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Calculate the value of the vector current density: (a)in rectangular coordinates at P₂(2,3,4) if H = x²z ay-y²x a₂ (b)in cylindrical coordiantes at Pg(1.5, 90deg, 0.5) if H = 2/p (cos 0.24) ap (c)in spherical coordinates at P. (2, 30deg. 20deg) if H = 1/(sine) a

Calculate the value of the vector current density: (a)in rectangular coordinates at P₂(2,3,4) if H = x²z ay-y²x a₂ (b)in cylindrical coordiantes at Pg(1.5, 90deg, 0.5) if H = 2/p (cos 0.24) ap (c)in spherical coordinates at P. (2, 30deg. 20deg) if H = 1/(sine) a

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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A:+W Solution turbospool.xyz Powered by Skippy % : Question: Calculate the value of the vector current density: (a)in rectangular coordinates at PA(2,3,4) if H = x²z ay-y²x a₂ (b)in cylindrical coordiantes at Pg(1.5, 90deg, 0.5) if H = 2/p (cos 0.24) ap (c)in spherical coordinates at P.(2, 30deg, 20deg) if H= 1/(sine) a Show work Answer: J-vector current density (a) For rectangular see Curl(H) = V XH y²)) ay above equation we get J=-16ax + 9ay + 16az (b) X and we get J=0.055az (c) when applying curl(H) we get J=(-2yx-2xz)ax + (2xz) az + (0-(- therefore for points (2,3,4) in the For cylindrical see curl (H) = V XH when applying curl(H) we get (0-² (-sin0.20 × 0.2))az x Then 0.4sin(0.20) jaz P now we insert the points (1.5,90°, 0.5) For spherical see Curl(H) = V XH when applying curl(H) we get ((r.)-0) ap then = (())a sin(0) аф rsin(0) now inserting point (2, 30°, 20°) we get J=ckp
Transcribed Image Text:A:+W Solution turbospool.xyz Powered by Skippy % : Question: Calculate the value of the vector current density: (a)in rectangular coordinates at PA(2,3,4) if H = x²z ay-y²x a₂ (b)in cylindrical coordiantes at Pg(1.5, 90deg, 0.5) if H = 2/p (cos 0.24) ap (c)in spherical coordinates at P.(2, 30deg, 20deg) if H= 1/(sine) a Show work Answer: J-vector current density (a) For rectangular see Curl(H) = V XH y²)) ay above equation we get J=-16ax + 9ay + 16az (b) X and we get J=0.055az (c) when applying curl(H) we get J=(-2yx-2xz)ax + (2xz) az + (0-(- therefore for points (2,3,4) in the For cylindrical see curl (H) = V XH when applying curl(H) we get (0-² (-sin0.20 × 0.2))az x Then 0.4sin(0.20) jaz P now we insert the points (1.5,90°, 0.5) For spherical see Curl(H) = V XH when applying curl(H) we get ((r.)-0) ap then = (())a sin(0) аф rsin(0) now inserting point (2, 30°, 20°) we get J=ckp
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