ECE221: Electric and Magnetic Fields

Transcription

1 0 The Edward S. Rogers Sr. Department of Electrical & Computer Engineering ;.. UNIVERSITY OF TORONTO ECE221: Electric and Magnetic Fields Final Exam - Tuesday, April 25, 2017 Instructors: Tome Kosteski (LEC01), Li Qian (LECO2), Sean V. Hum (LEC03) Lastname'... Firstname'... Studentnumber'... Instructions Duration: 2 hours 30 minutes (14:00 to 16:30) Exam Paper Type: A. Closed book. Only the aid sheets provided at the end of this booklet are permitted. The aid sheets may be carefully removed from this booklet if you wish. Calculator Type: 2. All non-programmable electronic calculators are allowed. Only answers that are fully justified will be given full credit! Problem I Problem 2 Problem 3 Problem 4 TOTAL /20 /20 /20 20 /80

2 ECE221 Winter 2017 PROBLEM #1. [20 POINTS] Two perfectly-conducting coaxial cylinders, one with a radius of a = 1 mm and the other with a radius of b = 4 mm, are placed along the z axis as shown in Figure 1. The inner conductor is solid, and both cylinders are of length = 100 mm. A DC voltage source is connected to the inner and outer surfaces so that the inner conductor is at +20 V with respect to the outer conductor. Ignore fringing effects at z = 0 and z = f. 20 V Figure 1 Coaxial arrangement of cylinders a) Derive the voltage V(r) as a function of radial position r within the cylinder, for a < r < b. [5 points]

3 Winter 2017 ECE221 Derive the electric field vector E as a function of radial position r within the cylinder for a < r < b. [3 points] For parts (c) - (e) below, the space between the cylinders, a < r < b, is filled with a conducting material with a = 0.1 S/rn. Determine the conduction current flowing between the two cylinders. [4 points]

4 ECE221 Winter 2017 Determine the displacement current flowing between the two cylinders. [1 point] Determine the resistance of the coaxial resistor. [2 points]

5 Winter 2017 ECE221 1) The conducting material between the cylinders is removed, and the DC voltage source is replaced with an AC voltage source with radian frequency w, as shown in Figure 2. The electric field in the region a < r < b,0 <z < Lis given as E = 10 cos(3 x 109t - 10z) Determine the magnetic field vector H in the region a <r <b, if the region has a magnetic permeability = M. [5 points] Z Figure 2 Coaxial arrangement of cylinders ru

6 ECE221 Winter 2017 PROBLEM #2. [20 POINTS] A parallel-plate capacitor connected to a DC voltage source is shown in Figure 3. The surface area, S, of the top plate and bottom plate is 4 x 10-2 m2 and the plate separation, d, is 0.01 m. Inserted between the plates are two perfect dielectric media, each with thickness of d/2. The relative permittivity for each dielectric is given in Figure 3. top plate DC voltage source ttom plate Figure 3 Parallel-plate capacitor a) Derive the capacitance for the parallel-plate capacitor using electrostatic principles. [5 points]

7 Winter 2017 ECE221 If the DC voltage source is 100 V, determine the total charge (and its polarity) on the top plate and on the bottom plate. 3 points] Using electrostatic principles, determine the electric field, E. (magnitude and direction) between the top plate and the bottom plate. 4 points] 6

8 ECE221 Winter 2017 The DC voltage source from part (b) is disconnected after the capacitor is fully charged. Both dielectrics are then carefully removed such that there is no charge removed from either plate. Determine the potential difference between the top and bottom plate. [3 points] The charged parallel-plate capacitor from pan (d) is connected in parallel with an uncharged capacitor. = 100 pf. as shown in Figure4. Determine the energy stored in each capacitor. [5 points] tcr dl C x Figure 4 Parallel-plate capacitor with a capacitor, C'3,

9 Winter 2017 ECE221 PROBLEM #3. [20 POINTS] In Figure 5, a transformer is comprised of a ferromagnetic toroidal core (hr = 1000) and a primary side with Ni = 200 turns and a secondary side with N2 = 400 turns. The magnetic field density in the core is given by,\tj B= [T] (1) 221r where r is the radius to the centre of the ferromagnetic toroidal core and d is the cross-sectional diameter of the core. For this question, r = 20 cm and d = 1 cm. Voltage source Figure 5 Toroidal transformer a) Derive the expression for the magnetic field density as given by Equation (1). [5 points]

10 ECE221 Winter 2017 h) The direction of positive current on the primary side is indicated on Figure 5. On the same figure draw the direction of positive current on the secondary side and justify your result using fundamental principles. [5 points] c) A voltage source of sin() V is applied to the primary side and the peak power absorbed by the load resistor is 2 W. Determine the maximum magnetic field density in the core. [3 points]

11 Winter 2017 ECE221 Assume that the ferromagnetic material has no hysteresis. The B-H curve for the magnetic core is shown below. Determine the slope of the curve. [3 points] I- 1 Assume that the core is non-ideal and the maximum magnetic flux density (B) in the core is 0.4 T. Determine the maximum magnetic flux in the core and the maximum current that will flow in the secondary side. [4 points] 10

12 ECE221 Winter 2017 PROBLEM #4. [20 POINTS] Consider two circular coils of radius R, each perpendicular to the axis of symmetry, with their centers located at z = ±1/2. There is a steady current I flowing in the same direction around each coil, as shown. The medium is air with i = /io. X Figure 6 Helmholtz coils a) Find the magnetic field intensity (direction and magnitude) along the z-axis due to the lower coil. [5 marks] 11

13 Winter 2017 ECE221 Find the magnetic field intensity (direction and magnitude) along the z-axis due to the upper coil [1 mark]. Find the total magnetic field intensity (direction and magnitude) along the z-axis. [1 mark] A charge +q at the origin is moving at a uniform velocity v in the +z direction, What is the magnetic force (direction and magnitude) on the charge? [2 marks] A charge +q at the origin is moving at a uniform velocity v in the +x direction. What is the magnetic force (direction and magnitude) on the charge? [2 marks] 12

14 ECE221 Winter 2017 A small square current loop with a current i1 and having an area a2, is placed at the origin. What is the torque (direction and magnitude) on the loop? Assume a < 1?, so that the field can be considered uniform over the loop area. [2 marks] A small square current loop with a current i and having an area a2. is placed at the origin. What is the torque (direction and magnitude) on the loop? Assume a < R. so that the field can be considered uniform over the loop area.. [2 marks] 13

15 Winter 2017 ECE221 h) A small circular conducting loop with a radius b and a resistance Rb is at the point (0, 0, 1/2). The loop is centered on and perpendicular to the z-axis. If it moves with a velocity vb in the +z direction. what are the direction and the amount of current induced in the loop? Assume b < R, so that the field can be considered uniform over the loop area. 115 marks] 14

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17 ECE221 Winter Coordinate Systems 1.1 Cartesian coordinates (z) P(xy,z) /::::J/ (x) Y Position vector: R = x + ys' + z Differential length elements: dl = *dx, d1 = dy. dlz = id-, Differential surface elements: dsx = *dydz, ds dxdz, ds = idxdy Differential volume element: dv = dxdydz 1.2 Cylindrical coordinates (z) r P(r4z) 7V, (y) 4~0 (x)

18 Winter 2017 ECE221 Position vector: ft = rf + z Differential length elements: dl, = f'dr, d1 5 = d1 = 2dz Differential surface elements: dsr = j+dçdz, ds = drdz, dsz = irddr Differential volume element: dv = rdrddz 1.3 Spherical coordinates (z) P(RO,) (y) (x) Position vector: R = RB. Differential length elements: dlj? = RdR, d19 = ORdO. d10 = R sin Od Differential surface elements: dsr = AR sin 6d6dç. ds0 = OR sin OdRdo. dsq = RdRdO Differential volume element: dv = R2 sin OdRd9d6

19 ECE221 Winter Relationship between coordinate variables - Cartesian Cylindrical Spherical x x 'r cos th R sin cos y y rsin R sin G sin th z z ft Cos O r /x 2+y2 r Rsin0 0 tan c) 2 Z R Cos O R /y2+j2± 2 sill 0 1 r 0 COS-1 tan - z R 0 tan 1 C) 3 Dot products of unit vectors cos oj sin 6 0 sin 0Cos 0 Cos O Cos ç sinç sin Cos 0 sin 0 sin ó Cos 0 sin ó cos C) cos 6 - sin 0 0 cosc sin sin Cos 0 -sin cosçi cos 6 - sin 0 0 R. sill 0 Cos ø sin Gsin Cos sin 0 Cos Cos G Cos çb Cos 0sill 6 sing Cos 0 sing ;;- -sinc) Cos

20 Winter 2017 ECE221 4 Relationship between vector components = Cartesian A A - - A, Cylindrical Ar cos çaó sin O Ar sin 6+A cos th ]Spherical 4R sin 9 cos c'±a6 cos O cos Ad, sin ó AR sin O sin 45+A9 cos 8 sin ø A ) 4Rc0sG Ae sin 8 - AR sin O+ A9 cos & Lizil Az A: Ar A cos ±A sin c f_ -A sin O± A Cos O A A. A A- - AR cos 8 - A9 sin 8 - AR A sin 8 cos c+a sin O sin d± Ar sin O+A: cos O AR cos 0 A9 A1 cos 0 cos +A cos 8 sin ó A cos 0 - A sin 0 il A sin 0 AC) A sin c+a Cos 5 All, - 5 Differential operators 5.1 Gradient av OV LIV LIV 101"- 0V LIV, 101z 1 ov 11 VV= ---x+ --z=jjr+ R 00 R sin 8LIç 5.2 Divergence 1 3(sin0A6 ) 1 3A11, Ox Ug Oz r Or r 35 az R2 OR R sin 30 RsinO LIth 5.3 Laplacian O21LI2 732V ia/w 132V32 V 13/ 2 3V'\ 1 LII. DV\ v-i! = R s1 n0)+ = 3x2LIy20z2 =.) JR2sjiig /- R2 sin2 0 3(2

21 ECE221 Winter Curl (DAZ DA / DA DA (A DAX'\ VxA = I-----ix+l Dy Dzj \ Ox Ix+I------Ix / Dx Dy) (1 OA, DA(D4r DAz 1(D(rA) OA, = \r Dth Oz ) ' a-- Dr ) r \ Dr D) 1 (D(sinOA) D(RA)\ - -OAo - + ( sin - R sin 9 ao a) R 0 96 DR ) + 1 (D(RA0) DAR Rk OR DO 6 Electromagnetic formulas F = Table 1 Electrostatics Q1Q2 1 [(R R') _JRR3(R2 _Rl) E=_jIRR,I3dQ V x E = 0 V.D=rp ic E - dl = 0 E = VV v = I Is D.dS=Q V=V(P2)_V(Pi)=_=_f Ed1 Q i'wi D = 0E + P =EE = XeOE n 2/ / 1 Eit Q = CV = QV We = 2 f f pvdv = D voi 2 Edv vol V - (VV) = V (VV) = 0

22 Winter 2017 ECE221 Table 2 Magnetostatics Fir = qu x B VxB=jJ B dl = p1 Jc JLI B f dl' x (R-R') - 4J7 IR R'13 B = yo (H + M) = ph = M x a B1, = B2 VXH=zJ L Nb 2Wm I 12 F7 = I! x B f V.B=0 B.dS =0 = /B -ds M = x1,h Jrn = V x M a0 x (H2 - H1 ) = J = H.dl= I f B Hdv - vol L19 = N212 = ]\'2 f B1. ds2 Table 3 Faraday's law. Ampere-Maxwell law Vernfyc(UXB).d1_j-ds cc 5-T ic IH-dl=fJ.dS+fD.dS V x H = J + ad dt Table 4 Currents I=JJ.dS at J=pu=E P= (E.J)dv 1,01 Ji,n = 12,n 92J1,t = 1J2,t 1 1 R= =PeI-1e Ors p Table 5 Constants = x 1012 F/rn P0 = 4 x 10 Wm q = X 10 C