Question

An 800 turn coil is wound on the centre limb of the frame which has two similar outer limbs in parallel with each other. 

Answer 1

Suppose, 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 1 mwb if a saw-tooth (an air-gap) of 1 mm is cut in the centre limb. The relative permeability of steel is 1000.

Given that,

Number of turns = 800

Gap length = 0.1 cm

Outer limb length = 70 cm

Outer limb area =  20 cm²

Gap area = 40 cm²

Central limb length = 25 cm

Flux = 1 mwb

We need to calculate the coil current

Using formula of flux

$\phi_{g}=\dfrac{F_{m}}{R}$

$\phi_{g}=\dfrac{NI}{\dfrac{R_{c}}{2}+R_{l}+R_{g}}$

$I=\dfrac{\phi}{N}(\dfrac{R_{c}}{2}+R_{l}+R_{g})$

$I=\dfrac{\phi}{\mu_{0}\mu N}(\dfrac{L_{c}}{2A_{c}}+\dfrac{\mu L_{g}}{A_{g}}+\dfrac{L_{l}}{A_{g}})$

Put the value into the formula  

$I=\dfrac{1\times10^{-3}}{4\pi\times10^{-7}\times1000\times800}(\dfrac{70\times10^{-2}}{2\times20\times10^{-4}}+\dfrac{1000\times0.1\times10^{-2}}{40\times10^{-4}}+\dfrac{25\times10^{-2}}{40\times10^{-4}})$

$I=0.48\ A$

Hence, The coil current is 0.48 A.