--- title: "Cross-field magnetron with 60° phase difference" subject: "RF and Microwave Engineering" chapter: 5 key: "semester_7/rf_and_microwave_engineering:crossfield-magnetron-60-degree" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/crossfield-magnetron-60-degree site: BCT Engineering --- # Cross-field magnetron with 60° phase difference RF and Microwave Engineering · Chapter 5: Microwave Generators - Asked in: 2081 Baisakh, 8 marks - Video: [Magnetron zero mode, Magnetron pi mode, Magnetron mode jumping & strapping of Magnetron](https://www.youtube.com/watch?v=ty8H9_OGKsE) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/crossfield-magnetron-60-degree - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-5 ## Question Explain the working principles of a cross-field magnetron having 60° of phase differences among the adjacent cavities. ## Short answer With 60° between adjacent cavities, N=6n (e.g., 6 cavities, n=1). Radial E and axial B above Hull cutoff cause E×B drift. By Hartree condition, drift velocity matches the rotating RF wave; phase focusing gathers electrons into one spoke that continuously transfers energy to the cavities, generating microwave oscillations. ## Answer A **cross-field (cavity) magnetron** is an M-type tube in which a radial DC electric field and an axial DC magnetic field act at right angles on the electrons. The electrons give their potential energy to a travelling RF wave carried by the cavities of the anode. **Mode and number of cavities for 60° phase difference** For an anode with N cavities, the phase difference between adjacent cavities in mode n is: ```text φ = 2πn / N 60° = 360°·n / N => N = 6n N = 6 -> n = 1 (one spoke) N = 12 -> n = 2 (two spokes) ``` So a 6-cavity anode in its n = 1 mode (or a 12-cavity anode in its n = 2 mode) has 60° between adjacent cavities. The RF field pattern repeats every 360°/60° = 6 cavities. The number of electron spokes equals n. (The usual π-mode has 180° between cavities and N/2 spokes.) ```text 6-cavity anode, 60° between cavities 0° 300° _ 60° \ / \ / ----( cath )---- B0 axial (out of page) / \_/ \ E radial 240° 120° 180° one spoke rotating at ω (angular) ``` **Working principle** 1. The heated cathode emits electrons. The anode voltage V₀ gives a radial E field. An axial B₀ above the Hull cut-off value, B₀c = √(8mV₀/e) / [b(1 − a²/b²)], bends the electron paths so that, without RF, they curve back to the cathode. 2. Noise excites the cavities. In this mode each cavity's fringing field is 60° behind its neighbour, so the RF pattern forms a wave that rotates around the cathode at angular velocity ω/n. 3. The DC fields make electrons drift around the cathode at the E × B speed. V₀ and B₀ are chosen (Hartree condition) so that this drift **matches the wave's angular velocity**. 4. **Phase focusing:** electrons that meet a tangential retarding RF field lose energy to the wave, slow down and move outward step by step to the anode. Electrons that meet an accelerating field gain energy and are returned to the cathode, which removes them early. 5. The useful electrons form **n rotating spokes** (one spoke for N = 6). Each spoke passes the cavity slots in step with the retarding phase, so energy flows into the cavities every cycle. The oscillation builds up and RF is taken out with a coupling loop. **Remarks** - The different modes lie close in frequency, so mode control is needed. **Strapping** and the rising-sun anode are designed mainly to hold the π-mode; for the 60° mode, V₀ and B₀ are set to the Hartree condition of that mode. - Efficiency is about 40–70%, because the electrons hit the anode with little leftover kinetic energy. - If an RF signal is injected into the cavity chain (as in a cross-field amplifier), the same interaction amplifies it. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The phase difference between adjacent cavities in mode n of an N-cavity magnetron is: A. φ = 2πn/N B. φ = πN/n C. φ = n/N D. φ = 2πN/n Answer: A. The total phase around the anode must be 2πn. (Key point: Phase relation) 2. For 60° between cavities, N = 6n. How many cavities and spokes for n = 1? A. N = 12, 2 spokes B. N = 6, 1 spoke C. N = 6, 3 spokes D. N = 3, 1 spoke Answer: B. 60° = 360°n/N gives N = 6n; the number of spokes equals n. (Key point: Cavity number) 3. What synchronism is chosen by V0 and B0? A. E×B drift matches the angular velocity of the rotating RF wave B. The cavities resonate at DC C. Zero drift D. Cathode speed matches sound Answer: A. This is the Hartree condition. (Key point: Synchronism) 4. Why does energy flow into the cavities? A. Spokes pass each slot when the field is retarding, so electrons lose potential energy to the wave B. Electrons are accelerated C. The magnet supplies it D. Anode heating Answer: A. Phase-focused electrons give energy every cycle. 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