltage (see the equivalent circuit above):

$$\Phi_g=\frac{F_M}{R}=\frac{NI}{R_c/2+R_l+R_g},\\$$

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