HOME WORK # 2 – SOLUTIONS
1.
Draw an analog electric schematic for the parallel magnetic circuit of the
following figure for both cases of μ = ∞ and μ ≠ ∞.
I2
1
I1
1

l
2
h
N1
N2
h
w
2
w
SOLUTION:
As    , reluctances of the ferromagnetic paths approach zero, giving the simple,
decoupled equivalent circuit below.
+
Φ1
N2 I2
_
Φ2
Rg1
+
N 1 I1
Rg2
_
If lumped parameter reluctance values are formed for each of the branches and
if leakage is neglected, the magnetic equivalent circuit for    is drawn below.
+
N2 I2
_
Φ1
Φ2
Rg1
+
R3
Rg2
N1 I1
_
Φ1 -Φ2
R1
2.
R2
For the parallel magnetic circuit of the following figure, assume that the core
material is infinitely permeable. Let 1 = 3 mm and  2 = 2 mm. The
thickness of all core members is l =50 mm. The core has a uniform depth
into the page of 75 mm. N1 = 2 N 2 = 100 turns. Neglect air gap fringing.
(a)
(b)
If I 2 = 0 and 1 = 15 mWb, find the value of I1 .
If I1 = 10 A and I 2 = 20 A, determine 1 and 2 .
SOLUTION:
Since the core material is infinitely permeable, the first figure of Problem 1 is
applicable.

0.003
R g1  1 
 6.366  105 H 1
7
o A1 4  10  0.050  0.075
R g2 
2
0.002

 4.244  105 H 1
7
o A2 4  10  0.050  0.075
(a) Summation of mmf’s around the decoupled left-hand branch gives


N1I1  1R g1   0.015 6.366 105  9549 A-t
I1 
9549 9549

 95.49 A
N1
100
(b) Since the two branches are decoupled,
1 
N1I1 100 10 

 1.57 mWb
R g1 6.366  105
-  
 50  20   2.36 mWb , or  = 2.36 mWb.
N2 I2

2
R g 2 4.244  105
3.
In the magnetic circuit shown below, the relative permeability of iron is. The
length of the air-gap is g=3 mm and the total length of the iron is 0.4 m. For the
magnet B510=rμr = 1.07 T, Hc=800 kA/m and the length of the magnet lm is 5 cm.
Assuming the cross section is uniform, what is the flux density in the air gap?
SOLUTION:
Since the cross section is uniform, B is the same everywhere and there is no current:
Hiron .liron  H m .lm  H g .lg  0
Bi  Bm  Bg  B
B
 ri
.liron  B
H c lm
B
 H c lm  g  0
Br

 800 103 
 3 103 
0.4


2
3
2
B

B

5

10

800

10

5

10

B
0



7
5 
7 
 4 10 *10 
 1.07 
 4 10 

Solving for B:
B  1.005T
