ADDIS ABABA UNIVERSITYADDIS ABABA INSTITUTE OF TECHNOLOGY

DEPARTEMENT OF ELECTRICAL AND COMPUTER ENGINEERING

Course Title: Introduction to Electrical Machine

ASSIGNMENT - 01

Submitted to Instructor: Dawit Habtu

Name: ID No.:

March 15, 2013

1, GivenWooden ring Ac = 300mm2

Lc = 200mm

N = 800 turns

Required

i. Filed strength HC when I = 2A assume μr =1ii. Magnetic flux density BC

iii. Current reqiered I when B=0.02wb/m2

Solution

i. HC= NI/ LC by Ampere's law F=NI=HC* LC

= 800 turns * 2A200mm

= 1600/0.2 = 8,000 wb/m

ii. BC= HC*μ μ=μ0*μr = 4π*10-7*1 = 4 π* 10 -7 = 8,000 wb/m*4π*10-7

= 32π*10-4

= 10.053mwb/m 2

iii. F=NI=HC* LC HC = B/μ =0.02wb/m 2 = 15.915kwb/m

4π*10-7

:. I = HC* LC

N = 15.915kwb/m* 0.2m

800 turns = 0.3978A

2, Given

Flux øc = 0.38mWb

Iron-ring diameter LC = 58cm=0.58m

Cross-sectional area Ac =3cm2= 0.0003m

Required

Ampere-turn NI

Solution

BC = øc/Ac = 0.38mWb = 1.27 wb/m2

0.0003m

If BC =1.27 wb/m2 from the table we can get μr

1.4-1.27 = 1000-x x= 0.13*500 + 1000 =13251.4-1.2 1000-1500 0.2

.: μr =1325

F=NI=HC* LC = BC * LC μ

F=NI = 1.27 wb/m 2 * 0.58m 1325

= 5.559At

B (wb/m2)

0.5 1.0 1.2 1.4

μr 2500 2000 1500 1000

3, Given

Horse-shoe magnet LC = 45.7cm=0.457m

Cross-sectional area Ac =6.45cm2= 0.000645mN = 500 turns each limb connected in series

Mass= 68kg

Negligible Relactance

μr = 700

Required

Current I =?

Solution

Force =m*g = 68kg*9.8m/s2 = 666.4NF=NI

I= F/2N each limb connected in series = 666.4 N 2* 500 turns = 0.6664 A