ENGINEERING ELECTROMAGNETICS NOTES - Architecture
10EC36ENGINEERING ELECTROMAGNETICSNOTES
10EC36Introduction to vectorsThe behavior of a physical device subjected to electric field can be studied either by Fieldapproach or by Circuit approach. The Circuit approach uses discrete circuit parameters likeRLCM, voltage and current sources. At higher frequencies (MHz or GHz) parameters would nolonger be discrete. They may become non linear also depending on material property andstrength of v and i associated. This makes circuit approach to be difficult and may not give veryaccurate results.Thus at high frequencies, Field approach is necessary to get a better understanding ofperformance of the device.The ‗Vector approach‘ provides better insight into the various as ects of Electromagneticphenomenon. Vector analysis is therefore an essential tool for the study of .The ‗Vector Analysis‘ comprises of ‗Vector Algebra‘ and ‗Vect r Calculus‘.Any physical quantity may be ‗Scalar quantity‘ or ‗Vector quantity‘. A ‗Scalar quantity‘ isspecified by magnitude only while for a ‗Vector quantity‘ requires both magnitude and directionto be specified.Examples :Scalar quantity : Mass, Time, Charge, D nsity, Pote tial, Energy etc.,Represented by alphab ts – A, B, q, t etcVector quantity : Electric field, force, velocity, acceleration, weight etc., represented byalphabets with arrow on top.A, B, E, Betc.,Vector algebra : If A, B, Care vectors and m, n are scalars then(1) AdditionA B B ACommutativ lawcitystudentsgroupA ( B C) (A B) C Associative law(2) SubtractionA - B A (- B)(3) Multiplication by a scalarmA AmCommutative lawm (n A) n (m A)Associative law(m n) A m A n ADistributive lawm (A B) m A m BDistributive lawDepartment of ECEPage 4
10EC36A ‗vector‘ is represented graphically by a directed line segment.A ‗Unit vector‘ is a vector of unit magnitude and directed along ‗that vector‘.aˆ Ais a Unit vector along the direction of A .Thus, the graphical representation of A and aˆ A areAUnit vector aˆ AVector AAlso aˆ A A / A or A aˆ A AcitystudentsgroupProduct of two or more vectors :(1) Dot Product ( . )A . B A ( B COS θ OR {A COS θ } B , 0 θ πBB A Cos θ AB Cos θAA.B B.A(A Scalar quantity)(2) CROSS PRODUCT (X)C A x B A B SIN θ ˆEx ,where ' θ ' is angle between A and B ( 0 θ π)and nˆ is unit vector perpendicular to plane of A and Bdirected such that A B C form a right handed system of vectorsA x B -BxAA x ( B C) A x B Department of ECEAxCPage 5
10EC36CO-ORDINATE SYSTEMS :For an explicit representation of a vector quantity, a ‗co-ordinate system‘ is essential.Different systems used Co-ordinate variablesx, y, zρ, , zr, , Unit vectorsax , ay , azaρ , a , azar , a , a These are ‗ORTHOGONAL‗ i.e., unit vectors in such system of co-ordinates are mutuallyperpendicular in the right circular way.citystudentsgroupi.e., x y z , z , r RECTANGULAR CO-ORDINATE SYSTEM :Zx 0 planeazpy 0planeaxXYayz 0 planeax.ay ay.az az.ax 0axxay azayxaz axazxax ayaz is in direction of ‗advance‘ of a right circular screw as is turned from ax to ayCo-ordinate variable ‗x‘ is intersection of planes OYX and OXZ .e, z 0 & y 0Location of point P :If the point P is at a distance of r from O, thenIf the components of r along X, Y, Z are x, y, z thenr x a x y a y z az r arDepartment of ECEPage 6
10EC36Equation of Vector AB :If OA A Ax a x Ay a y Az a zBand OB B Bx a x By a y Bz a z thenA AB B or AB B - A0where As , Ay & Az are components ofand Bs , By & Bz are components ofABBAAA along X, Y and ZB along X, Y and ZDot and Cross Products :A . B (Ax a x Ay a y Az a z ) . (Bx a x By a y Bz a z ) Ax Bx Ay By Az CzcitystudentsgroupA x B (Ax a x A ayy Azaz ) x (Bx a x By a y Bz a z )Taking 'Cross products' term by term and grouping, we getax ayA x B AxBxA y AzBy BzAxA . (B x C ) azAyAzxyzCxCyCzBBBIf A, B and C are non zero vectors,(i) A . B 0 then Cos θ 0 .e., θ 900A and B are perpendicular
A x B 0 then Sin θ 0θ 0 A and B are parallel(ii) A . ( B x C) represents the volume of a parallelopoid of sides A , B and CUnit Vector along ABa AB ABABwhereVector length AB ABDepartment of ECE (AB . AB)Page 7
10EC36Differential length, surface and volume elements in rectangular co-ordinate systemsr x aˆ x y aˆ y z aˆ z rdr x r rdx y dy zdzdr dx aˆ x dy aˆ y dz aˆ zDifferential length dr [ dx222 1/2 dy dz ]-----1Differential surface element, ds1. rto z : dxdy aˆ z2. to z : dxdy aˆ zr3. to z : dxdy aˆ zr------ 2citystudentsgroupDifferential Volume elementdv dx dy dz------ 3zp‘dxpdyrdzr dr0yxOther Co-ordinate sy tems :Depending on the geometry of problem it is easier if we use the appropriate co-ordinate systemthan to use the Car esian co-ordinate system always. For problems having cylindrical symmetrycylindrical -ordinate system is to be used while for applications having spherical symmetryspherical o-ordinate system is preferred.Cylindrical Co-ordiante systems :zP(ρ, , z)rρaprazx ρ Cos y ρ Sin z z22ρ x y0 y-1φ tan y / xz zρxDepartment of ECEPage 8
10EC36r x aˆ x y aˆ y z aˆ zr ρ Cos aˆ x ρ Sin aˆ y z aˆ z r r rdr ρ d ρ d z dz------1 r r ρ Cos aˆ x Sin aˆ y ρ aˆ hρ aˆρ r - ρ Sin aˆ ρ Cos aˆ r xyaˆ h;ρ r ρ 1 ρ aˆ ; h r ρ r aˆz zh r 1z zcitystudentsgroupThus unit vectors in (ρ, , z) systems can be expressed in (x,y,z) system asaρ Cos a x Sin a ya x Cos a Sin a a - Sin a x Cos a ya y Sin a Cos a az az;a , a and a z are orthogonalFurther , dr d ρ aˆ ρ d aˆ dz aˆ z2and dr------2 d ρ (ρ d ) (dz)222Differential areas :ds aˆz (d ρ) (ρ d ) . aˆzds aˆ (dz) (ρ d ) . aˆ -------3ds aˆ (d ρ dz) aˆ Differential volume :d (d ρ) (ρ d ) (dz)or d ρ d ρ dzSpherical Co-ordinate Systems :Z----- 4zX r Sin Cos Y r Sin Sin Z r Cos pR 0xry Yr Sin XDepartment of ECEPage 9
10EC36R r Sin Cos aˆ x r Sin Sin aˆ y r Cos aˆ zaˆr R / R Sin Cos aˆ x Sin Sin aˆ y Cos aˆ z r r RR Cos Cos aˆ x Cos Sin aˆ y Sin aˆ z/aˆ RR - Sin aˆ x Cos aˆ y/aˆ dR R rdr Rd Rd dR dr aˆ r r d aˆ r Sin d aˆ 2d S r r Sin d d 2d S r Sin dr d d S r dr d 2dv rSin dr d d General Orthogonal Curvilinear Co-ordinates :z u1 a3u3a1u2a2yxCo-ordinate Variables : (u1 , u2, u3); Hereu1 is Intersection of surfaces u2 C & u3 Cu2 is Intersection of surfaces u1 C & u3 Cu3 is Intersection of surfaces u1 C & u2 Caˆ1 , aˆ 2 , aˆ3 are ubnit vectors tangential to u1 , u2 & u3 Systemis Orthogonal if aˆ1 . aˆ 2 0 , aˆ 2 . aˆ3 0 & aˆ3 . aˆ1 0If R x aˆ x y aˆ y z aˆ z & x, y, z are functions of u1 , u2 & u3then d R R du R du R du u1 1 u2 u323 h1 du1 aˆ1 h2 du2 aˆ 2 h3 du3 aˆ3where h1 , h2 , h3 are scale factors ;h 1 R , h 2 R u1 u2,h3 R u3
Department of ECEPage 10
10EC36Co-ordinate Variables, unit Vectors and Scale factors in different systemsSystemsCo-ordinate Variables Unit VectorGeneralu1Scale factorsu2u3a1a2a3h1h2h3Rectangular xyzaxayaz111Cylindrical ρ zaρa az1ρ1Spherical ara a 1rr sin rTransformation equations (x,y,z interms of cylindrical and spherical co-ordinate systemvariables)Cylindrical : x ρ Cos , y ρ Sin , z z ; ρ 0, 0 2 - z Sphericalx r Sin Cos , y r Sin Cos , z r Sin r 0 , 0 , 0 2 V 1 v aˆ 1 v aˆ 1 v aˆ12h2h3 u 3 u 2h 1 u 11 . A (h 2 h 3 A 1 ) (h 1 h 3h hh 1 2 3 u1 u2hhh 1 aˆ 123aˆ 3aˆ 2 1 x A h 1 h 2 h 3 u uh A11h A21where V V ( u 1 , u 2 , u 3 )223A 2 ) u3 (h 1 h 2 A 3 ) uh A33a Scalar3field& A A 1 aˆ 1 A 2 aˆ 2 A 3 aˆ 3 is a Vector field where A1 A1 (u1 , u2 , u3 )A2 A2 (u1 , u2 , u3 ) and A3 A3 (u1 , u2 , u3 )Department of ECEPage 11
10EC36Vector Transformation from Rectangular to Spherical :Rectangula r : AR Ax aˆ x Ay aˆ y Az aˆ z ( AR aˆ r ) aˆ r (AR aˆ ) aˆ (AR . aˆ ) aˆ Spherical : AS Ar aˆ r A aˆ A aˆ where Ar , A , A are related to Ax , Ay , Az as A r A aˆ xaˆ. aˆr aˆ x . aˆ aˆx. aˆA. aˆryaˆ . aˆyaˆ y A x aˆ z . aˆ aˆ . aˆ . aˆ aˆ z . aˆ r z A y A z A ‗field‘ is a region where any object experiences a force. The study of performance in thepresence of Electric field (E) , Magnetic field ( ) is the essence of EM Theory.P1 : Obtain the equation for the line between the points P(1,2,3) and Q (2,-2,1)PQ ax - 4 a y - 2 azP2 : Obtain unit vector from the origin to G (2, -2, 1)Problems on Vector AnalysisExamples :1. Obtain the vector equation for the line PQ between the points P (1,2,3)m and Q (2, -2, 1)mZPQP (1,2,3)Q(2,-2,-1)0YXThe vector PQ (xq - xp ) aˆ x (yq - yp ) aˆ y (zq - zp ) aˆz (2 - 1) aˆ x (-2 - 2) aˆ y (-1 - 3) aˆz (aˆ x - 4 aˆ y - 2 aˆz ) 2. Obtain unit vector from origin to G (2,-2,-1)GG0Department of ECEPage 12
10EC36The vector G (xg - 0) aˆ x (yg - 0) aˆ y (zg - 0) aˆ z (2 aˆ x - 2 aˆ y - aˆ z )ˆ The unit vector , ag2GG2G 22 (-2) (-1) 3 aˆg (0.667 aˆ x - 0.667 aˆ y - 0.333 aˆ z )3. GivenA 2 aˆ x - 3 aˆ y aˆzB - 4 aˆ x - 2 aˆ y 5aˆzfind (1) A . B and (2) A x BSolution :(1) A . B (2 ax - 3 a y az ) . (-4 ax - 2 a y 5 az ) -8 6 5 3Since ax . ax ay . ay az . az 0 and ax ay ay az az ax 0(2) A x B axa y az2 4 3 1 2 5 (-13 ax -14 ay - 16 az)4. Find the distance between A( 2, /6, 0) and B ( 1, /2, 2)Soln : The points are given in Cylindrical Co-ordinate (ρ, , z). To find the distance betweentwo points, the co-ordinates are to be in Cartesian (rectangular). The correspondingrectangular co-ordinates are (ρ Cos , ρ Sin , z) A 2 Cos& B Cos aˆ6aˆ6x Sinx 2 Sin aˆ 1.73 aˆ aˆ6yxaˆ 2 aˆ aˆ 2 aˆ2yzyyz AB (Bx - Ax ) aˆ x (By - Ay ) aˆ y (Bz - Az ) aˆ z -1.73 aˆ x (1 -1) aˆ y (2 - 0) aˆ z - 1.73 aˆ x 2 aˆ z 2 (AB) 1.732 2 2.645. Find the distance between A( 1, /4, 0) and B ( 1, 3 /4, )Soln : The specified co-ordinates (r, , ) are spherical. Writing in rectangular, they are (rSin Cos , r Sin Sin , r Cos ).Therefore, A & B in rectangular co-ordinates,Department of ECEPage 13
10EC36A (1 Sin Cos 0 aˆ 1 Sin4 ( 0.707 aˆ x Sin 0 aˆ y 1 Cos aˆ z )x44 0.707 aˆ y )B ( Sin 3 Cos aˆ x Sin 3 Sin aˆ y Cos 3 aˆ z )44 ( 0.707 aˆ x 0.707 aˆ y )4AB (Bx - Ax ) aˆ x (By - Ay ) aˆ y (Bz - Az ) aˆz - 1.414 aˆ x (- 0.707) aˆ y (-0.707) aˆ zAB ( AB . AB )1/2 (2 0.5 0.5)1/2 1.732 6. Find a unit vector along AB in Problem 5 above.ˆABaAB AB1 [ - 1.414 ax (-0.707) ay (-0.707) az] 1.732 ( - 0.816 aˆx - 0.408 aˆ y 0.408 aˆz )7. Transform F (10 aˆx - 8 aˆ y 6 aˆz ) into F in Cylindrical Co - ordinates.Soln :FCyl (F . aˆp ) aˆp (F . aˆ ) aˆ (F . aˆz ) aˆz [(10 aˆ x - 8 aˆ y 6 aˆ z ) . (Cos aˆ x Sin aˆ y )] aˆ [ (10 aˆ x - 8 aˆ y 6 aˆ z ) . (- Sin aˆ x Cos aˆ y )] aˆ [ (10 aˆ x - 8 aˆ y 6 aˆ z ) . (aˆ x )] aˆ xρ (10 Cos - 8 Sin ) aˆ (-10 Sin - 8 Cos ) aˆ 6 aˆ z x Cos xy Sin tan2-12 y 12.81yxy0 - 38.66xFCyl [ 10 Cos (- 38.66) - 8 Sin ( - 38.66) ] aˆp [- 10 Sin (- 38.66) - 8 Cos (- 38.66)] aˆ 6 aˆz (12.8 aˆ 6 aˆz )8. Transform B y aˆx - x aˆ y z aˆz into Cylindrical Co-ordinates.Department of ECEPage 14
10EC36x Cos , y Sin B Sin aˆ x - Cos aˆ y z aˆ zBCyl (B. aˆ ) aˆ (B. aˆ ) aˆ (B . aˆ z ) aˆ z [ ( Sin aˆ x - Cos aˆ y z aˆ z ). (Cos aˆ x Sin aˆ y )] aˆ [ ( Sin aˆ x - Cos aˆ y z aˆ z ). (- Sin aˆ x Cos aˆ y )] aˆ z aˆ z [ Sin Cos - Sin Sin ] aˆ [ - Sin - Cos ] aˆ z aˆ z22 - aˆ z aˆ z 9. Transform 5 aˆxinto Spherical Co-ordinates.ASph (A. aˆ r ) aˆ r (A. aˆ ) aˆ (A. aˆ ) aˆ [ 5 aˆ x . (Sin Cos aˆ x Sin Sin aˆ y Cos aˆ z )] aˆ r [ 5 aˆ x . (Cos Cos aˆ x Cos Sin aˆ y - Sin aˆ z ] aˆ [ 5 aˆ x . (- Sin aˆ x Cos aˆ y )] aˆ 5 Sin Cos aˆ r 5 Cos Sin aˆ 5 Sin aˆ 10. Transform to Cylindrical Co-ordinates G (2 x y) aˆx - (y - 4x) aˆ y at Q ( , , z)Soln :GCyl (G. a ) aˆ (G. a ) aˆ (G. a z ) aˆ zGCyl [ (2 x y) aˆ x - (y - 4x) aˆ y ] . [ Cos aˆ x Sin aˆ ] aˆ [ (2 x y) aˆ x - (y - 4x) aˆ y ] . [ - Sin aˆ x Cos aˆ y ] aˆ 0 [ ( 2x y) Cos - (y - 4x) Sin ] aˆ [ - (2 x y) Sin - (y - 4x ) Cos ] aˆ x Cos , y Sin GCyl [ ( 2 Cos Sin ) Cos - ( Sin - 4 Cos ) Sin ] aˆ [ - ( 2 Cos Sin ) Sin - ( Sin - 4 Cos ) Cos ] aˆ [ 2 Cos2 Sin Cos - Sin 2 4 Sin Cos ] aˆ [ - 2 Sin Cos - Sin 2 - Sin Cos 4 Cos2 ] aˆ ( 2 Cos2 5 Sin Cos - Sin 2 ) aˆ ( 4 Cos2 Sin 2 - 3 Sin Cos ) aˆ 11. Find a unit vector from ( 10, 3 /4, /6) to (5, /4, )Soln :A(r, , ) expressed in rectangular co-ordinatesDepartment of ECEPage 15
10EC36OA r Sin Cos aˆ x r Sin Sin aˆ y r Cos aˆ z 10 Cos A 10 SinB 5 Sin 3 Cos4 aˆx6Cos aˆ4x 10 Sin 5 Sin 4A 6.12 aˆ x 3.53 aˆ y - 7.07 aˆ z3 Sin 3 aˆ46ySin aˆ 5 Cosy 4aˆ4aˆzzB - 3.53 aˆ x 3.53 aˆ zAB B - A - 9.65 aˆ x - 3.53 aˆ y 10.6 aˆ z2AB 9.65aˆ AB2 3.53 10.6 AB (- 0.65 aˆxAB2 14.77- 0.24 aˆ y 0.72 aˆ z )12. Transform F 10 aˆx - 8 aˆ y 6 aˆz into F in Spherical Co-orindates.aˆ r Sin Cos aˆ x Sin Sin aˆ y Cos aˆ zaˆ Cos Cos aˆ x Cos Sin aˆ y - Sin aˆ zaˆ - Sin aˆ x Cos aˆ y FSph (F . aˆ r ) aˆ r (F . aˆ ) aˆ (F . aˆ ) aˆ (10 Sin Cos - 8 Sin Sin 6 Cos ) aˆ r (10 Cos Cos - 8 Cos Sin - 6 Sin ) aˆ (- 10 Sin - 8 Cos ) aˆ r 102 82 62 200 ; Cos-1z Cos-1r6 64.890200-10 tan - 8 - 38.6610Sin Sin 64.69 0.9Sin Sin (-38.66) - 0.625Cos Cos 64.69 0.42 Cos Cos (-38.66) 0.781F (10 x 0.9 x 0.781 - 8 x 0.9 x (-0.625)) aˆ r ( 10 x 0.42 x 0.781 - 8 x 0.42 x (0.625)) aˆ (-10 x - 0.625 - 8 x 0.781) aˆ F (11.529 aˆ r 5.38 aˆ 0.783 aˆ )Line IntegralsIn general orthogonal Curvilinear Co-ordinate systemdl h1 du1 aˆ1 h2 du2 aˆ 2 h3 du3 aˆ3F F1 aˆ1 F2 aˆ2 F3 aˆ33 CF . dl h1C F1 du1 h2 FC2 du2 h3 F3 duCDepartment of ECEPage 16
10EC36 Conservative Field A field is said to be conservative if it is such that . dl 0C . dl bd (b) - (a) (does not depend on the path !). If is electrostatic flux, thenaE - represent the electric field intensity and b . dl represent the potential between b and a and is zero if it is taken around a closed contour.a i.e., . dl 0Therefore ES flux field is ‗Conservative‘.EXAMPLES : I a . dl13. Evaluatelineintegral2along y x from A (1,1) to B (4,2)wherea (x y) aˆ x (y - x) aˆ ySoln : dl dx aˆx dy aˆ ya . dl (x y) dx (y - x )2dy y x or 2 dy dy dx22 a . dl (y y) 2y dy (y - y ) dy221 1 (2 y3 2 y2 y - y2 ) dy (2 y21 23 y2 y) dy1 2 y4 4 2 2 4 32 y 2 3 28 8 y323 3 2 2 4 2 - 22 1 1- 1 2314. Evaluate the Integral I E . ds 133-1 3 1 2 132 12 11where E x aˆx and S is hunisphere of radius aSSoln:If S is hemisphere of radius a, then S is defined byDepartment of ECEPage 17
10EC36x22 y z2 a2,z 0;ds (a d ) (a Sin ) d aˆ rds a Sin d d aˆ r2E (E . aˆ r ) aˆ r (E. aˆ ) aˆ (E. aˆ ) a Er x Sin Cos aˆ r; x a Sin Cos E . ds E r . ds a ( Sin Cos ) aˆ r . a Sin d d 22E . ds a Sin Cos d d 0 / 2 , 0 2 3 /22 E . ds a Sin d 3230 0 Cos2 d a3x23x 2 a33where r, r1 , r2 . rm are the vector distances of q, q1 , qm from origin, 0.r - rm is distance between charge qm and q.aˆ m is unit vector in the direction of line joining qm to q.Department of ECEPage 18
10EC36Unit-1a. Coulomb’s Law and electric field intensity: Experimental law of Coulomb Electric field intensity Field due to continuous volume charge distribution Field of a line charge b. Electric flux density, Gauss’ law and divergence: Electric flux density Gauss‘ law Divergence, Maxwell‘s First equation(Electrostatics) vector operator and divergence theorem Department of ECEPage 19
10EC36Electric field is the region or vicinity of a charged body where a test charge experiences aforce. It is expressed as a scalar function of co-ordinates variables. This can be illustrated bydrawing ‗force lines‘ and these may be termed as ‗Electric Flux‘ represented by and unitis coulomb (C).Electric Flux Density (D) is the measure of cluster of ‗electric lines of force‘. It is thenumber of lines of force per unit area of cross section.ψ2i.e., D A c/mor ψ D nˆ ds Cwhere nˆ is unit vector normal to surfaceSElectric Field Intensity (E) at any point is the electric force on a unit ve charge at thatpoint.qi.e., EF1 2q4 raˆ1N/c0 1 1 q aˆ 4 r2 0 1 D N / c 11N / c or D E C in vacuum00In any medium other than vacuum, the field Intensity at a point distant r m from Q C isQE 4 r2 aˆ r N / c ( or V / m)0rQand D 0 r E Cor D 4 r2aˆr CThus D is independent of medium, while E depends on the property of medium.r QCEq 1 C (Test Charge)Source chargeE E 0r,mElectric Field Intensity E for different charge configurations1. E due to Array of Discrete chargesLet Q, Q1 , Q2 , Qn be ve charges at P, P1 , P2 , . Pn . It is required to findE at P.Department of ECEPage 20
10EC36Q1r 1EnP1Q2Pr2E2E1P2Qnr10rnPrEr 4 1Qm 2aˆm V / mr-r0m2. E due to continuous volume charge distributionaˆ RRPρv C / m3The charge is uniformly distributed within in a closed surface with a volume charge densitydQ3of ρv C / m i.e, Q V dvand V dvV Q E aˆ4 R 2 R0Er V (r ) 1 aˆ R01214 (r - r )V V V24 Raˆ R N / C0aˆ R is unit vector directed from ‗source‘ to ‗filed point‘.3. Electric field intensity E due to a line charge of infinite length with a line chargeˆadensity of ρl C / mRPRρl C / mdlL Ep 41 Rldl2aˆR N / C0 LDepartment of ECEPage 21
10EC364. E due to a surface charge with density of ρS C / m2ˆa RP (Field point)dsR(Source charge)Ep 1 4 S ds R2aˆRN/C0 SElectrical Potential (V) The work done in moving a unit ve charge from Infinity to that iscalled the Electric Potential at that point. Its unit is volt (V).Electric Potential Difference (V12) is the work done in moving a unit ve charge from onepoint to (1) another (2) in an electric field.Relation between E and VIf the electric potential at a point is expressed as a Scalar function of co-ordinate variables(say x,y,z) then V V(x,y,z)dV -f - E . dldl- - - - - - - - (1)qAlso, dV V dx V dy V dz x y zdV V . dl- - - - - - - - - (2)From (1) and (2) E - VDetermination of electric potential V at a point P due to a point charge of Q Caˆ lR dR 0 QAt point P, E RPaˆ RQ2 aˆR N / C4 R0Therefore, the force f on a unit charge at P.Department of ECEPage 22
10EC36Q 4 R 2 aˆR N f 1 x Ep0The work done in moving a unit charge over a distance dl in the electric field isdV - f . dl - E . dlRQ dl Vp - 4 R 2 (aˆ R . aˆl ) VP0Q 4 R 2 VoltR Q4 2R dR0(a scalar field)0Electric Potential Difference between two points P & Q distant Rp and Rq from 0 isVpq 1 RQ (Vp - Vq ) 4 0 p-1 voltRq Electric Potential at a point due to different charge configurations.1. Discrete charges. Q1.Q2QmPRmV
The ‗Vector approach‘ provides better insight into the various as ects of Electromagnetic phenomenon. Vector analysis is therefore an essential tool for the study of . The ‗Vector Analysis‘ comprises of ‗Vector Algebra‘ and ‗Vect r Calculus‘. Any physical quantity may be ‗Scalar quantity‘ or ‗Vector quantity‘.
surveys given to the students at the end of each MATLAB assignment. The Electromagnetics Concept Inventory was also used. KEYWORDS computers and education, computer-assisted instruction and learning, computer exploration in electromagnetics, electromagnetics teac
Chapter 1 Electromagnetics and Optics 1.1 Introduction In this chapter, we will review the basics of electromagnetics and optics. We will briefly discuss various laws of electromagnetics leading to Maxwell's equations. The Maxwell's equations will be used to derive the wave equation which forms the basis for the study of optical fibers in .
1. Numericals based on solving Engg. Electromagnetics problems using MATLAB for tutorials are available in CD accompanied with the book of “Fundamentals of Engineering Electromagnetics by Sunil Bhooshan” 2. Matlab Experiments manual for Electromagnetics by Dr. M.H. Bakr SR. NO. LIST OF PRACTICALS 1. INTR
Domain Method for Electromagnetics with MATLAB Simulations, ACES Series on Computational Electromagnetics and Engineering, SciTech Publishing Inc. an Imprint of the IET, Second Edition, Edison, NJ, 2015. [2] R.C. Booton, Computational Methods for Electromagneticsand Microwaves, Wiley, 1992, pp. 59-73
electromagnetics education and surveys and experiences in teaching/learning electromagnetic fields and waves are pre-sented in [1]–[6]. Generally, there is a great diversity in the teaching of undergraduate electromagnetics courses, in content, scope, and pedagogical philosophy. Some electromagnetics courses
Electromagnetics Inductors and the Wave Equation Inductors Capacitors deal with voltage and store energy in an electric field. Inductors, on the other hand, deal with . To see how the wave equation behaves, let's use Matlab to simulate an LC circuit. To start with, let's
Cruz-Pol, Electromagnetics UPRM Cruz-Pol, Electromagnetics UPRM Uniform plane em wave approximation Maxwell Equations in General Form Differential form Integral Form Gauss’s Law for E field. Gauss’s Law for H field. Nonexistence of monopole Faraday’s Law t Ampere’s Circuit Law Cruz-Pol,
Electromagnetics Chapter 1 Introduction . 1.3 The Nature of Electromagnetism (EM) . There are three branches of electromagnetics: 1- Electrostatics 2- Magnetostatics 3- Dynamics (Time-varying fields). The electric field E is governed by charge q.
