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ele401week7

# ele401week7 - d ~ H = Id ~ l × ~ r-~ r 4 π | ~ r-~ r | 3...

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ELE401 - Field Theory: Tutorial Problems Week 7 1. Given that L H · dl = S ( ∇ × H ) · dS = 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 ) a x + ( x + y ) 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 ∇ ×

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Unformatted text preview: d ~ H = Id ~ l × ( ~ r-~ r ) 4 π | ~ r-~ r | 3 [ A/m ] (3) (a) Solve for the magnetic ﬁeld intensity vector ~ H at point P (0 , , z ) if the current I ﬂows along the path shown in Figure 2. (i.e. the path is a quarter of a circle where ρ = a ). (b) Solve for ~ H , if the same geometry as in (2a), the current path is half a circle (from φ = 0 to φ = π ). 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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ele401week7 - d ~ H = Id ~ l × ~ r-~ r 4 π | ~ r-~ r | 3...

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