Electromagnetism: Tutorial JUAS 2013: Electromagnetism Exercises H. Henke

Tut-Ex 1 Given is a conducting hollow sphere carrying a charge Q. What is the field inside and outside and what is the stored energy? 𝐷 ⋅𝑑 𝐹 = ρ 𝑑𝑉 a Q → 𝐸 = 0 𝑄 4π ϵ 0 𝑟 2 𝑒 𝑟 𝐷 ⋅𝑑 𝐹 = 0𝑟≤𝑎 𝑄𝑟≥𝑎 JUAS 2013: Electromagnetism Exercises H. Henke

Stored energy 𝑊 𝑒 = 1 2 𝐸 ⋅ 𝐷 𝑑𝑉= 𝑄 2 8π ε 0 𝑎 ∞ 𝑑𝑟 𝑟 2 = 𝑄 2 8π ε 0 𝑎 𝑊 𝑒 = 1 2 𝐸 ⋅ 𝐷 𝑑𝑉= 𝑄 2 8π ε 0 𝑎 ∞ 𝑑𝑟 𝑟 2 = 𝑄 2 8π ε 0 𝑎 𝑊 𝑒 = 1 2 ρ Φ𝑑𝑉= 1 2 ρ 𝑆 Φ𝑑𝐹 = 1 2 𝑄 4π 𝑎 2 𝑄 4π ϵ 0 𝑎 𝑑𝐹= 𝑄 2 8π ϵ 0 𝑎 JUAS 2013: Electromagnetism Exercises H. Henke

Tut-Ex 2 A capacitor C is filled with a lossy dielectric and charged to a voltage V. What is the time constant for discharge? A y εr κ V, C 𝐶𝑐𝑎𝑟𝑟𝑖𝑒𝑠𝑎𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑐ℎ𝑎𝑟𝑔𝑒 𝑞 𝑆 = 𝐶𝑉 𝐴 =𝐷 ∇ ⋅ ∇ × 𝐻 =𝐽+ ∂ 𝐷 ∂𝑡 → κ ϵ ∇ ⋅ 𝐷 + 𝑑 𝑑𝑡 ∇ ⋅ 𝐷 =0 κ ε 𝐷 + 𝑑 𝐷 𝑑𝑡 =0→ 𝐷 = 𝐷 0 e −𝑡 𝑇 ε , 𝑇 ε = ε κ JUAS 2013: Electromagnetism Exercises H. Henke

Tut-Ex 3 Given is a 1-dimensional planar diode. What is the current density at saturation? 𝑃𝑜𝑖𝑠𝑠𝑜𝑛𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑑 2 Φ 𝑧 𝑑𝑧 2 = 𝑒 𝑛 𝑒 𝑧 ϵ 0 𝑎𝑡𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛: 𝐸 𝑧=0 = 𝑑ϕ 𝑑𝑧 =0 JUAS 2013: Electromagnetism Exercises H. Henke

𝑘𝑖𝑛𝑒𝑡𝑖𝑐𝑒𝑛𝑒𝑟𝑔𝑦:𝑒Φ 𝑧 = 1 2 𝑚 0 𝑣 2 𝑧 𝑘𝑖𝑛𝑒𝑡𝑖𝑐𝑒𝑛𝑒𝑟𝑔𝑦:𝑒Φ 𝑧 = 1 2 𝑚 0 𝑣 2 𝑧 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑑𝑒𝑛𝑠𝑖𝑡𝑦:𝐽=𝑒 𝑛 𝑒 𝑒 𝑣 𝑧 =𝑒 𝑛 𝑒 𝑧 2𝑒Φ 𝑧 𝑚 0 𝑃𝑜𝑖𝑠𝑠𝑜𝑛𝑒𝑞𝑢.: 𝑑 2 Φ 𝑑𝑧 2 = 𝑚 0 2𝑒Φ 𝐽 ϵ 0 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔𝑤𝑖𝑡ℎ2𝑑 Φ 𝑑𝑧 and𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑑Φ 𝑑𝑧 2 = 4 ϵ 0 𝐽 𝑚 0 2𝑒 Φ +𝐶 JUAS 2013: Electromagnetism Exercises H. Henke

𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑡𝑧=0: Φ=0,𝐸=𝑑 Φ 𝑑𝑧 =0→𝐶=0 𝑑Φ 𝑑𝑧 =2 𝐽 ϵ 0 1 2 𝑚 0 2𝑒 Φ 1 4 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑡𝑧=0: Φ=0,𝐸=𝑑 Φ 𝑑𝑧 =0→𝐶=0 𝑑Φ 𝑑𝑧 =2 𝐽 ϵ 0 1 2 𝑚 0 2𝑒 Φ 1 4 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛: 4 3 Φ 3 4 =2 𝐽 ϵ 0 1 2 𝑚 0 2𝑒 1 4 𝑧+𝐶 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠:Φ 𝑧=0 =0,Φ 𝑧=𝑑 = 𝑉 0 𝐽= 4 9 ϵ 0 2 𝑒 𝑚 0 𝑉 0 3 2 𝑑 2 𝐶ℎ𝑖𝑙𝑑−𝐿𝑎𝑛𝑔𝑚𝑢𝑖𝑟𝑙𝑎𝑤 JUAS 2013: Electromagnetism Exercises H. Henke

Derive the magnetic vector potential for a given current density. Tut-Ex 4 Derive the magnetic vector potential for a given current density. ∇ ⋅ 𝐵 =0→ 𝐵 = ∇ × 𝐴 ∇ × 𝐵 = ∇ ∇ ⋅ 𝐴 − ∇ 2 𝐴 =μ 𝐽 ∇ ⋅ 𝐴 =0→ ∇ 2 𝐴 =−μ 𝐽 ∇ 2 𝐴 𝑖 =−μ 𝐽 𝑖 ,𝑖=𝑥,𝑦,𝑧 1 𝑃𝑜𝑖𝑠𝑠𝑜𝑛𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: ∇ 2 Φ=− ρ ϵ 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙𝑜𝑓𝑝𝑜𝑖𝑛𝑡𝑐ℎ𝑎𝑟𝑔𝑒:Φ= 𝑞 4πϵ𝑟 JUAS 2013: Electromagnetism Exercises H. Henke

𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙𝑜𝑓𝑐ℎ𝑎𝑟𝑔𝑒𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛: Φ 𝑟 = 1 4πϵ ρ 𝑟 ′ ∣ 𝑟 − 𝑟 ′∣ 𝑑𝑉′ 2 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙𝑜𝑓𝑐ℎ𝑎𝑟𝑔𝑒𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛: Φ 𝑟 = 1 4πϵ ρ 𝑟 ′ ∣ 𝑟 − 𝑟 ′∣ 𝑑𝑉′ 2 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠𝑜𝑓 1 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑎𝑠 2 Φ→ 𝐴 𝑖 , 1 ϵ →μ,ρ→ 𝐽 𝑖 𝐴 𝑟 = μ 4π 𝐽 𝑟 ′ ∣ 𝑟 − 𝑟 ′∣ 𝑑𝑉′ JUAS 2013: Electromagnetism Exercises H. Henke

What is the end energy and the total accelerating voltage? Tut-Ex 5 Given is a non-relativistic cyclotron with a constant magnetic induction B and maximum radius R. What is the end energy and the total accelerating voltage? 𝑐𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙𝑓𝑜𝑟𝑐𝑒𝑒𝑞𝑢𝑎𝑙𝑠𝐿𝑜𝑟𝑒𝑛𝑡𝑧𝑓𝑜𝑟𝑐𝑒: 𝑚 0 𝑣 2 𝑟 =𝑞𝑣𝐵→ 𝑣 𝑚𝑎𝑥 = 𝑞𝐵 𝑚 0 𝑅 JUAS 2013: Electromagnetism Exercises H. Henke

Example : R=0.5m, B=1.5T, deuterium (p, n) with q=e, m0=3.34 10-27kg 𝑚𝑎𝑥𝑖𝑚𝑢𝑚𝑘𝑖𝑛𝑒𝑡𝑖𝑐𝑒𝑛𝑒𝑟𝑔𝑦: 𝐸 𝑘𝑖𝑛 = 1 2 𝑚 0 𝑣 𝑚𝑎𝑥 2 = 1 2 𝑞 2 𝑚 0 𝐵 2 𝑅 2 𝑡𝑜𝑡𝑎𝑙𝑒𝑥𝑝𝑒𝑟𝑖𝑒𝑛𝑐𝑒𝑑𝑣𝑜𝑙𝑡𝑎𝑔𝑒: 𝑉= 𝐸 𝑘𝑖𝑛 𝑞 = 1 2 𝑞 𝑚 0 𝐵 2 𝑅 2 𝑐𝑦𝑐𝑙𝑜𝑡𝑟𝑜𝑛𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦and𝑅𝐹−𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦: ω 𝐶 =2π𝑓= 2π 𝑇 = 𝑣 𝑟 = 𝑞𝐵 𝑚 0 , ω 𝑅𝐹 =2 ω 𝐶 Example : R=0.5m, B=1.5T, deuterium (p, n) with q=e, m0=3.34 10-27kg Ekin=13.5 MeV, fRF=23 MHz JUAS 2013: Electromagnetism Exercises H. Henke

Tut-Ex 6 A long dipole magnet is excited by a coil with n windings and current I0. Calculate the magnetic field in the air gap. 𝐻 ⋅𝑑 𝑠 = 𝐽 ⋅𝑑 𝐴 𝐻 𝑖 𝑙+ 𝐻 𝑜 𝑔=𝑛 𝐼 0 l n×I0 g 𝐵 𝑖 = 𝐵 𝑜 →μ 𝐻 𝑖 = μ 0 𝐻 𝑜 𝐻 𝑜 = 𝑛 𝐼 0 𝑔+ μ 0 𝑙 μ ≈ 𝑛 𝐼 0 𝑔 μ JUAS 2013: Electromagnetism Exercises H. Henke

What is the time-averaged radiated power density? Tut-Ex 7 Give the E- and H-field of a z-polarized plane wave which propagates in x-direction. What is the time-averaged radiated power density? 𝐸 = 𝐸 0 e 𝑖 ω𝑡−𝑘𝑥 𝑒 𝑧 ,𝑘= ω 𝑐 𝑍 𝐻 = 𝑒 𝑥 × 𝐸 =− 𝐸 0 e 𝑖 ω𝑡−𝑘𝑥 𝑒 𝑦 ,𝑍= μ ϵ 𝑆 𝑐 = 1 2 𝐸 × 𝐻 ∗ = 1 2Z ∣ 𝐸 0 ∣ 2 𝑒 𝑥 JUAS 2013: Electromagnetism Exercises H. Henke

What is the equation for the separation constants? Tut-Ex 8 Derive the longitudinal vector potential for TM-waves in a rectangular waveguide. What is the equation for the separation constants? ∂ 2 𝐴 𝑧 ∂ 𝑥 2 + ∂ 2 𝐴 𝑧 ∂ 𝑦 2 + ∂ 2 𝐴 𝑧 ∂ 𝑧 2 + 𝑘 2 𝐴 𝑧 =0,𝑘= ω 𝑐 𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖𝑎𝑛𝑠𝑎𝑡𝑧: 𝐴 𝑧 𝑥,𝑦,𝑧 =𝑋 𝑥 𝑌 𝑦 𝑍 𝑧 1 𝑋 𝑑 2 𝑋 𝑑𝑥 2 ⏟ − 𝑘 𝑥 2 + 1 𝑌 𝑑 2 𝑌 𝑑𝑦 2 ⏟ − 𝑘 𝑦 2 + 1 𝑍 𝑑 2 𝑍 𝑑𝑧 2 ⏟ − 𝑘 𝑧 2 + 𝑘 2 =0 𝐷𝑖𝑠𝑝𝑒𝑟𝑠𝑖𝑜𝑛𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑒𝑞𝑢.𝑜𝑓𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 : 𝑘 2 = 𝑘 𝑥 2 + 𝑘 𝑦 2 + 𝑘 𝑧 2 JUAS 2013: Electromagnetism Exercises H. Henke

𝑓𝑜𝑟𝑌 𝑦 ,𝑍 𝑧 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔𝑙𝑦. 𝑑 2 𝑋 𝑑𝑥 2 + 𝑘 𝑥 2 𝑋=0→𝑋= 𝐶 1 cos 𝑘 𝑥 𝑥 + 𝐶 2 sin 𝑘 𝑥 𝑥 = cos 𝑘 𝑥 𝑥 sin 𝑘 𝑥 𝑥 or e 𝑖𝑘 𝑥 𝑥 e − 𝑖𝑘 𝑥 𝑥 𝑓𝑜𝑟𝑌 𝑦 ,𝑍 𝑧 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔𝑙𝑦. 𝑔𝑒𝑛𝑒𝑟𝑎𝑙𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝐴 𝑧 𝑥,𝑦,𝑧 = cos 𝑘 𝑥 𝑥 sin 𝑘 𝑥 𝑥 cos 𝑘 𝑦 𝑦 sin 𝑘 𝑦 𝑦 e 𝑖𝑘 𝑧 𝑧 e − 𝑖𝑘 𝑧 𝑧 e 𝑖ω𝑡 𝑇𝑀−𝑤𝑎𝑣𝑒𝑠: 𝐻 = ∇ ×𝐴 𝑒 𝑧 𝐸 𝑧 =− 1 𝑖ωϵ 1 ρ ∂ ∂ρ ρ ∂ 𝐴 𝑧 ∂ρ + 1 ρ 2 ∂ 2 𝐴 𝑧 ∂ φ 2 = 𝑘 𝑥 2 + 𝑘 𝑦 2 𝑖ωϵ 𝐴 𝑧 ∼ 𝐴 𝑧 JUAS 2013: Electromagnetism Exercises H. Henke

𝐸 𝑧 𝑥=0,𝑎 = 𝐸 𝑧 𝑦=0,𝑏 =0: 𝐴 𝑧 =𝐶𝑠𝑖𝑛 𝑘 𝑥 𝑥 sin 𝑘 𝑦 𝑦 e 𝑖 ω𝑡− 𝑘 𝑧 𝑧 , 𝑘 𝑥 =𝑚 π 𝑎 , 𝑘 𝑦 =𝑛 π 𝑏 𝑘 2 = ω 𝑐 2 = 𝑚 π 𝑎 2 + 𝑛 π 𝑏 2 + 𝑘 𝑧 2 Tut-Ex 9 Give the longitudinal wavelength and phase and group velocity of a TE10-mode in a rectangular waveguide. 𝑚=1,𝑛=0: 𝑘 𝑧 = 2π λ 𝑧 = 𝑘 2 − π 𝑎 2 → λ 𝑧 = λ 0 1− λ 0 2a 2 JUAS 2013: Electromagnetism Exercises H. Henke

E, H y TE10-mode b a x 𝑣 𝑝ℎ = ω β = ω 𝑘 𝑧 = 𝑐 1− λ 0 2 𝑎 2 >𝑐 𝑣 𝑝ℎ = ω β = ω 𝑘 𝑧 = 𝑐 1− λ 0 2 𝑎 2 >𝑐 𝑣 𝑔 = ∂ω ∂β = ∂ω ∂ 𝑘 𝑧 =𝑐 1− λ 0 2 𝑎 2 <𝑐 → 𝑣 𝑝ℎ 𝑣 𝑔 = 𝑐 2 y TE10-mode b E, H a x JUAS 2013: Electromagnetism Exercises H. Henke

What is the lowest mode in a circular waveguide? Tut-Ex 10 What is the lowest mode in a circular waveguide? Show the field pattern. In which frequency range is mono-mode operation possible? 𝑇𝐸 11 −𝑚𝑜𝑑𝑒: 𝐾 11 = 𝑗 11 ′ 𝑎 = 𝑘 2 − 𝑘 𝑧 2 , 𝑗 11 ′=1.841 𝑇𝑀 01 −𝑚𝑜𝑑𝑒: 𝑘 𝑐 = 𝑗 01 𝑎 , 𝑗 01 =2.405 𝑓 𝑐11 = 𝑗 11 ′ 𝑐 2 π𝑎→ 𝑓 𝐶01 = 𝑗 01 𝑐 2 π𝑎 JUAS 2013: Electromagnetism Exercises H. Henke

JUAS 2013: Electromagnetism Exercises H. Henke

Tut-Ex 11 Calculate the accelerating voltage, shunt impedance and R-upon-Q of a TM010-mode pill-box cavity. 𝑜𝑛𝑎𝑥𝑖𝑠: 𝐸 𝑧 =−𝑖 2 ωϵ 𝑗 01 𝑎 2 𝐷 010 e 𝑖ω𝑡 𝑉 𝑚 = ∣ 0 𝑔 𝑎 𝑚 𝑒 𝑚 ⋅ 𝑒 𝑧 e 𝑖ω𝑡 𝑑𝑧∣ = ∣ 0 𝑔 𝐸 𝑧 e 𝑖ω 𝑧 𝑣 𝑑𝑧∣ ,𝑧=𝑣𝑡 = 2 ωϵ 𝑗 01 𝑎 2 𝐷 010 𝑔𝑇,𝑇= sin ω 𝑔 2v ω 𝑔 2v JUAS 2013: Electromagnetism Exercises H. Henke

𝑅 𝑠ℎ = 𝑉 2 𝑃 𝑑 = 4 π 𝑍 δ 𝑆 𝑔 2 𝑎+𝑔 𝑇 2 𝑗 01 𝐽 1 2 𝑗 01 𝑃 𝑑 = 4π κ δ 𝑠 𝑗 01 2 1+ 𝑔 𝑎 ∣ 𝐷 010 ∣ 2 𝐽 1 2 𝑗 01 𝑅 𝑠ℎ = 𝑉 2 𝑃 𝑑 = 4 π 𝑍 δ 𝑆 𝑔 2 𝑎+𝑔 𝑇 2 𝑗 01 𝐽 1 2 𝑗 01 𝑊 = 2π𝑔 ω 2 ϵ 𝑗 01 4 𝑎 2 ∣ 𝐷 010 ∣ 2 𝐽 1 2 𝑗 01 𝑅 𝑠ℎ 𝑄 0 = 𝑉 2 ω 𝑊 = 2 π 𝑍 𝑔 𝑎 𝑇 2 𝑗 01 𝐽 1 2 𝑗 𝑜1 JUAS 2013: Electromagnetism Exercises H. Henke