<p> Chapter 3 Additional Problems</p><p>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</p><p> numerical value of voltage vc as the conductor passes Point 1 and as the conductor passes Point 2.</p><p>Voltages v1 and v2 have the polarities indicated on Fig. X3.1 as determined by Lenz's law. Based on [3.10],</p><p>Point 1: vcv 2  v 1  Binℓ 2 v  B out ℓ 1 v 1 1 1  1 2 1   1 V</p><p>Point 2: vc v2  v 1 1 1 1  1 1 1  0 V</p><p>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.</p><p>1 (a) Place plus sign to left of e1 and below e2 .</p><p>(b) Place plus sign to right of e1 and above e2 .</p><p>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.</p><p>The arrow should point upward.</p><p>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</p><p> 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.</p><p>Based on [3.4], I1 and I2 establish oppositely-directed H-fields at point P with values</p><p>I1 3 5 H1    A / m 2R1 2  0.3 </p><p>I2 I 2 I 2 H2    A / m 2R2 2  0.2 0.4 </p><p>Equate H1 and H2 to find I2  2 A .</p><p>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 direc- tion of force acting on the conductor.</p><p>3 Since flux exits the north pole of the PM, the B-field in the region of the con- ductor is directed to the right. Based on [3.2], the force on the conductor is directed downward.</p><p>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?</p><p>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.</p><p>X3.7 Determine the apparent inductance at the point of operation for the series magnetic circuit of Example 3.4.</p><p>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</p><p>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.</p><p>The flux flowing through this added coil is r  3 mWb . Based on [3.30],</p><p>N2r 500 0.003 M 21    335.6 mH i1 4.47</p><p>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.</p><p>Since the core mmf drop is known from Example 3.3,</p><p>F c 2826.9 5 -1 R c   7.067  10 H T 0.004</p><p>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</p><p>(a) vS  t  , (b) i t , and (c) coil inductance L.</p><p>(a) By Faraday's law, d v t  N 100 0.15 10 cos 10 t  150cos 10 t V S dt</p><p>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</p><p> i t  238.7sin 10 t mA</p><p>(c) Based on [3.22], N  t 100 0.15sin 10 t  L    62.84 H i t 0.2387sin 10 t</p><p>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</p><p> labeled mean length paths are ℓA  42 cm , ℓB  54 cm , and ℓC 13.95 cm . (a) If the</p><p> 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).</p><p>(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</p><p> for BA  0.6 to read H A  90.5 A-t . Then,</p><p>6 F AH Aℓ A 90.5 0.42  38.01 A-t</p><p>Now, FB F A . The magnetic field intensity of the right leg is</p><p>F B 38.01 HB    70.39A-t / m ℓB 0.54</p><p>Enter Fig. X3.9 to find BB ; 0.45 T .</p><p> BA A A SF 0.6 0.04 0.05 0.96  1.152 mWb</p><p>BB B A B SF 0.45 0.04 0.05 0.96  0.864 mWb</p><p>C  A   B 1.152  0.864  2.016 mWb</p><p>3 C 2.016 10 Bg B C   1.008 T AC 0.04 0.05</p><p>From Fig. X3.9, HC H g ; 230 A-t / m .</p><p>F CH Cℓ C 230 0.1395  32.08 A-t</p><p>1.008 0.5 103 Bg    H      401.07 A-t F g g 7 o 4  10 Summing mmfs,</p><p>N1 I 1 FC  F g  F A or, 32.08 401.07  38.01 I   9.42 A 1 50</p><p>7 (b) Based on [3.30], 3 N  30 0.864 10  M 2 B   2.75 mH I1 9.42</p><p>2 2 X3.12 For the magnetic circuit of Fig. X3.10, A1  20 cm , A2 10 cm , ℓ1  400 cm ,</p><p>ℓ2 100 cm , and N 100 turns . The ferromagnetic core material is 24 ga. M-27 ESS.</p><p>The stacking factor can be considered unity. F 2 (the mmf drop along mean length</p><p> path ℓ2 ) is known to be 300 A-t . Determine the value of coil current I.</p><p>F 2 300 H2    300 A-t / m ℓ2 1</p><p>From Fig. X3.9, B2 ; 1.18 T . 4  B2 A 2 1.18 10  10  1.18 mWb</p><p> 1.18 103 B    0.59 T 1 4 A1 20 10</p><p>From Fig. X3.9, H1 ; 80 A-t / m .</p><p>F1H 1ℓ 1 80 4  320 A-t Summing mmfs,</p><p>NI F1  F 2 or, 320 300 I   6.2 A 100</p><p>8 X3.13 Draw the schematic for the magnetic circuit of Fig. X3.8 if I2  0 .</p><p>See Fig. X3.11.</p><p>X3.14 For the toroidal-shaped magnetic circuit of Fig. X3.12, the outside diameter</p><p>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.</p><p>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</p><p>From Fig. X3.9, Hc 135 A-t / m .</p><p>9 D1 D 2  0.15 0.10  ℓc         0.001  0.392 m 2   2 </p><p>F cH cℓ c 135 0.392  52.88 A-t</p><p>Bg 0.8 0.001 H      636.62 A-t F g g 7 o 4  10</p><p>Fc F g 52.88 636.62 I    3.45 A N 200</p><p>X3.15 For the (equilateral) triangular-shaped magnetic circuit of Fig. X3.13, N1  50 turns</p><p> 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</p><p> the core is known to be Hc  240 A-t / m , (a) determine flux  and (b) the coil current I.</p><p>(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</p><p> Bc A c  B c ht 1 0.02165 0.06  1.299 mWb</p><p>10 (b) The mean length flux path is</p><p>a b  0.075  0.15  3 ℓm 3    3    0.2  10  0.3373 m 2   2 </p><p>F cH cℓ m 240 0.3373  80.95 A-t</p><p>1 0.2 103 Bg      159.15 A-t F g 7 o 4  10 Noting that the coils are connected subtractive,</p><p>Fc F g 80.95 159.15 I    4.80 A N2 N 1 100  50</p><p>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</p><p> 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.</p><p>The minor modifications to Cckt1.m are indicated by boxes in the code below.</p><p>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.</p><p>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</p><p> force Fd .</p><p>The problem is set up as though the air gap is reduced by a distance x, then the result is evaluated for x  0 .</p><p>oA  o wd P   ℓ   x</p><p>2 2 o N wd L N P    x</p><p>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.</p><p>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</p><p> force Fd valid for i  0 and a x  a  b . Neglect air gap fringing.</p><p>1ℓ 1 g R    Ao w x a</p><p>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</p><p>14</p>