GATE EC Electromagnetics Formula Sheet

Organised by the official GATE 2027 EC syllabus, Section 8: Electromagnetics. Each worked example was checked in code. Practise these: GATE EC Electromagnetics previous year questions.

Maxwell's equations

Syllabus: Differential and integral forms and their interpretation, boundary conditions, wave equation, Poynting vector.

Maxwell's equations and displacement current

∇⋅D=ρv,∇⋅B=0,∇×E=−∂B∂t,∇×H=J+∂D∂t\nabla \cdot \mathbf{D} = \rho_v,\quad \nabla \cdot \mathbf{B} = 0,\quad \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t},\quad \nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}
∇⋅D=ρv,∇⋅B=0,∇×E=−∂B∂t,∇×H=J+∂D∂t\begin{aligned} \nabla \cdot \mathbf{D} &= \rho_v, \\ \nabla \cdot \mathbf{B} &= 0, \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t}, \\ \nabla \times \mathbf{H} &= \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t} \end{aligned}
Symbols
  • D=εE\mathbf{D} = \varepsilon\mathbf{E}: electric flux density (C/m²)
  • B=μH\mathbf{B} = \mu\mathbf{H}: magnetic flux density (T)
  • ∂D/∂t\partial\mathbf{D}/\partial t: displacement current density (A/m²)
Applies when
Linear, isotropic media. In phasor form ∂/∂t becomes jω, so the displacement current density is jωεE.
Common mistake
Forgetting the displacement current term, which is what lets current 'flow' through a capacitor.
Worked example
E = 10 sin(2π × 10⁹ t) V/m in free space: peak displacement current density: 0.556 A/m²

Boundary conditions

Et1=Et2,Dn1−Dn2=ρs,Ht1−Ht2=K,Bn1=Bn2E_{t1} = E_{t2}, \qquad D_{n1} - D_{n2} = \rho_s, \qquad H_{t1} - H_{t2} = K, \qquad B_{n1} = B_{n2}
Et1=Et2,Dn1−Dn2=ρs,Ht1−Ht2=K,Bn1=Bn2\begin{aligned} E_{t1} &= E_{t2}, \\ D_{n1} - D_{n2} &= \rho_s, \\ H_{t1} - H_{t2} &= K, \\ B_{n1} &= B_{n2} \end{aligned}
Symbols
  • Et,HtE_t, H_t: tangential components
  • Dn,BnD_n, B_n: normal components
  • ρs,K\rho_s, K: surface charge (C/m²) and surface current (A/m)
Applies when
At an interface between two media. For a charge-free dielectric interface, tan θ₁ / tan θ₂ = ε₁ / ε₂ for the field lines.
Common mistake
Making the normal E continuous: it is the normal D (= εE) that is continuous without surface charge.
Worked example
E meets a boundary from εr = 1 into εr = 4 at 45° to the normal: the angle in the second medium: 75.96 ° from the normal

Poynting vector

S=E×H,Savg=∣E0∣22η\mathbf{S} = \mathbf{E} \times \mathbf{H}, \qquad S_{avg} = \frac{|E_0|^2}{2\eta}
S=E×H,Savg=∣E0∣22η\begin{aligned} \mathbf{S} &= \mathbf{E} \times \mathbf{H}, \\ S_{avg} &= \frac{|E_0|^2}{2\eta} \end{aligned}
Symbols
  • E0E_0: peak electric field amplitude (V/m)
  • η\eta: intrinsic impedance, 377 Ω in free space
Applies when
The time average of a uniform plane wave in a lossless medium. With RMS values there is no factor of 2.
Common mistake
Using E₀²/η with a peak amplitude: the average needs the ½.
Worked example
E0E_{0} = 100 V/m in free space: 13.27 W/m²

Plane waves and properties

Syllabus: Reflection and refraction, polarization, phase and group velocity, propagation through various media, skin depth.

Wave parameters

β=ωμε,vp=1με=cμrεr,η=με=377εr Ω (μr=1)\beta = \omega\sqrt{\mu\varepsilon}, \qquad v_p = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{\sqrt{\mu_r\varepsilon_r}}, \qquad \eta = \sqrt{\frac{\mu}{\varepsilon}} = \frac{377}{\sqrt{\varepsilon_r}}\ \Omega\ (\mu_r = 1)
β=ωμε,vp=1με=cμrεr,η=με=377εr Ω (μr=1)\begin{aligned} \beta &= \omega\sqrt{\mu\varepsilon}, \\ v_p &= \frac{1}{\sqrt{\mu\varepsilon}} \\ &= \frac{c}{\sqrt{\mu_r\varepsilon_r}}, \\ \eta &= \sqrt{\frac{\mu}{\varepsilon}} \\ &= \frac{377}{\sqrt{\varepsilon_r}}\ \Omega\ (\mu_r = 1) \end{aligned}
Symbols
  • β\beta: phase constant (rad/m)
  • η\eta: intrinsic impedance (Ω)
Applies when
Lossless, non-magnetic medium for the last form. A lossless medium with constant εr\varepsilon_r is non-dispersive, so vgv_{g} = vpv_{p}.
Common mistake
Dividing 377 by εr\varepsilon_r instead of its square root.
Worked example
εr = 4, µr = 1: intrinsic impedance: 188.4 Ω

Normal-incidence reflection

Γ=η2−η1η2+η1,τ=2η2η2+η1=1+Γ\Gamma = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}, \qquad \tau = \frac{2\eta_2}{\eta_2 + \eta_1} = 1 + \Gamma
Γ=η2−η1η2+η1,τ=2η2η2+η1=1+Γ\begin{aligned} \Gamma &= \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}, \\ \tau &= \frac{2\eta_2}{\eta_2 + \eta_1} \\ &= 1 + \Gamma \end{aligned}
Symbols
  • Γ,τ\Gamma, \tau: reflection and transmission coefficients for E
  • η1,η2\eta_1, \eta_2: intrinsic impedances, incident side first
Applies when
A plane wave hitting a plane boundary head-on. The reflected power fraction is |Γ|².
Common mistake
Taking τ = 1 − Γ: the tangential E is continuous, so 1 + Γ = τ.
Worked example
From air into εr = 4: -0.333 (11.1% of the power reflects)

Snell's law, Brewster and critical angles

n1sin⁡θ1=n2sin⁡θ2,tan⁡θB=n2n1,sin⁡θc=n2n1 (n1>n2)n_1\sin\theta_1 = n_2\sin\theta_2, \qquad \tan\theta_B = \frac{n_2}{n_1}, \qquad \sin\theta_c = \frac{n_2}{n_1}\ (n_1 > n_2)
n1sin⁡θ1=n2sin⁡θ2,tan⁡θB=n2n1,sin⁡θc=n2n1 (n1>n2)\begin{aligned} n_1\sin\theta_1 &= n_2\sin\theta_2, \\ \tan\theta_B &= \frac{n_2}{n_1}, \\ \sin\theta_c &= \frac{n_2}{n_1}\ (n_1 > n_2) \end{aligned}
Symbols
  • n=εrn = \sqrt{\varepsilon_r}: refractive index
  • θB\theta_B: Brewster angle: no reflection for parallel polarisation
  • θc\theta_c: critical angle for total internal reflection
Applies when
Non-magnetic dielectrics. The Brewster angle exists only for parallel (TM) polarisation.
Common mistake
Using the ratio of permittivities directly in tan⁡θB\tan\theta_B: it is the ratio of the refractive indices, the square root.
Worked example
Air to εr = 4: the Brewster angle: 63.43 °

Skin depth

δ=1πfμσ,fields fall as e−z/δ\delta = \frac{1}{\sqrt{\pi f \mu \sigma}}, \qquad \text{fields fall as } e^{-z/\delta}
δ=1πfμσ,fields fall as e−z/δ\begin{aligned} \delta &= \frac{1}{\sqrt{\pi f \mu \sigma}}, \\ \text{fields fall as } e^{-z/\delta} \end{aligned}
Symbols
  • σ\sigma: conductivity (S/m)
  • ff: frequency (Hz)
Applies when
Good conductors, σ ≫ ωε. The field falls to 1/e (37%) in one skin depth.
Common mistake
Using ω in place of πf, which is off by √2.
Worked example
Copper, σ = 5.8 × 10⁷ S/m, at 1 MHz: 66.1 µm

Polarisation sense

E=x^Excos⁡(ωt−βz)+y^Eycos⁡(ωt−βz+ϕ):0<ϕ<π⇒left-handed, −π<ϕ<0⇒right-handed\mathbf{E} = \hat{x}E_x\cos(\omega t - \beta z) + \hat{y}E_y\cos(\omega t - \beta z + \phi):\quad 0 < \phi < \pi \Rightarrow \text{left-handed},\ -\pi < \phi < 0 \Rightarrow \text{right-handed}
E=x^Excos⁡(ωt−βz)+ y^Eycos⁡(ωt−βz+ϕ)0<ϕ<π:left-handed−π<ϕ<0:right-handed\begin{gathered} \mathbf{E} = \hat{x}E_x\cos(\omega t - \beta z) \\ +\, \hat{y}E_y\cos(\omega t - \beta z + \phi) \\ 0 < \phi < \pi: \text{left-handed} \\ -\pi < \phi < 0: \text{right-handed} \end{gathered}
Symbols
  • ϕ\phi: phase of EyE_{y} relative to ExE_{x}
Applies when
Wave travelling along +z, IEEE convention (judged looking along the direction of travel). φ = 0 or π is linear; |ExE_{x}| = |EyE_{y}| with φ = ±π/2 is circular.
Common mistake
Forgetting to flip the rule for a wave travelling along −z.
Worked example
ExE_{x} = 2, EyE_{y} = 1, φ = π/2: the axial ratio: 2 (left-handed elliptical)

Transmission lines

Syllabus: Equations, characteristic impedance, impedance matching, impedance transformation, S-parameters, Smith chart.

Characteristic impedance

Z0=R+jωLG+jωC  →  LC (lossless),vp=1LCZ_0 = \sqrt{\frac{R + j\omega L}{G + j\omega C}} \;\to\; \sqrt{\frac{L}{C}}\ (\text{lossless}), \qquad v_p = \frac{1}{\sqrt{LC}}
Z0=R+jωLG+jωC  →  LC (lossless),vp=1LC\begin{aligned} Z_0 &= \sqrt{\frac{R + j\omega L}{G + j\omega C}} \; \\ &\to\; \sqrt{\frac{L}{C}}\ (\text{lossless}), \\ v_p &= \frac{1}{\sqrt{LC}} \end{aligned}
Symbols
  • R,L,G,CR, L, G, C: per-unit-length line constants
Applies when
Lossless or low-loss lines for the simple form.
Common mistake
Using total inductance and capacitance of a length of line: Z₀ uses per-unit-length values, and it doesn't depend on length.
Worked example
L = 250 nH/m, C = 100 pF/m: 50 Ω (vpv_{p} = 2 × 10⁸ m/s)

Input impedance of a line

Zin=Z0 ZL+jZ0tan⁡βlZ0+jZLtan⁡βl,λ4: Zin=Z02ZL,λ2: Zin=ZLZ_{in} = Z_0\,\frac{Z_L + jZ_0\tan\beta l}{Z_0 + jZ_L\tan\beta l}, \qquad \frac{\lambda}{4}:\ Z_{in} = \frac{Z_0^2}{Z_L}, \qquad \frac{\lambda}{2}:\ Z_{in} = Z_L
Zin=Z0 ZL+jZ0tan⁡βlZ0+jZLtan⁡βl,λ4: Zin=Z02ZL,λ2: Zin=ZL\begin{aligned} Z_{in} &= Z_0\,\frac{Z_L + jZ_0\tan\beta l}{Z_0 + jZ_L\tan\beta l}, \\ \frac{\lambda}{4}:\ Z_{in} &= \frac{Z_0^2}{Z_L}, \\ \frac{\lambda}{2}:\ Z_{in} &= Z_L \end{aligned}
Symbols
  • βl\beta l: electrical length (rad)
  • ZLZ_L: load impedance (Ω)
Applies when
Lossless line. A shorted line gives jZ₀ tan βl, an open one −jZ₀ cot βl.
Common mistake
Measuring l from the generator: it is the distance from the load.
Worked example
A λ/4 line of 50 Ω ending in 100 Ω: 25 Ω

Reflection coefficient and VSWR

ΓL=ZL−Z0ZL+Z0,VSWR=1+∣Γ∣1−∣Γ∣,∣Γ∣=VSWR−1VSWR+1\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0}, \qquad \text{VSWR} = \frac{1 + |\Gamma|}{1 - |\Gamma|}, \qquad |\Gamma| = \frac{\text{VSWR} - 1}{\text{VSWR} + 1}
ΓL=ZL−Z0ZL+Z0,VSWR=1+∣Γ∣1−∣Γ∣,∣Γ∣=VSWR−1VSWR+1\begin{aligned} \Gamma_L &= \frac{Z_L - Z_0}{Z_L + Z_0}, \\ \text{VSWR} &= \frac{1 + |\Gamma|}{1 - |\Gamma|}, \\ |\Gamma| &= \frac{\text{VSWR} - 1}{\text{VSWR} + 1} \end{aligned}
Symbols
  • ΓL\Gamma_L: load reflection coefficient
Applies when
Lossless line, so |Γ| is the same everywhere on it. Voltage minima are λ/2 apart, with a maximum λ/4 from each.
Common mistake
Reading a VSWR below 1: it is always 1 or more.
Worked example
100 Ω on a 50 Ω line: 2 (Γ = 1/3)

Quarter-wave transformer

Zq=Z0ZLZ_q = \sqrt{Z_0 Z_L}
Symbols
  • ZqZ_q: characteristic impedance of the λ/4 section (Ω)
Applies when
Real load, at the design frequency only. A complex load first needs a length of line to make it real.
Common mistake
Using the arithmetic mean of the two impedances.
Worked example
Matching 200 Ω to 50 Ω: 100 Ω

Waveguides, optical fibres and antennas

Syllabus: Rectangular and circular waveguides, light propagation in optical fibers, dipole and monopole antennas, linear antenna arrays.

Rectangular waveguide

fc=c2(ma)2+(nb)2,vp=c1−(fc/f)2,vpvg=c2f_c = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}, \qquad v_p = \frac{c}{\sqrt{1 - (f_c/f)^2}}, \qquad v_p v_g = c^2
fc=c2(ma)2+(nb)2,vp=c1−(fc/f)2,vpvg=c2\begin{aligned} f_c &= \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}, \\ v_p &= \frac{c}{\sqrt{1 - (f_c/f)^2}}, \\ v_p v_g &= c^2 \end{aligned}
Symbols
  • a,ba, b: inner width and height, a > b (m)
  • m,nm, n: mode indices; TE₁₀ is the dominant mode
Applies when
Air-filled guide. Only modes with fcf_{c} below the operating frequency propagate.
Common mistake
Expecting vpv_{p} ≤ c inside a guide: the phase velocity exceeds c, and the group velocity is below it.
Worked example
WR-90 guide, a = 2.286 cm: TE₁₀ cut-off: 6.56 GHz

Optical fibre numerical aperture

NA=sin⁡θa=n12−n22,V=2πaλNA (single mode if V<2.405)NA = \sin\theta_a = \sqrt{n_1^2 - n_2^2}, \qquad V = \frac{2\pi a}{\lambda} NA \ (\text{single mode if } V < 2.405)
NA=sin⁡θa=n12−n22V=2πaλNA(single mode if V<2.405)\begin{gathered} NA = \sin\theta_a = \sqrt{n_1^2 - n_2^2} \\ V = \frac{2\pi a}{\lambda} NA \\ \text{(single mode if } V < 2.405) \end{gathered}
Symbols
  • n1,n2n_1, n_2: core and cladding indices
  • θa\theta_a: acceptance half-angle in air
  • aa: core radius
Applies when
Step-index fibre, light entering from air.
Common mistake
Using n₁ − n₂ without the squares.
Worked example
n1n_{1} = 1.48, n2n_{2} = 1.46: 0.2425

Dipole radiation resistance

Rr=80π2(dlλ)2 (short dipole),Rr≈73 Ω (λ/2),Rr≈36.5 Ω (λ/4 monopole)R_r = 80\pi^2\left(\frac{dl}{\lambda}\right)^2\ (\text{short dipole}), \qquad R_r \approx 73\ \Omega\ (\lambda/2), \qquad R_r \approx 36.5\ \Omega\ (\lambda/4 \text{ monopole})
Short dipole:Rr=80π2(dlλ)2λ/2 dipole: Rr≈73 Ωλ/4 monopole: Rr≈36.5 Ω\begin{gathered} \text{Short dipole:} \\ R_r = 80\pi^2\left(\frac{dl}{\lambda}\right)^2 \\ \lambda/2 \text{ dipole: } R_r \approx 73\ \Omega \\ \lambda/4 \text{ monopole: } R_r \approx 36.5\ \Omega \end{gathered}
Symbols
  • dldl: length of a short (Hertzian) dipole, dl ≪ λ
Applies when
Free space, uniform current for the Hertzian dipole. A monopole over a ground plane has half the dipole's radiation resistance.
Common mistake
Using the Hertzian formula for a half-wave dipole, whose current isn't uniform.
Worked example
dl = λ/50: 0.316 Ω

Uniform linear array

AF(ψ)=sin⁡(Nψ/2)Nsin⁡(ψ/2),ψ=βdcos⁡θ+αAF(\psi) = \frac{\sin(N\psi/2)}{N\sin(\psi/2)}, \qquad \psi = \beta d\cos\theta + \alpha
AF(ψ)=sin⁡(Nψ/2)Nsin⁡(ψ/2),ψ=βdcos⁡θ+α\begin{aligned} AF(\psi) &= \frac{\sin(N\psi/2)}{N\sin(\psi/2)}, \\ \psi &= \beta d\cos\theta + \alpha \end{aligned}
Symbols
  • NN: number of elements
  • dd: element spacing
  • α\alpha: progressive phase shift
  • θ\theta: angle from the array axis
Applies when
Identical, equally spaced, equal-amplitude elements. α = 0 gives a broadside beam; α = −βd gives end-fire.
Common mistake
Measuring θ from broadside when the formula measures it from the array axis.
Worked example
N = 4, d = λ/2, broadside: the first null from the axis: 60 ° (cos θ = 1/2)

Practise these: GATE EC Electromagnetics previous year questions

Source for the section headings: the official GATE 2027 EC syllabus, IIT Madras. The formulas, examples and notes are Studyloaf's own.

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