Current density is given in cylindrical coordinates

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Current density is given in cylindrical coordinates as J = 10 6 z 1 . 5 a z A/m 2 in the region 0 ρ 20 μ m; for ρ 20 μ m, J = 0. ( a ) Find the total current crossing the surface z = 0 . 1 m in the a z direction. ( b ) If the charge velocity is 2 × 10 6 m/s at z = 0 . 1 m, find ρ ν there. ( c ) If the volume charge density at z = 0 . 15 m is 2000 C/m 3 , find the charge velocity there. Solution: (a) I = J.ds s = ( -10 6 z 1.5 az).Pd dza s 20 2 = -10 6 z 1.5 Pdp d 0 0 20 2 = -10 6 (0.1) 1.5 P 2 /2 .
Professor Aatif Saeed Preston University Islamabad Page 30 0 0 = - 10 -6 (0.0316). ½(20 2 0 2 ).(2 - 0) = - 10 -6 (0.0316).(400).( ) m 2 = - 10 -6 (0.0316).(400).(3.14) 10 -6 10 -6 = -39.7 A (b) J = v V v = J / V = - 10 6 z 1.5 a z 2 10 6 = - 10 6 (0.1) 1.5 a z 2 10 6 = -(0.0316) 2 = - 0.0158 c/m 3 = - 15.8 mc/m 3 (c) J = v V V = J / v = - 10 6 (0.15) 1.5 a z - 2000 = 10 6 (0.058) 2 10 3 = 10 3 (0.058) 2 = 58 2 = 29m/s ANSWER D5.3. Find the magnitude of the current density in a sample of silver for which σ = 6 . 17 × 10 7 S/m and μ e = 0 . 0056m 2 /V s if ( a ) the drift velocity is 1 . 5 μ m/s ; ( b ) the electric field intensity is 1 mV/m; ( c ) the
sample is a cube 2.5 mm on a side having a voltage of 0.4 mV between opposite faces; ( d ) the sample is a cube 2.5 mm on a side carrying a total current of 0.5 A. Solution: d e ) 5 3 2 E 2 2 2 2 E 7 7 7 2
Professor Aatif Saeed Preston University Islamabad Page 31
(d) J = I / S = 0.5 (2.5mm) 2 = 0.5 (2.5 10 -3 m) 2 = 0.5 2.5 2.5 10 -3 10 -3 = 2.5 10 6 6.25 = 0.08 10 6 A/m 2 = 80 10 3 A/m 2 = 80 k A/m 2 ANSWER D5.4. A copper conductor has a diameter of 0.6 in. and it is 1200 ft long. Assume that it carries a total dc current of 50 A. ( a ) Find the total resistance of the conductor. ( b ) What current density exists in it? ( c ) What is the dc voltage between the conductor ends? ( d ) How much power is dissipated in the wire? Solution: m r 2 2 S 2
Professor Aatif Saeed Preston University Islamabad Page 32
Professor Aatif Saeed Preston University Islamabad Page 33 ( c ) V = I R = 50 A 0.035 = 1.75 volts (d) P = V I = (1.75v)(50A) = 87.5 w ANSWER Drill 8.1: (a) H 2 = I 1 L 1 × a R 12 4 R 2 12 Here, a R 12 = (4 0) a x +(2 0) a y +(0 p 2) a z 4 2 +2 2 +2 2 = 0 . 816 a x + 0 . 408 a y 0 . 408 a z R 2 12 = 4 2 + 2 2 + 2 2 = 24 so, H 2 = I 1 L 1 × a R 12 4 R 2 12 = 2 a z ×(0 . 816 a x +0 . 408 a y 0 . 408 a z ) μ 301 . 59 = 5 . 12 a y 2 . 56 a x μ 301 . 59 = 8 . 5 a x + 17 . 0 a y nA/m (b) As, H 2 = I 1 L 1 × a R 12 4 R 2 12 Here, a R 12 = (4 0) a x p+(2 2) a y +(3 0) a z 4 2 +0 2 +3 2 = 0 . 8 a x + 0 . 6 a z R 2 12 = 4 2 + 0 2 + 3 2 = 25 so, H 2 = I 1 L 1 × a R 12 4 R 2 12 = 2 a z ×(0 . 8 a x +0 . 6 a z ) μ 100 = 5 . 02 a y μ 100 = 16 a y nA/m (c) As H 2 = I 1 L 1 × a R 12 4 R 2 12 Here, a R 12 = ( 3 1) a x p+( 1 2) a y +(2 3) a

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