ele401week7

ele401week7 - d ~ H = Id ~ l ( ~ r-~ r ) 4 | ~ r-~ r | 3 [...

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ELE401 - Field Theory: Tutorial Problems Week 7 1. Given that I L ~ H · d ~ l = Z S ( ~ ∇ × ~ H ) · d ~ S = I ENC . (1) Solve both sides of the Stokes’ Theorem for the close path and surfaces indicated in Figure 1 when ~ H is given by: (in mixed coordinates) ~ H = r ρ [( x - y ) c a x + ( x + y ) c a y ] [ A/m ] (2) The path “L” is given by x 2 + y 2 = 1; z = 3. There are 3 surfaces which are enclosed by the path. (a) Surface ~ S 1 the dome formed by a portion of a sphere of radius r = 2 [ m ] encircled by the path L . (b) Surface ~ S 2 the flat disk formed by the area of the circle x 2 + y 2 = 1 when z = 3. (c) Surface ~ S 3 (indicated as - ~ S 3 on the diagram) formed by the conical surface where tip is at the origin and the circle x 2 + y 2 = 1 at z = 3 forms the opening or lip of the cone. Note: Recall the R.H.L. for the direction of the surface vector w.r.t. the direction of the path. Note that ~ ∇ × ~ H = ~ J [ A/m 2 ] 2. Given that the incremental form of Biot-Savart’s Law is:
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Unformatted text preview: d ~ H = Id ~ l ( ~ r-~ r ) 4 | ~ r-~ r | 3 [ A/m ] (3) (a) Solve for the magnetic eld intensity vector ~ H at point P (0 , , z ) if the current I ows along the path shown in Figure 2. (i.e. the path is a quarter of a circle where = a ). (b) Solve for ~ H , if the same geometry as in (2a), the current path is half a circle (from = 0 to = ). 1 x y z L r = 2-S 3 z = 3 S 2 S 1 Figure 1: A close path and surfaces (c) Solve for ~ H , if the current path form a complete circle. (Do not omit any component due to symmetry arguments when calculating the solution.) x y z P(0,0,z) a z' a a Idl Figure 2: Current I 2...
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This note was uploaded on 04/17/2008 for the course ELE 401 taught by Professor Forgot during the Spring '08 term at Ryerson.

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ele401week7 - d ~ H = Id ~ l ( ~ r-~ r ) 4 | ~ r-~ r | 3 [...

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