Chapter 3 Additional Problems

X3.1 The conductor of Fig. X3.1 is being forced to move downward with a speed of 1 m/s as indicated. The conductor is everywhere perpendicular to the two B-fields enclosed by dashed lines. Each B-field is uniform in magnitude and time invariant. Find the

numerical value of voltage vc as the conductor passes Point 1 and as the conductor passes Point 2.

Voltages v1 and v2 have the polarities indicated on Fig. X3.1 as determined by Lenz's law. Based on [3.10],

Point 1: vcv 2  v 1  Binℓ 2 v  B out ℓ 1 v 1 1 1  1 2 1   1 V

Point 2: vc v2  v 1 1 1 1  1 1 1  0 V

X3.2 For the magnetic circuit of Fig. X3.2, clearly label the polarities of voltages e1 and e2 if (a) current I is increasing in value and (b) if current I is decreasing in value.

1 (a) Place plus sign to left of e1 and below e2 .

(b) Place plus sign to right of e1 and above e2 .

X3.3 In Fig. X3.3, the constant B-field is directed into the page within the area enclosed by dashed lines. The B-field is zero outside of this area. The conductor is being moved by an external force and has a velocity as indicated. Place an arrow beside the current symbol (I) to correctly show the direction of current flow through resistor R.

The arrow should point upward.

2 X3.4 The two parallel conductors of Fig. X3.4 have negligible diameter and carry currents in opposite directions as indicated. Both conductors and point P lie in the plane of the

page. If I1  3 A , determine the numerical value of current I2 such that the H-field perpendicular to the page at point P has a zero value.

Based on [3.4], I1 and I2 establish oppositely-directed H-fields at point P with values

I1 3 5 H1    A / m 2R1 2  0.3 

I2 I 2 I 2 H2    A / m 2R2 2  0.2 0.4 

Equate H1 and H2 to find I2  2 A .

3.5 A conductor is placed near the north pole of a permanent magnet as shown in Fig. X3.5. A dc current directed into the page flows through the conductor. Determine the direction of force acting on the conductor.

3 Since flux exits the north pole of the PM, the B-field in the region of the conductor is directed to the right. Based on [3.2], the force on the conductor is directed downward.

X3.6 A conductor is moving toward the left through the air gap region of a C-shaped magnetic core as shown in Fig. X3.6. The coil current I 10 A . If a path were provided for current flow in the conductor due to induced voltage, what would be the direction of that current?

By the right-hand rule, a B-field is established upward through the coil; thus, the B-field is downward in the air gap region where the conductor is located. Complete a loop with leads connected to the conductor as shown in Fig. 3.5 of the textbook. This would be a view looking downward from the top of the C-shaped core. Since the downward directed flux through this created loop is decreasing as the conductor moves to the left, Lenz's law requires that a current would flow out of the page of Fig. X3.6 through the conductor, if it could, to establish a B-field downward on the left side of the conductor, thereby opposing the decrease in flux through the loop.

X3.7 Determine the apparent inductance at the point of operation for the series magnetic circuit of Example 3.4.

The total flux through the coil was found to be T  4.364 mWb . Based on [3.22], 500 4.364 103 NT     L   162.6 mH i 13.42

4 X3.8 If a 500-turn coil (open circuit) were wound around the core of Fig. 3.18 in the lower right-hand window and the magnetic circuit is operating as described in Example 3.10, determine the value of the mutual inductance associated with this added coil. Let subscript 1 denote the original coil and subscript 2 denote the added coil.

The flux flowing through this added coil is r  3 mWb . Based on [3.30],

N2r 500 0.003 M 21    335.6 mH i1 4.47

X3.9 Calculate the value of reluctance of the ferromagnetic core at the point of operation for the series magnetic circuit of Example 3.3.

Since the core mmf drop is known from Example 3.3,

F c 2826.9 5 -1 R c   7.067  10 H T 0.004

X3.10 The magnetic structure of Fig. X3.7 has a ferromagnetic core that can be treated as infinitely permeable. Air gap fringing is negligible. Let N 100 turns , A  0.1 m2 , and  2  105 m . The coil is excited by a sinusoidal voltage source that establishes a flux given by  t  0.15sin 10 t Wb . Eddy currents and hysteresis are negligible so that  t and i t are in time-phase. Coil resistance can be neglected. Determine

(a) vS  t  , (b) i t , and (c) coil inductance L.

(a) By Faraday's law, d v t  N 100 0.15 10 cos 10 t  150cos 10 t V S dt

5 (b) Since the core is infinitely permeable, the mmf drop of the air gap must equal the coil mmf at any instant in time. 1  Ni t R g t    t  o A or,  2 105 i t t  0.15sin 10 t     7   o AN 4  10 0.1 100

i t  238.7sin 10 t mA

(c) Based on [3.22], N  t 100 0.15sin 10 t  L    62.84 H i t 0.2387sin 10 t

X3.11 The magnetic circuit of Fig. X3.8 has negligible leakage and air gap fringing. The laminated core is 24 ga. M-27 ESS with a stacking factor SF 0.96 . Values of the

labeled mean length paths are ℓA  42 cm , ℓB  54 cm , and ℓC 13.95 cm . (a) If the

flux density along the left-hand leg is BA  0.6 T , determine the coil current I1 . (b) Calculate the mutual inductance of coil 2 for the conditions of part (a).

(a) In order to obtain a B-H curve suitable for use in this problem, the first 10 stored data points from the arrays of hm27.m are plotted to give Fig. X3.9. Enter this curve

for BA  0.6 to read H A  90.5 A-t . Then,

6 F AH Aℓ A 90.5 0.42  38.01 A-t

Now, FB F A . The magnetic field intensity of the right leg is

F B 38.01 HB    70.39A-t / m ℓB 0.54

Enter Fig. X3.9 to find BB ; 0.45 T .

 BA A A SF 0.6 0.04 0.05 0.96  1.152 mWb

BB B A B SF 0.45 0.04 0.05 0.96  0.864 mWb

C  A   B 1.152  0.864  2.016 mWb

3 C 2.016 10 Bg B C   1.008 T AC 0.04 0.05

From Fig. X3.9, HC H g ; 230 A-t / m .

F CH Cℓ C 230 0.1395  32.08 A-t

1.008 0.5 103 Bg    H      401.07 A-t F g g 7 o 4  10 Summing mmfs,

N1 I 1 FC  F g  F A or, 32.08 401.07  38.01 I   9.42 A 1 50

7 (b) Based on [3.30], 3 N  30 0.864 10  M 2 B   2.75 mH I1 9.42

2 2 X3.12 For the magnetic circuit of Fig. X3.10, A1  20 cm , A2 10 cm , ℓ1  400 cm ,

ℓ2 100 cm , and N 100 turns . The ferromagnetic core material is 24 ga. M-27 ESS.

The stacking factor can be considered unity. F 2 (the mmf drop along mean length

path ℓ2 ) is known to be 300 A-t . Determine the value of coil current I.

F 2 300 H2    300 A-t / m ℓ2 1

From Fig. X3.9, B2 ; 1.18 T . 4  B2 A 2 1.18 10  10  1.18 mWb

 1.18 103 B    0.59 T 1 4 A1 20 10

From Fig. X3.9, H1 ; 80 A-t / m .

F1H 1ℓ 1 80 4  320 A-t Summing mmfs,

NI F1  F 2 or, 320 300 I   6.2 A 100

8 X3.13 Draw the schematic for the magnetic circuit of Fig. X3.8 if I2  0 .

See Fig. X3.11.

X3.14 For the toroidal-shaped magnetic circuit of Fig. X3.12, the outside diameter

D2 15 cm , the inside diameter D1 10 cm , the thickness t  5 cm , and the air gap length  1 mm . If  1 mWb and the core is solid ferromagnetic material with the B-H characteristic of 24 ga. M-27 ESS, find the current flowing in the 200-turn coil. Neglect leakage and fringing.

The core flux density is   0.001 B     0.8 T c D D 0.15  0.10 Ac 2 1 t 0.05 2 2

From Fig. X3.9, Hc 135 A-t / m .

9 D1 D 2  0.15 0.10  ℓc         0.001  0.392 m 2   2 

F cH cℓ c 135 0.392  52.88 A-t

Bg 0.8 0.001 H      636.62 A-t F g g 7 o 4  10

Fc F g 52.88 636.62 I    3.45 A N 200

X3.15 For the (equilateral) triangular-shaped magnetic circuit of Fig. X3.13, N1  50 turns

and N2 100 turns . Let   0.2 mm ,   7.5 cm , b 15 cm , and t  6 cm . Neglect any fringing and leakage. The ferromagnetic core is built up with 24 ga. M-27 ESS, but the stacking factor can be treated as unity. If the magnetic field intensity in the legs of

the core is known to be Hc  240 A-t / m , (a) determine flux  and (b) the coil current I.

(a) From Fig. X3.9 with Hc  240 A-t / m , Bc 1 T . Using the formula for a circle inscribed within an equilateral triangle, 3 3 h b  a 0.15  0.075  0.02165 m 6 6

 Bc A c  B c ht 1 0.02165 0.06  1.299 mWb

10 (b) The mean length flux path is

a b  0.075  0.15  3 ℓm 3    3    0.2  10  0.3373 m 2   2 

F cH cℓ m 240 0.3373  80.95 A-t

1 0.2 103 Bg      159.15 A-t F g 7 o 4  10 Noting that the coils are connected subtractive,

Fc F g 80.95 159.15 I    4.80 A N2 N 1 100  50

X3.16 A series magnetic circuit has the C-shape of Fig. 3.11 with ℓ1 ℓ 2  ℓ 3  ℓ 4  0.05 m , h d  0.10 m , w  0.15 m , and   0.002 m . The coil has N  500 turns . The solid ( SF 1.0 ) ferromagnetic core is cast steel. Fringing and leakage are negligible. A function-file Hcs.m is available at www.engr.uky.edu/ee/faculty/cathey to handle linear interpolation for the cast steel B-H curve of Fig. 3.12. Assume that the flux to

be produced in the core material is T  7 mWb . Edit Cckt1.m to give a MATLAB program to solve for the required coil current and generate a -i plot for this magnetic circuit.

The minor modifications to Cckt1.m are indicated by boxes in the code below.

11 Execution of X3_16.m gives a screen display showing I  7.254 A for the operating point current. The resulting -i plot is displayed by the graph below.

12 X3.17 Although the magnetic structure of Fig. X3.15 is rigid, there is a force acting to attempt to close the air gap if i  0 . Find an expression in symbols that describes this

force Fd .

The problem is set up as though the air gap is reduced by a distance x, then the result is evaluated for x  0 .

oA  o wd P   ℓ   x

2 2 o N wd L N P    x

13 By [3.69], 2 2 12dL 1 2oN wd 1 2  o N wd Fd i   i   i 2 dx 2  x 2 2  2 x0   x0 where the negative sign indicates that the force acts opposite to the direction of x, or in the direction to close the air gap.

X3.18 For the magnetic structure of Fig. X3.16, the slug can only move in the x-direction. The brass shim prevents any vertical movement. Find an expression for the x-directed

force Fd valid for i  0 and a x  a  b . Neglect air gap fringing.

1ℓ 1 g R    Ao w x a

N 2  N2 w x a L   o R g By [3.69], 2 12dL 1 2  N w F i  i o d 2 dx2 g