Topics covered in this week: • Concept of displacement current and displacement current density 𝐼

1 | P a g e Umm Al-Qura Universtiy, Makkah

Department of Electrical Engineering Electromagnetics (2) - (8022320)

Term 1: 2021/2022

Solution Homework 2 Dr. Waheed Ahmad Younis

Do not submit this homework. There will be a quiz from this homework on Wednesday, 15 Sep 2021.

Topics covered in this week:

• Concept of displacement current and displacement current density 𝐼_𝑑 = ∫_𝑠 ^𝜕𝐷_𝜕𝑡^⃗⃗ . 𝑑𝑆 𝐽⃗⃗⃗ =_𝑑 ^{𝜕𝐷⃗⃗}

𝜕𝑡

• Modified Maxwell’s equations (adding the displacement current)

1. ∮ 𝐸⃗ . 𝑑𝑙 = − ∫_𝑠 ^𝜕𝐵_𝜕𝑡^⃗ . 𝑑𝑆 Faraday’s law 2. ∮ 𝐻⃗⃗ . 𝑑𝑙 = ∫ 𝐽 . 𝑑𝑆 _𝑠 + ∫_𝑠 ^𝜕𝐷_𝜕𝑡^⃗⃗ . 𝑑𝑆 = ∫ (𝐽 +_𝑠 ^𝜕𝐷_𝜕𝑡^⃗⃗ ) . 𝑑𝑆 = 𝐼 Ampere’s law

3. ∮ 𝐷⃗⃗ . 𝑑𝑆 = ∫ 𝜌 𝑑𝑣_𝑣 = 𝑄 Gauss’s law

4. ∮ 𝐵⃗ . 𝑑𝑆 = 0 No isolated magnetic pole

• 2 theorems from vector calculus (for any vector field 𝐴 )

1. ∮ 𝐴 . 𝑑𝑆 = ∫ (∇. 𝐴 _𝑣 ) 𝑑𝑣 Divergence theorem (relating the close surface integral to the volume integral

2. ∮ 𝐴 . 𝑑𝑙 = ∫ (∇ × 𝐴 ). 𝑑𝑆 _𝑠 Stokes’ theorem (relating the close line integral to the surface integral

• Point form of Maxwell’s equations (using divergence and Stokes theorems) 1. ∇ × 𝐸⃗ = −^𝜕𝐵⃗

𝜕𝑡

2. ∇ × 𝐻⃗⃗ = 𝐽 +^{𝜕𝐷⃗⃗}

𝜕𝑡

3. ∇. 𝐷⃗⃗ = 𝜌 4. ∇. 𝐵⃗ = 0

• Maxwell’s equations for free space (using 𝐽 = 0, 𝜌 = 0, 𝐷⃗⃗ = 𝜖₀𝐸⃗ 𝑎𝑛𝑑 𝐵⃗ = 𝜇₀𝐻⃗⃗ ) 1. ∇ × 𝐸⃗ = −𝜇₀^{𝜕𝐻⃗⃗}

𝜕𝑡

2. ∇ × 𝐻⃗⃗ = 𝜖₀^𝜕𝐸⃗

𝜕𝑡

3. ∇. 𝐸⃗ = 0 4. ∇. 𝐻⃗⃗ = 0

• Phasor notation:

𝑣(𝑡) = 𝑉₀cos(𝜔𝑡 + 𝜃) = 𝑉₀ 𝑅𝑒{𝑒^𝑗𝜔𝑡𝑒^𝑗𝜃} ⇔ 𝑉_𝑝ℎ = 𝑉_𝑟𝑚𝑠𝑒^𝑗𝜃 = 𝑉_𝑟𝑚𝑠∠𝜃 𝑑𝑣(𝑡)

𝑑𝑡 = −𝜔𝑉₀sin(𝜔𝑡 + 𝜃) = 𝜔𝑉₀cos (𝜔𝑡 + 𝜃 +𝜋

2) = 𝜔𝑉₀ 𝑅𝑒 {𝑒^𝑗𝜔𝑡𝑒^𝑗𝜃𝑒^𝑗𝜋2}

= 𝜔𝑉₀ 𝑅𝑒{𝑒^𝑗𝜔𝑡𝑒^𝑗𝜃𝑗}

⇔ 𝑑(𝑉_𝑝ℎ)

𝑑𝑡 = 𝑗𝜔𝑉_𝑟𝑚𝑠𝑒^𝑗𝜃 = 𝑗𝜔𝑉_𝑝ℎ

2 | P a g e

• Maxwell’s equations for sinusoidal fields in free space 1. ∇ × 𝐸⃗⃗⃗⃗ = −𝑗𝜔𝜇_{𝑠 0}𝐻⃗⃗⃗⃗ _𝑠

2. ∇ × 𝐻⃗⃗⃗⃗ = 𝑗𝜔𝜖_{𝑠 0}𝐸⃗⃗⃗⃗ _𝑠 3. ∇ ∙ 𝐸⃗⃗⃗⃗ = 0 _𝑠 4. ∇ ∙ 𝐻⃗⃗⃗⃗ = 0 _𝑠

• Derivation of Helmholtz equation from Maxwell’s equations Maxwell’s equations for sinusoidal fields in free space

∇ × 𝐸⃗⃗⃗⃗ = −𝑗𝜔𝜇_{𝑠 0}𝐻⃗⃗⃗⃗ ∇ × 𝐻_𝑠 ⃗⃗⃗⃗ = 𝑗𝜔𝜖_{𝑠 0}𝐸⃗⃗⃗⃗ ∇ ∙ 𝐸_𝑠 ⃗⃗⃗⃗ = 0 ∇ ∙ 𝐻_𝑠 ⃗⃗⃗⃗ = 0 _𝑠 Taking curl of the first equation

∇ × (∇ × 𝐸⃗⃗⃗⃗ ) = −𝑗𝜔𝜇_{𝑠 0}(∇ × 𝐻⃗⃗⃗⃗ ) _𝑠

Apply the vector identity [∇ × (∇ × 𝐴 ) = ∇(∇ ∙ 𝐴 ) − ∇²𝐴 ]

∇(∇ ∙ 𝐸⃗⃗⃗⃗ ) − ∇_𝑠 ²𝐸⃗⃗⃗⃗ = −𝑗𝜔𝜇_{𝑠 0}(∇ × 𝐻⃗⃗⃗⃗ ) _𝑠

Inserting the 2^nd and the 3^rd Maxwell’s equations in the above

∇(0) − ∇²𝐸⃗⃗⃗⃗ = −𝑗𝜔𝜇_{𝑠 0}(𝑗𝜔𝜖₀𝐸⃗⃗⃗⃗ ) _𝑠

∇²𝐸⃗⃗⃗⃗ = 𝑗𝜔𝜇_{𝑠 0}(𝑗𝜔𝜖₀𝐸⃗⃗⃗⃗ ) _𝑠

∇²𝐸⃗⃗⃗⃗ = −𝜔_𝑠 ²𝜇₀𝜖₀𝐸⃗⃗⃗⃗ _𝑠

The above equation is known as Helmholtz equation.

• Solution of the 2^nd order differential equation:

𝑑²𝑦

𝑑𝑥² = −𝑘²𝑦 is 𝑦(𝑥) = 𝑌𝑒^{−𝑗𝑘𝑥} Proof:

𝑦 = 𝑌𝑒^{−𝑗𝑘𝑥} 𝑦^′= −𝑗𝑘 𝑌𝑒^{−𝑗𝑘𝑥} 𝑦^′′= (−𝑗𝑘)²𝑌𝑒^{−𝑗𝑘𝑥} = −𝑘²𝑌𝑒^{−𝑗𝑘𝑥} = −𝑘²𝑦

• Solution of the Helmholtz equations.

Assume that the electric field exists along x-axis only. Writing Helmholtz equation for this case

∇²𝐸_𝑥𝑠 = −𝜔²𝜇₀𝜖₀𝐸_𝑥𝑠

𝜕²𝐸_𝑥𝑠

𝜕𝑥² +𝜕²𝐸_𝑥𝑠

𝜕𝑦² +𝜕²𝐸_𝑥𝑠

𝜕𝑧² = −𝜔²𝜇₀𝜖₀𝐸_𝑥𝑠

Assume that the electric field varies along z-axis only (the derivatives along other axes would be zero)

𝑑²𝐸_𝑥𝑠

𝑑𝑧² = −𝜔²𝜇₀𝜖₀𝐸_𝑥𝑠

This is a 2^nd order differential equation. A solution of this equation is 𝐸_𝑥𝑠(𝑧) = 𝐸_𝑥0 𝑒^{−𝑗𝜔√𝜇0𝜖0 𝑧}

Converting phasor into time-domain function

𝐸_𝑥(𝑧, 𝑡) = 𝐸_𝑥0 𝑅𝑒 [𝑒^𝑗𝜔𝑡𝑒^{−𝑗𝜔√𝜇0𝜖0 𝑧}] = 𝐸_𝑥0 𝑅𝑒 [𝑒^{𝑗(𝜔𝑡−𝜔√𝜇0𝜖0 𝑧)}] = 𝐸_𝑥0cos(𝜔𝑡 − 𝜔√𝜇₀𝜖₀ 𝑧) Above expression represents a travelling wave of electric field. It is directed along x-axis and travelling along z-axis.

Wavelength = 𝜆 = ^2𝜋

𝜔√𝜇0𝜖0 Period= 𝑇 =^2𝜋

𝜔 Speed = 𝑓𝜆 = ¹

√𝜇0𝜖0

____________________________________________________________________________________

3 | P a g e Q1. For the electric flux density given in Q5 of HW-1:

𝐷⃗⃗ = (2𝑥𝑦 𝑎⃗⃗⃗⃗ + 𝑥_𝑥 ²𝑎⃗⃗⃗⃗ ) 𝐶 𝑚_𝑦 ⁄ ²

Find the total charge inside the close surface of the parallelepiped formed by the planes x=0 and 1, y=0 and 2, and z=0 and 3. Use divergence theorem.

Solution:

According to divergence theorem: ∮ 𝐷⃗⃗ ∙ 𝑑𝑆 = ∫ (∇. 𝐷⃗⃗

𝑣 ) 𝑑𝑣

𝐶ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝜌 = ∇. 𝐷⃗⃗ =𝜕𝐷_𝑥

𝜕𝑥 +𝜕𝐷_𝑦

𝜕𝑦 +𝜕𝐷_𝑧

𝜕𝑧 = 2𝑦 + 0 + 0 = 2𝑦 𝐶 𝑚⁄ ² 𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒 = ∫ 𝜌

𝑣

𝑑𝑣 = ∫ (∇. 𝐷⃗⃗

𝑣

) 𝑑𝑣 = ∭ 2𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑧

𝑥=1,𝑦=2,𝑧=3

𝑥=0,𝑦=0,𝑧=0

= 12 𝐶

Q2. For the vector field given in Q6 of HW-1:

𝐷⃗⃗ = 6𝜌 sin^𝜑

2 𝑎⃗⃗⃗⃗ + 1.5𝜌 cos_𝜌 ^𝜑

2𝑎⃗⃗⃗⃗ _𝜑

Find the surface integral for the close surface bounded by 𝜌 = 2, 𝜑 = 0 𝑎𝑛𝑑 𝜋, 𝑧 = 0 𝑎𝑛𝑑 5.

Use divergence theorem.

Solution:

According to divergence theorem: ∮ 𝐷⃗⃗ ∙ 𝑑𝑆 = ∫ (Div 𝐷⃗⃗

𝑣 ) 𝑑𝑣

For cylindrical coordinates, Div 𝐷⃗⃗ = 1

𝜌

𝜕(𝜌𝐷_𝜌)

𝜕𝜌 +1 𝜌

𝜕𝐷_𝜑

𝜕𝜑 +𝜕𝐷_𝑧

𝜕𝑧 =1 𝜌

𝜕 (𝜌 6𝜌 sin𝜑 2)

𝜕𝜌 +1

𝜌

𝜕 (1.5𝜌 cos𝜑 2)

𝜕𝜑 = 11.25 sin𝜑 2

∫ (Div 𝐷⃗⃗

𝑣

) 𝑑𝑣 = ∭ 11.25 sin𝜑

2 𝜌 𝑑𝜌 𝑑𝜑 𝑑𝑧

𝜌=2,𝜑=𝜋,𝑧=5

𝜌=0,𝜑=0,𝑧=0

= 11.25 ∫ sin𝜑 2 𝑑𝜑

𝜋

𝜑=0

∫ 𝜌 𝑑𝜌

2

𝜌=0

∫ 𝑑𝑧

5

𝑧=0

= 11.25 |−2 cos𝜑 2|

0 𝜋

|𝜌² 2|

0 2

|𝑧|₀⁵ = −22.5(0 − 1)(2 − 0)(5 − 0) = 225

Q3. For the magnetic field given in Q7 of HW-1:

𝐻⃗⃗ = (6𝑥𝑦 𝑎⃗⃗⃗⃗ − 3𝑦_𝑥 ²𝑎⃗⃗⃗⃗ ) 𝐴 𝑚_𝑦 ⁄

Find the total current enclosed by the rectangular path formed by the lines 2 ≤ 𝑥 ≤ 5 and −1 ≤ 𝑦 ≤ 1. Use Stokes’ theorem.

Solution:

According to Stokes’ theorem: ∮ 𝐻⃗⃗ . 𝑑𝑙 = ∫ (∇ × 𝐻⃗⃗ _𝑠 ) . 𝑑𝑆

Current density = 𝐽 = ∇ × 𝐻⃗⃗ = ||

𝑎_𝑥

⃗⃗⃗⃗ 𝑎⃗⃗⃗⃗ _𝑦 𝑎⃗⃗⃗⃗ _𝑧

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧 𝐻_𝑥 𝐻_𝑦 𝐻_𝑧

|| = ||

𝑎_𝑥

⃗⃗⃗⃗ 𝑎⃗⃗⃗⃗ _𝑦 𝑎⃗⃗⃗⃗ _𝑧

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧 6𝑥𝑦 −3𝑦² 0

||

= 𝑎⃗⃗⃗⃗ (0 − 0) − 𝑎_𝑥 ⃗⃗⃗⃗ (0 − 0) + 𝑎_𝑦 ⃗⃗⃗⃗ (0 − 6𝑥) = −6𝑥 𝑎_𝑧 ⃗⃗⃗⃗ 𝐴𝑚𝑝𝑒𝑟𝑒/𝑚_𝑧 ² 𝑑𝑆 = 𝑑𝑥 𝑑𝑦 𝑎⃗⃗⃗⃗ _𝑧

(∇ × 𝐻⃗⃗ ). 𝑑𝑆 = −6𝑥 𝑑𝑥 𝑑𝑦

4 | P a g e

∫ 𝐽

𝑠

. 𝑑𝑆 = ∫ (∇ × 𝐻⃗⃗

𝑠

) . 𝑑𝑆 = −6 ∬ 𝑥 𝑑𝑥 𝑑𝑦

𝑦=1,𝑥=5

𝑦=−1,𝑥=2

= −6 ∫ 𝑥 𝑑𝑥

5

𝑥=2

∫ 𝑑𝑦

1

𝑦=−1

= −126 𝐴𝑚𝑝𝑒𝑟𝑒

Q4. For the magnetic field given in Q8 of HW-1:

𝐻⃗⃗ = 20𝜌²𝑎⃗⃗⃗⃗ 𝐴/𝑚: _𝜑

Fine the total current passing through the circular surface 0 ≤ 𝜌 ≤ 1, 𝑧 = 0. Use Stoke’s theorem.

Solution:

Curl 𝐻⃗⃗ = [¹

𝜌

𝜕𝐻𝑧

𝜕𝜑 −^𝜕𝐻𝜑

𝜕𝑧 ] 𝑎⃗⃗⃗⃗ + [_𝜌 ^𝜕𝐻𝜌

𝜕𝑧 −^𝜕𝐻𝑧

𝜕𝜌] 𝑎⃗⃗⃗⃗ +_𝜑 ¹

𝜌[^{𝜕(𝜌𝐻𝜑)}

𝜕𝜌 −^𝜕𝐻𝜌

𝜕𝜑] 𝑎⃗⃗⃗⃗ _𝑧 = [¹

𝜌

𝜕(0)

𝜕𝜑 −^{𝜕(20𝜌2)}

𝜕𝑧 ] 𝑎⃗⃗⃗⃗ + [_𝜌 ^𝜕(0)

𝜕𝑧 −^𝜕(0)

𝜕𝜌] 𝑎⃗⃗⃗⃗ +_𝜑 ¹

𝜌[^{𝜕(20𝜌3)}

𝜕𝜌 −^𝜕(0)

𝜕𝜑] 𝑎⃗⃗⃗⃗ _𝑧 =¹

𝜌

𝜕(20𝜌³)

𝜕𝜌 𝑎⃗⃗⃗⃗ = 60𝜌 𝑎_𝑧 ⃗⃗⃗⃗ 𝐴/𝑚_𝑧 ²

𝑑𝑆 = 𝜌 𝑑𝜑 𝑑𝜌 𝑎⃗⃗⃗⃗ (Curl 𝐻_𝑧 ⃗⃗ ). 𝑑𝑆 = 60𝜌² 𝑑𝜑 𝑑𝜌 𝐼 = ∮ 𝐻⃗⃗ . 𝑑𝑙 = ∫(∇ × 𝐻⃗⃗

𝑠

) . 𝑑𝑆 = ∬ 60𝜌² 𝑑𝜑 𝑑𝜌

𝜌=1,𝜑=2𝜋

𝜌=0,0,𝜑=0

= 60 ∫ 𝜌²𝑑𝜌

1

0

∫ 𝑑𝜑

2𝜋

0

= 60 (1 3) (2𝜋)

= 40𝜋 𝐴

Q5. Find the amplitude of displacement current density around a car antenna where the field strength of an FM signal is 𝐸⃗ = 80 cos(6.277 × 10⁸𝑡 − 2.092𝑦) 𝑎⃗⃗⃗⃗ 𝑉 𝑚_𝑧 ⁄ .

Solution:

𝐷⃗⃗ = 𝜖₀𝐸⃗ = 80𝜖₀cos(6.277 × 10⁸𝑡 − 2.092𝑦) 𝑎⃗⃗⃗⃗ 𝐶 𝑚_𝑧 ⁄ ² Displacement current density =^{𝜕𝐷⃗⃗}

𝜕𝑡

= −80𝜖₀(6.277 × 10⁸) sin(6.277 × 10⁸𝑡 − 2.092𝑦) 𝑎⃗⃗⃗⃗ 𝐶 𝑠. 𝑚_𝑧 ⁄ ²

= −0.445 sin(6.277 × 10⁸𝑡 − 2.092𝑦) 𝑎⃗⃗⃗⃗ 𝐴 𝑚_𝑧 ⁄ ² Hence, amplitude = 0.445 𝐴 𝑚⁄ ²

Q6. Express the following time function as phasor:

𝐸_𝑦 = 100 cos (10⁸𝑡 − 0.5𝑧 +^𝜋

6) Solution:

𝐸_𝑦 = 100 cos (10⁸𝑡 − 0.5𝑧 +^𝜋

6) = 100𝑅𝑒 [𝑒^{𝑗(108𝑡−0.5𝑧+}

𝜋 6)

] = 100𝑅𝑒 [𝑒^𝑗108𝑡𝑒^{𝑗(−0.5𝑧+}

𝜋 6)

] 𝐸_𝑦𝑠 =¹⁰⁰

√2 𝑒^{−𝑗0.5𝑧+𝑗}

𝜋

6 = 70.71∠ (−0.5𝑧 +^𝜋

6)

Q7. Consider the field intensity vector expressed as phasor. Recover the time-domain expression:

𝐸_𝑠

⃗⃗⃗⃗ = 100∠30° 𝑎⃗⃗⃗⃗ + 20∠ − 50° 𝑎_𝑥 ⃗⃗⃗⃗ + 40∠210° 𝑎_𝑦 ⃗⃗⃗⃗ V/m _𝑧 Solution:

5 | P a g e 𝐸_𝑠

⃗⃗⃗⃗ = 100𝑒^𝑗30° 𝑎⃗⃗⃗⃗ + 20𝑒_𝑥 ^−𝑗50° 𝑎⃗⃗⃗⃗ + 40𝑒_𝑦 ^𝑗210° 𝑎⃗⃗⃗⃗ _𝑧 𝐸_𝑠

⃗⃗⃗⃗ (𝑡) = (100𝑒^𝑗30° 𝑎⃗⃗⃗⃗ + 20𝑒_𝑥 ^−𝑗50° 𝑎⃗⃗⃗⃗ + 40𝑒_𝑦 ^𝑗210° 𝑎⃗⃗⃗⃗ )𝑒_𝑧 ^𝑗𝜔𝑡

𝐸⃗ (𝑡) = 𝑅𝑒(100√2𝑒^𝑗𝜔𝑡𝑒^𝑗30° 𝑎⃗⃗⃗⃗ + 20√2𝑒_𝑥 ^𝑗𝜔𝑡𝑒^−𝑗50° 𝑎⃗⃗⃗⃗ + 40√2𝑒_𝑦 ^𝑗𝜔𝑡𝑒^𝑗210° 𝑎⃗⃗⃗⃗ ) _𝑧

= 100√2 cos(𝜔𝑡 + 30°) 𝑎⃗⃗⃗⃗ + 20√2 cos(𝜔𝑡 − 50°) 𝑎_𝑥 ⃗⃗⃗⃗ + 40√2 cos(𝜔𝑡 + 210°) 𝑎_𝑦 ⃗⃗⃗⃗ V/m _𝑧 Q8. Convert the following phasor into time-domain expression:

𝐻_𝑠

⃗⃗⃗⃗ = 20𝑒−(0.1+𝑗20)𝑧 𝑎⃗⃗⃗⃗ A/m _𝑥 Solution:

𝐻⃗⃗ (𝑡) = 20√2 𝑅𝑒[𝑒−(0.1+𝑗20)𝑧𝑒^𝑗𝜔𝑡]𝑎⃗⃗⃗⃗ = 20√2 𝑅𝑒[𝑒_𝑥 −0.1𝑧+𝑗𝜔𝑡−𝑗20𝑧]𝑎⃗⃗⃗⃗ _𝑥

= 20√2𝑒^−0.1𝑧 𝑅𝑒[𝑒^{𝑗(𝜔𝑡−20𝑧)}]𝑎⃗⃗⃗⃗ = 20√2 𝑒_𝑥 ^−0.1𝑧cos(𝜔𝑡 − 20𝑧) 𝑎⃗⃗⃗⃗ A/m _𝑥