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### HW1_Solution

Course: ECE 000, Spring 2009
School: UCSB
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Word Count: 1865

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201A ECE Problem set 1 solution 1. (a) ( AB (b) A) = ( C = B (A A) ( C) (A ) A = ( A) 2 A B) C ( E H ) = ( Ec H ) + ( E H c ) This is the application of the chain rule. Subscript c means that in the differentiation the term with that subscript will be treated as a constant. Then ( Ec H ) = ( H Ec ) = ( H Ec ) = ( H ) E rule #1 rule #2 since doesn't operate on...

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201A ECE Problem set 1 solution 1. (a) ( AB (b) A) = ( C = B (A A) ( C) (A ) A = ( A) 2 A B) C ( E H ) = ( Ec H ) + ( E H c ) This is the application of the chain rule. Subscript c means that in the differentiation the term with that subscript will be treated as a constant. Then ( Ec H ) = ( H Ec ) = ( H Ec ) = ( H ) E rule #1 rule #2 since doesn't operate on Ec ( E H c ) = E H c = ( E ) H rule #1 since doesn't operate on H c Combining the two ( E H ) = ( E ) H ( H ) E 2. (a) In rectangular coordinates (u1 , u2 , u3 ) = ( x, y , z ) In cylindrical coordinates (u1 , u2 , u3 ) = ( , , z ) In spherical coordinates (u1 , u2 , u3 ) = ( r, , ) Z z r (x,y,z) (,,z) (r,,) y x X Y (b) In rectangular coordinates ( h1 , h2 , h3 ) = (1,1,1) In cylindrical coordinates ( h1 , h2 , h3 ) = (1, ,1) In spherical coordinates ( h1 , h2 , h3 ) = (1, r, r sin ) (c) According to the definition of gradient ( )1 = so, 1 = 1 h1 u1 = 1 1 1 a1 + a2 + a3 h1 u1 h2 u2 h3 u3 In rectangular coordinates: = ax + ay + az x y z In cylindrical coordinates: = 1 a + a + az z 1 1 r + + r r r sin In spherical coordinates: = l (d) F =Vim 0 F dS V u3 V u2 h3du3 h1du1 u1 h2du2 0 Let F = a1 F1 + 2 F2 + 3 F3 Flux coming out of volume, V, has 6 components, which are: F dS = F u + 1 1 du1 du , u2 , u3 F1 u1 1 , u2 , u3 h2 h3du2 du3 2 2 du du + F2 u1 , u2 + 2 , u3 F2 u1 , u2 2 , u3 h1h3du1du3 2 2 du du 3 + F3 u1 , u2 , u3 + 3 F3 u1, u 2, u 3 h1h2 du1du2 2 2 Note that h1, h2 and h3 are functions of u1,u2 and u3. So we should write: du1 du1 h2 h3 F1 u1 + 2 , u2 , u3 h2 h3 F1 u1 2 , u2 , u3 du2 , du3 + similar terms Dividing this by the volume and taking the limit as V 0, for the first term we get: du1 du1 h2 h3 F1 (u1 + 2 , u2 , u3 ) h2 h3 F1 (u1 2 , u2 , u3 ) 1 lim lim du1 0 du1 0 h1h2 h3 du1 du1 1 = ( h2 h3 F1 ) h1h2 h3 u1 Other terms are evaluated similarly and the result becomes: F = 1 h1h2 h3 u ( h2 h3 F1 ) + u ( h1h3 F2 ) + u ( h1h2 F3 ) 2 3 1 In rectangular coordinates: F = Fx Fy Fz + + x y z In cylindrical coordinates: F = 1 ( pF ) F ( Fz ) 1 ( F ) 1 F F z + + + + = z z Appropriate expressions can also be generated for the spherical coordinates. 3. (a) I z a y x At DC current is distributed uniformly over the cross section of the cylinder and flows along the cylinder, i.e., in z direction. This can be seen using Maxwell's equations at DC, i.e., for = 0 . Furthermore current and current density are in z direction. Through ohm's law J = E . This makes electric field in z direction as well. Then E = j H = 0 . Written explicitly a + a + a a E = a E z + a E z = 0 z z z z But due to cylindrical symmetry there is no variation hence Furthermore =0. This implies E z =0. we take the divergence of both sides E z = 0 . Therefore E z has no z variation either. H 0 = J = E . Hence E = z This is also obvious from the fact that since both curl and divergence of the electric field is zero it must be a constant vector. Similarly J also must be a constant vector. Hence H = j E + J = E . If ( ) I = J idS = Jaz iaz d d = J a 2 . 00 a 2 This yields J= I a a2 z But inside the cylinder = J E , hence E= I az for 0 a . a 2 There is electric field outside the cylinder due to varying potential on the outside surface of the cylinder. This electric field should also be a constant vector outside since both its curl and divergence are zero. Due to continuity of the tangential electric field on the surface of the conductor its value is the same as the electric field inside the conductor. In other words E= I az for 0 . a 2 This may sound counter intuitive but results from the fact that we are considering an infinitely long cylindrical conductor. The resistance of this conductor approaches infinity and an infinitely large voltage needs to be applied to maintain a constant current I through it. As a result an electric field that extends to infinity with constant amplitude is generated. It will be shown later that this solution still satisfies the Poyntings theorem. Magnetic field is found using Amperes law in integral form, which is H idl = J idS In this case H is directed and does not depend on due to symmetry. So H = H a and dl = d a H idl = 2 0 H a ia d = H d =H d = 2 H 0 0 2 2 Inside the cylinder J idS = So 2 H = 2 I I a 2 az iaz d d = a 2 00 2 d d = 00 I 1 I2 2 2 = 2 a a2 2 I2 I . Hence H = for 0 a 2 2 a 2 a Outside the cylinder J idS = I hence H = I 2 for a . (b) Power flux leaving the cylinder per unit length is P= E H idS Let's evaluate the flux over the cylinder. In other words closed surface is the surface of the conducting cylinder of unit length and the volume is the volume of this cylinder. So a i a z = 0 a i a z = 0 a i a =1 P = Ez H ( az a )iaz dS + Ez H ( az a )i( az ) dS + Ez H ( az a )ia dS Flux through the top surface=0 Flux through the bottom surface=0 Flux through the side surface 1 2 1 2 2 P = Ez H ad dz = 00 00 I I a 2 2 a ad dz = I2 I2 dz d = 2 2 2 a 2 0 a 0 1 (c) I Pd = E i JdV = E dV = 2 a 2 2 dV Volume of the cylinder=1. a 2 = I2 a 2 Pd = I2 1 = RI 2 where R = 2 is the resistance of the cylindrical conductor of unit length. 2 a a (d) The instantaneous Poyntings theorem can be written as ( E i J i + H i M i dV = ) Power supplied by the sources in the volume E H idS + t 2 E Power leaving the volume 1 2 2 1 + H dV + E i JdV 2 Rate of increase of the stored energy in the volume Power dissipated in the volume Clearly there are no sources and = 0 . So t E H idS + E i JdV = 0 But E H idS = I2 and a 2 E i JdV = I2 as found above and the Poyntings theorem is a 2 satisfied. The same result is obtained if another cylindrical surface of radius larger then a is used. This is because although E remains unchanged, H reduces with as we move away from the cylinder. In other words P= E H idS = Ez H d dz = 00 00 2 1 2 1 2 I I a 2 2 d dz = I2 I2 dz d = 2 2 2 a 2 0 a 0 1 2 2 I Pd = E i JdV = E dV = 2 a dV Volume of the cylinder=1. a 2 = I2 a 2 In Pd calculation we end up with the volume of the conducting cylinder rather than the cylindrical volume we are considering since there is no conduction current outside the conducting cylinder. Outside the conducting cylinder there is a constant electric field but the conduction current is zero so the dissipated power density is zero. Obviously the Poyntings theorem is satisfied.
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