Electronic Science and Technology Rabindra Kishore Mishra Lecture 5- Hertzian Dipole (Part 2) EST-E-335: Radiation System 3 Credits Electronic Science and Technology Rabindra Kishore Mishra Outline β’Recap β’Hertzian Dipole: Example β’Lines of Forces β’π¨ π = Electronic Science and Technology Rabindra Kishore Mishra Recap 1 1 β«Χ¬β¬ 4π π π π′ π −πππ π π3π′ β’π΅. π¨ = −πππππ β’π΅2π¨ π + π 2 π¨ π = −π π π •πΊ π = ππ₯ π −πππ 4ππ π π π′ πΊ(π • π¨ π =β«Χ¬β¬ − π′)π3π′ 1 β’π― = π΅ × π¨ π β’π¬ = 1 ππππ (π΅ π΅. π¨ + π 2 π¨) ππ· β’ππππ = , ππππ = −π΅. π· ππ‘ β’P(r)=pπΏ 3 π β’π(r)=ππππΏ 3 π , π(r)= − p.∇πΏ 3 π β’π¨ π = πππ0 π πΊ π π΅.π¨ β’π π = πππ0 π0 β’π― = −πππ × π΅πΊ π β’π¬ = 1 π0 π΅ π β π΅ + π2π πΊ π β’π― = ππ ππ + 1 Electronic Science and Technology Rabindra Kishore Mishra Recap 2 1 π (π × π)πΊ ΖΈ π 1 β’π¬ = π΅ × ππ ππ + (π × ΰ΅¬ πππ0 π π)πΊ ΖΈ π ΰ΅° β’π¬ = 1 π0 ππ + 2 1 ΖΈ −π 3πΖΈ πβπ π π π + πΖΈ × (π × π)πΊ ΖΈ π π0 πΊ π β’π―πππ π ππ₯ π −πππ π) = 4ππ β’π¬πππ π ππ₯ π −πππ πΖΈ ) 4ππ = β’π0 π―πππ = πΖΈ × π¬πππ π2 π0 π2 π0 π0 (πΖΈ × πΖΈ × (π × β’A historically important problem Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole: Example (1/4) Explicit expressions for the real-valued electric and magnetic fields of an oscillating z-directed dipole p(t)= p πΰ· cosωt. • considered first by Hertz in 1889 • H. Hertz, Electric Waves, Dover Publications, New York, 1962 • Available in google books: • https://books.google.co.in/books?id=8GkOAAAA IAAJ&q=Electric+Waves&dq=Electric+Waves&hl= en&sa=X&ved=2ahUKEwjG2uaLnvXqAhUXxzgGH bRyA_YQ6AEwAHoECAQQAg β’π― = ππ ππ Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole: Example (2/4) β’π¬ = 1 π0 1 +π ππ + 1 π (π × π)πΊ ΖΈ π ΖΈ −π 3πΖΈ πβπ π πΊ π + π2 π0 πΖΈ × (π × π)πΊ ΖΈ π β’ Replace • π = ππ§π ΖΈ πππ‘ ππ₯ π −πππ ππ₯ π −π(ππ−ππ‘) 1 × π πππ‘ = = cos ππ − ππ‘ 4ππ 4ππ 4ππ 1 π ππ ππ + = −ππ + π π π 1 1 ππ ππ + πΊ π, π‘ = cos ππ − ππ‘ − πsin ππ − ππ‘ −ππ π 4ππ 1 sin(ππ−ππ‘) π π(ππ ππ + πΊ π, π‘ ) = ππ −ππππ ππ − ππ‘ + π π 1 1 cos(ππ−ππ‘) π π( ππ + πΊ π, π‘ ) = π ππ ππ ππ − ππ‘ + π0 π π • πΊ π, π‘ = • • • • − πsin ππ − ππ‘ +π π π β’π¬ π = π ππ ππ ππ − ππ‘ Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole: Example (3/4) β’π― π = ππ −ππππ ππ − cos(ππ−ππ‘) 3π πβπ§ −π§ + π 4ππ0 π 2 sin(ππ−ππ‘) π§×π ππ‘ + π 4ππ β’Use: ΰ· β’π§ΖΈ = ππππ π ΖΈ − ππ πππ ΰ· β’π.ΖΈ π§ΖΈ = π.ΖΈ ππππ π ΖΈ − ππ πππ = πππ π ΰ· β’π§ΖΈ × πΖΈ = ππππ π ΖΈ − ππ πππ × πΖΈ = −ππ πππ ΰ· ΰ· β’πΖΈ × π§ΖΈ × πΖΈ = πΖΈ × −ππ πππ ΰ· = −ππ πππ + ππ 2 π×(π§×π) cos 4ππ0 π ππ − ππ‘ β’π¬π π = π ππ ππ ππ − ππ‘ Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole: Example (4/4) β’π¬π½ π = π ππ ππ ππ − ππ‘ ππ 2 π πππ cos 4ππ0 π cos(ππ−ππ‘) 2πππ π + π 4ππ0 π 2 cos(ππ−ππ‘) π πππ + π 4ππ0 π 2 − ππ − ππ‘ β’π―π π = ππ −ππππ ππ − ππ‘ + sin(ππ−ππ‘) π π πππ 4ππ β’Azimuthal symmetry in π Electronic Science and Technology Rabindra Kishore Mishra Lines of Forces (1/6) • Plane of interest π = 0 plane (i.e. xy-plane) β’E-field is tangential to its field lines • At any point βsmall displacement dr along the tangent to a line − parallel to E → dr×E = 0 • Significance β can be used to determine the lines ΰ·‘ πππ × πΰ· πΈπ + π½πΈ ΰ·‘ π β’ππ × π¬ = πΰ· ππ + π½ Electronic Science and Technology Rabindra Kishore Mishra Lines of Forces (2/6) ΰ· πππΈπ − ππππΈπ = 0 =π ππ ππΈπ ⇒ = ππ πΈπ • r is a function of θ βpolar representation of the line curve β’Solving the equation • Use dimensionless variables β’πΌ = ππ; π = ππ‘; −πΈ0 = Electronic Science and Technology Rabindra Kishore Mishra Lines of Forces (3/6) ππ 3 4ππ0 • (dimensionless variables) 2πππ π β’πΈπ = πΈ0 2 πΌ π πππ β’πΈπ = −πΈ0 πΌ π ππ πΌ − π πππ πΌ − π cos(πΌ−π) + πΌ cos(πΌ−π) sin(πΌ−π) − πΌ2 − πΌ β’Ψ πΌ = π ππ πΌ − π + Electronic Science and Technology Rabindra Kishore Mishra Lines of Forces (4/6) πΨ πΌ β’ ππΌ cos(πΌ−π) πΌ = Ψ′ πΌ = πππ πΌ − π 2πππ π β’πΈπ = πΈ0 2 Ψ πΌ πΌ π πππ β’πΈπ = −πΈ0 Ψ′ πΌ πΌ cos(πΌ−π) sin(πΌ−π) − πΌ2 − πΌ β’the equation for the lines in the variable α : Electronic Science and Technology Rabindra Kishore Mishra Lines of Forces (5/6) ππΌ = ππ ⇒ πΌπΈπ π ππ πΈπ = −2πππ‘π Ψ πΌ Ψ′ πΌ ππΨ πΌ π = −2πππ‘π = πππ ππ2 π ππ π ππ ππΨ πΌ π ππ2 π = 0 ⇒ Ψ πΌ π ππ2 π = π cos(πΌ − π) π ππ πΌ − π + π ππ2 π πΌ = απ ππ ππ − ππ‘ cos(ππ − ππ‘) + α π ππ2 π = π ππ β’Ideal method Electronic Science and Technology Rabindra Kishore Mishra Lines of Forces (6/6) • solve for r in terms of θ βclosed form impossible β’Alternative method • think of the lines as a contour plot at different values of the constant Q • Write a program for the contour plot