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# INTRO TO ENGINEERING ELECTROMAGNETICS EE 468G

UK

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This 11 page Class Notes was uploaded by Adaline Pollich on Friday October 23, 2015. The Class Notes belongs to EE 468G at University of Kentucky taught by Caicheng Lu in Fall. Since its upload, it has received 18 views. For similar materials see /class/228311/ee-468g-university-of-kentucky in Electrical Engineering at University of Kentucky.

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Date Created: 10/23/15

EE468G NOTES 1 Reading assignment Vectors Contents Unit Vectors Vectors Vector operations Vectors in three coordinate systems Vector A quantity that has magnitude and direction Examples Force velocity field intensity Notation and writing A 0r A or A Unit vector a vector Whose magnitude is 1 Examples ax x is a unit vector pointing in Xdirection y is a unit vector pointing in ydirection Vector representation in rectangular coordinate system component format A25 332 79 C2 2 2 In general a vector has three components Ax Ay AZ gt AAxfcij2A223 Component Ax is the projection of vector A on XaXis Calculation of magnitude Igll A 14 14 Examples 3233 79 1 3 32 7202 z 2 2 z J222212 J 3 Direction of a vector The direction of a vector can be described by three angles that the vector made with the xaxis yaxis and z axis respectively In the above examples 333 79 B 32 7202 3 It makes an angle of cos 1 3 J5 6680 with xaxis It makes an angle of COS 1 7 15680 with yaxis It makes an angle of cos 10 90 with zaxis In general if A Ax 14y 1422 the angles it makes with x y and zcomponents are 05x70 7 and 052 Where AD 09 cos 1Ax 062 cos 1 AZ 05 a and a2 3 x y Example Let 21 656 259 082 calculate A Solution 2 162 252 082 W 655 06x 2 cos 16655 2365 Qty 2 cos 125655 6756 a cos 1 08655 97020 2 Vector operations Addition and subtraction AzAx ij2A223 BBx ByjzBz Addition of 1a and E ZlBAX Bx Ay Byf2AzBz2 Subtraction of A and B Zl EAx Bx Ay By3Az Bz2 The results of addition and subtraction are vectors The addition and subtraction operation satisfy the commutative law and distributive law Example Let Z6 25f2 08 B 3c2j262 calculate 33 and 121 1 Solution 216 3 252 6 0823 453522 21 363 25 29 6o829 o59 682 Graphical representation of Z E and 11 B Vector operation Scalar product dot product ZAjcij2A223 BBxfcByjzB22 gt gt Scalar product A 39 B AxBx AyBy 14sz 39 The result of the dot product of two vectors is a scalar 39 The addition and subtraction operation satisfy the commutative law 1311 2 3131 and distributive law A7B 11 15 I If the two vectors Zr and E make an angle of HM then Hlelllglcosww 3E2j262 Example Let 216 c25f2 0 C and 216 21 39102 Calculate 2111 339 Solution 6125 3 0810 95 gm O H Two vectors A and B are said to be perpendicular to each other ifand onlyif 2130 A is perpendicular to E can also be written as ALE Example Let 216 cl5j2a 3 E 3 2 623 Show that When a21563wehave El LE Proof 2i if requires 0 Consider 2116x 315x2ax6 156a Let g39EZOJhen 61156 Q Q 39 Q J N In Cartesian coordinate system ax J y y Cross product A Axx Ayy Azz B BxxByyBzZ Cross product is defined as 21 X a 14sz Asz 2 1sz 14sz j AxBy AyBx 2 gtThe result of the cross product of two vectors is a vector gtwhich is perpendicular to both A and B RHRULE gtCross product satisfies distributive law 21 x E C 21 x 5 21 x C gtCross product does not satisfy commutative law 171 X E t E X 171 But 2x1 l gtlti In Cartesian coordinate system a x07 07 a x512le a x07 zay x y 2 y z x Hence the unit vectors satisfy the righthand rule axgtaygtaz TT U TTltltltltlt U a Calculation of cross product using determinants 2 y 2 A A A A AC A AxfazAxA Azz y Z yx 22 y y By 32 Bx 32 Bx By Bx By 32 Example Let 216 253 0823 E 3E2j6zAj 39102 ca1culate Axfa 3x57 Solution fc j 2 A A A 25 08 A 6 08 A 6 25 AgtltB 6 25 08x y Z 2 6 3 6 3 2 3 2 6 166E 336j2l952 chyZAz A 36A 32 BgtltC 3 2 6 x y Z 3 10 l 10 l 3 l 3 10 38fc36jz72 MATLAB COMMAND gtgta 6 25 08 Enterarowvector a gtgtb 3 2 6 Enterarowvector b gtgt c cross a b Do cross product 166000 336000 195000 Calculate Unit vectors Given 171 633 25 3 083 find 51A the unit vector points to the direction of 171 Solution 1 Step 1 21 62 252 082 m 655 2 Step 2 61A 2121 6fc25j2 082655 0916fc0382j2 01222 Example Let 21 6 c25j2 082 3 E 3E 2 62 3 C 2 3j2 102 3 Find the unit vector 51 that points to the direction of 11 gtlt f3 Solution From the example on the previous page 21 x3 1662 33633 1952 Hence a3 166 33652 19521662 3362 1952 2 039 0803 0462 Property of cross product I la gtltBl1allglsmg 6 is the acute angle between A and 13 Cylindrical coordinate System A point in 3D space is determined by three scalar quantities 0 The distance to zaXis positive The angle with XaXis 0 27 or 7I gt7r Z The zcoordinate value 00 lt Z lt 00 Unit vectors 0 in 0 direction 51 in direction 5 in Z direction Relationship 63p gt gt 512 gt 63p form RHS system Relationship of p 2 with x y Z Relationship with 63x 63y 512 apzaxcos ays1n axzapcos a s1n ampxsin ampycos or yzfzpsin l cos 5122512 5122512 The above relationship gives rise to Ca ap COS etc Example Given 171 451x 355 08512 find fl in cylindrical system Solution We need to find the components AP Ag and AZ Ap A7ap 451x 3ay 08 z 1p 4axap 3ayapo8azap 4cos 3sin 08gtlt0 A 415 451x 3ay 08 z z 4aXa 3ay amp 08 zz 1 4sin 3cos 08gtlt0 A Ap zp A z Az zz 4cos 3sin zp 4sin 3cos z 086zz Another method A 451x 3ay 0855 2 451p cow 51 sin 3 lp sin l cos 08 lz 4cos 3sin zp 4sin 3cos z 08 z Put the start point of the vector A at the origin of the coordinate system then the end point of this vector is at 4308 This gives rise to p 2 J42 32 5 3 tan 17 3687 Z 08 Using the above values we have 21 4cos3687 3sin3687 zp 4sin3687 3cos3687quot z 0825 04921 4986 08212

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