HW1_Solution

# HW1_Solution - ECE 201A Problem set 1 solution 1. (a) ( AB...

This preview shows pages 1–4. Sign up to view the full content.

ECE 201A Problem set 1 solution 1. (a) 2 ()() ()( ) () AA A A A AB C B A C C ∇× ∇× =∇ ∇⋅ − ∇⋅ ∇ =∇∇ ⋅ − ↑↑↑↑ ↑↑ ×× = (b) ( ) ( ) cc EH E H ∇⋅ × =∇⋅ × +∇⋅ × 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 since doesn't rule #1 rule #2 operate on ( ) ( ) c c E HE × =−∇⋅ × =−∇× =− ∇× since doesn't rule #1 operate on ( ) c H × =∇× ⋅ = ∇× Combining the two ( ) × − ∇× 2. (a) In rectangular coordinates 123 (, , ) (,,) uuu xyz = In cylindrical coordinates z ρφ = In spherical coordinates (, , ) ( ,,) r φθ = x y z r X Y Z φ θ ( x,y,z ) ( ρ , ,z) ( r, , ) (b) In rectangular coordinates ( , , ) (1,1,1) hhh = In cylindrical coordinates 1 ,, 1 ) = In spherical coordinates 1 ,,s i n) rr =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(c) According to the definition of gradient () 1 11 1 1 hu φ ∂∂ ∇= = A s o , 123 2 2 33 1 ˆˆˆ aaa φφ ∂∂∂ + + In rectangular coordinates: x yz x + + In cylindrical coordinates: 1 ˆˆ ˆ z aa a z ρφ ρρ + + In spherical coordinates: r â r + ââ sin rr θ θθ + (d) lim 0 V V Fd S F ∇• = G G G v u 2 u 1 u 3 h 1 du 1 h 2 du 2 h 3 du 3 0 V Let ˆ Fa F ++ = JG Flux coming out of volume, V, has 6 components, which are:
11 2 3 2 3 2 323 21 2 3 2 3 1 313 3 3 123 31 2 3 3 1 212 22 ,, du du Fd S F u uu uu hhd ud u du du Fuu u Fuu u h h d u d u du du F u u u F u u u h h du du ⎡⎤ ⎛⎞ ⋅= + ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ ++ G G v Note that h 1 , h 2 and h 3 are functions of u 1 ,u 2 and u 3 . So we should write: 2 3 1 1 23 2 3 2 3 , du du hhF u u u du du +− + …similar terms Dividing this by the volume and taking the limit as V Æ 0, for the first term we get: () 231 1 2 3 2 3 23 1 1 00 (, , ) , ) 1 lim lim 1 du du du du hhh du hhF hhh u →→ = Other terms are evaluated similarly and the result becomes: 231 13 2 12 3 1 2 3 1 Fh h F h h F h h F u u u ∂∂∂ ∇⋅ = + + G In rectangular coordinates: y x z F FF F x yz ∂∂ = + + G In cylindrical coordinates: ( ) 1 z z pF F F F F zz ρ φφ ρρ φ = + + = + + G Appropriate expressions can also be generated for the spherical coordinates.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/02/2009 for the course ECE 000 taught by Professor O during the Spring '09 term at UCSB.

### Page1 / 6

HW1_Solution - ECE 201A Problem set 1 solution 1. (a) ( AB...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online