Electromagnetic 1 Examination Solution 2010

Fourth Stage Electromagnetic Time: 75 minutes

Sulaimani University College of Sciences Physics Department Q1/ transform

into spherical coordinate and find its magnitude at point 3, 4, 0 .

Q2/ Verify Stoke’s theorem for a vector field defined by

2,

, and 0

Q3/ Determine the flux of Q4/ Show that the scalar function

Spherical Coordinate:

Cylindrical Coordinate:

ρ

·

·

, in the segment of cylindrical surface 3.

through the entire prism shown in Fig(2). , is harmonic at point √3, , 0 .

Sulaimani University - College of Sciences - Physics Department First Examination (5/11/2009) Electromagnetic Q1/ transform

into spherical coordinate and find its magnitude at point 3, 4, 0 .

· · ·

· · · · · ·

0

· · · 0 0 1

· · ·

· · ·

· · ·

0

√9

16 5 √25 5 0 90 5 4 53.13 3

5 sin 90 cos 53.13 5 0 cos 90 sin 53.13 5 cos 90 cos 53.13 5 0 sin 90 sin 53.13 3

0

0.018

.

Q2/ Verify Stoke’s theorem for a vector field 2 , 60

defined by

, in the segment of cylindrical surface

90 , and 0

3.

The mathematical representation of Stokes’s theorem is given by: r

r

r

∫ A ⋅ dl = ∫ ∇ × A ⋅ ds L

L

The line integral around a closed path defined by these bounded region is as follows: b r c r d r a r r B ⋅ dl = B ⋅ dl + B ⋅ dl + B ⋅ dl + ∫ ∫ ∫ ∫ ∫ B ⋅ dl L

a

b

c

d

b

r B ∫ ⋅ dl = ∫ B z aˆ z ⋅ ρ dφ aˆφ = 0

ρ

a

3 r cos φ cos 90 ˆ ˆ B ⋅ dl = B a ⋅ dz a = z ∫b ∫ z z ∫0 ρ dz = 2 × z c

3

=0

0

d

r B ∫ ⋅ dl = ∫ B z aˆ z ⋅ ρ dφ (−aˆφ ) = 0 c

3 r cos φ cos 60 ˆ ˆ B ⋅ dl = B a ⋅ dz − a = ( ) z ∫d ∫ z z ∫0 ρ dz = − 2 × z a

3 0

=−

3 4

r r The left hand side of the Stokes’s theorem is : ∫ ∇ × A ⋅ ds s

ρ aˆφ

aˆ ρ ∂ ρ ∂ρ

r r 1 ∇×B =

∂ ∂φ

0

0

aˆ z 1 ∂ = ρ ∂z cos φ

⎤ ⎡ sin φ cos φ ⎢aˆ ρ (− ρ + 0) + ρ aˆφ (0 + 2 ) + aˆ z (0 − 0)⎥ ρ ⎦ ⎣

ρ

r r 1 ∇ × B = 2 [cos φ aˆφ − sin φ aˆ ρ ]

ρ

Hence, π /2 3 r r 1 sin φ ˆ ˆ ˆ [ ] φ a φ a ρ dz d φ a ( ) cos sin ∇ × ⋅ = − ⋅ = B ds φ ρ ρ ∫ ∫ 2 ∫ ∫ − 2 ρ dz dφ s

s

ρ

π /3 0

r r 1 ( ∇ ∫ × B) ⋅ ds = − × (cos φ ) s

ρ

r r 3 1 ( ∇ ∫s × B) ⋅ ds = 2 (0 − 2 )

π /2 π /3

× ( z)

ρ

1 = × (cos π / 2 − cos π / 3) × (3 − 0) 0 2

3

r r 3 ⇒ ∫ (∇ × B) ⋅ ds = − 4 s

It is clearly seen that the left and right hand side has the same value, which indicates the validity of the Stokes’s theorem.

Q3/ Determine the flux of

through the entire prism shown in Fig(2).

z

c

The equation of surface (abc) is: 1 1, 6

3,

2

6

y

6

6

6

b

2

x a

The equation of line (ab) is:

3 3 3 3 To find flux through the entire prism we can use divergence theorem: . · · 3 .

3

.

.

3

.

27

.

27

1 3

1 4

3

6

6 3

3

1 5

H H H

2

6 3

3

2

3

3 2

7.65

Q4/ Show that if the scalar function

H

6

, is harmonic at point √3, , 0 .

1 r 1 r

1 ∂H 1 ∂ ∂H ∂ ∂ H r sinθ r sinθ ∂θ ∂θ r sin θ ∂φ ∂r ∂r ∂ ∂ 1 1 sin θ e 0 r sinθ 2 sin θ cosθ e ∂r r sinθ ∂θ r sin θ 2e sin θ r e sin θ 2 sin θ cos θ 2r e r r sinθ 2e sin θ 2 cos θ sin θ e r 2 r r

At point √3, , 0 ,

H

0

is not harmonic.