- 2067 Mangsir · 6 marks
Define TM mode; TM11 cut-off 5×3 cm
Short answer
Transverse magnetic (TM) mode has no longitudinal magnetic field (H_z=0). Using f_c=(c/2)√((m/a)²+(n/b)²) with m=n=1, a=5 cm, b=3 cm gives f_c≈5.83 GHz.
Define transverse magnetic mode. A rectangular waveguide has dimensions a = 5 cm and b = 3 cm. The medium within the waveguide has ε_r = 1, μ_r = 1, σ = 0 and conducting walls of the waveguide are perfect conductors. Determine the cutoff frequency for TM11 mode.
Transverse magnetic (TM) mode
A TM mode is a wave mode in which the magnetic field has no component along the direction of propagation (H_z = 0), while the electric field has a longitudinal component E_z ≠ 0. In a rectangular guide
E_z = E₀ sin(mπx/a) sin(nπy/b) e^{−jβz}
with m ≥ 1 and n ≥ 1 (if either is zero all the field components vanish). The lowest TM mode is TM₁₁.
Cut-off of TM₁₁ (a = 5 cm, b = 3 cm, air-filled, perfect walls)
f_c = (c/2) √[(m/a)² + (n/b)²]
(1/a)² = (1/0.05)² = 400
(1/b)² = (1/0.03)² = 1111.1
sum = 1511.1, √ = 38.87 m⁻¹
f_c(TM₁₁) = 1.5×10⁸ × 38.87 = 5.83×10⁹ Hz
Answer: f_c(TM₁₁) = 5.83 GHz (λ_c = 5.14 cm). For comparison the dominant TE₁₀ has f_c = 3 GHz.
✓ Answer checked by Claude Opus 5.5 (high effort)
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