Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 2.1

The magnetic circuit of Fig. 2.4(a) has dimensions:$A_{c} =4\times 4cm^{2}$,$l_{g} =0\cdot 06$cm, $l_{c} =40$ cm; $N=600$  turns. Assume the value of$\mu _{r} =6000$ for iron. Find the exciting current for$B_{c} =1\cdot 2$T and the corresponding flux and flux linkages.

## Verified Solution

From Eq. (2.9),

$Ni= H_{c} l_{c}+ H_{g}l_{g}$ ,

$Ni=\frac{B_{c} }{\mu _{c} }l_{c} + \frac{B_{g} }{\mu _{0} }l_{g}$

the ampere-turns for the circuit are given by $Ni=\frac{B_{c} }{\mu _{o}\mu _{r} }l_{c} + \frac{B_{g} }{\mu _{0} }l_{g}$

Neglecting fringing$A_{c}= A_{g}$ therefore $B_{c}= B_{g}$

Then $i=\frac{B_{c} }{\mu _{0}N } \left\lgroup\frac{l_{c} }{\mu _{r} }+ l_{g} \right\rgroup =\frac{1\cdot 2}{4\pi \times 10^{- 7}\times 600 } \left\lgroup\frac{40}{6000}+ 0\cdot 06 \right\rgroup \times 10^{- 2} =1\cdot 06$A

The reader should note that the reluctance of the iron path of 40 cm is only $\left\lgroup\frac{2/ 3}{6} \right\rgroup =0\cdot 11$ of the reluctance of the $0\cdot 06$ cm air-gap.

$\phi =B_{c}A_{c}=1\cdot 2\times 16\times 10^{-4} =19\cdot 2\times 10^{-4}$ Wb

Flux linkages, $\lambda =N\phi =600\times 19 \cdot 2\times 10^{-4} =1\cdot 152$Wb- truns

If fringing is to be taken into account, one gap length is added to each dimension of the air-gap constituting the area.
Then $A_{g} =\left(4+ 0\cdot 06\right) \left(4+ 0\cdot 06\right) =16\cdot 484 cm^{2}$

Effective $A_{g} \gt A_{c}$reduces the air-gap reluctance. Now

$B_{g} =\frac{19\cdot 2\times 10^{-4} }{16\cdot 484\times 10^{-4} } =1\cdot 165T$

From Eq. (i)

$Ni=\frac{B_{c} }{\mu _{o}\mu _{r} }l_{c} + \frac{B_{g} }{\mu _{0} }l_{g}$

$i=\frac{1}{\mu _{0}N } \left\lgroup\frac{B_{c}l_{c} }{\mu _{r} }+ B_{g}l_{g} \right\rgroup =\frac{1}{4\pi \times 10^{-7}\times 600 } \left\lgroup\frac{1\cdot 2\times 40\times 10^{-2} }{6000}+ 1\cdot 165\times 0\cdot 06\times 10^{-2} \right\rgroup =1\cdot 0332$A