---
title: "Maxwell's equations in phasor form"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:maxwell-equations-phasor-free-space"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/maxwell-equations-phasor-free-space
site: BCT Engineering
---

# Maxwell's equations in phasor form

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2076 Chaitra, 2 marks; 2075 Asoj, 2 marks (asked 2 times)
- Video: [Maxwell's Equations in Phasor Form](https://www.youtube.com/watch?v=HE8gz5H8nBA) — RTU Wallah
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/maxwell-equations-phasor-free-space
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Write Maxwell's equations in phasor form (for the time varying case in free space / a perfect dielectric).

## Short answer

In phasor form for free space: ∇×E_s = −jωμH_s, ∇×H_s = jωεE_s, ∇·D_s = ρ_vs, ∇·B_s = 0.

## Answer

For time-harmonic fields varying as e^(jωt), replace ∂/∂t by jω. In a perfect dielectric (σ = 0) or free space there is no conduction current (J = 0). Allowing a volume charge density ρ_v, and using D = εE and B = μH:

| Point form | Name |
|---|---|
| ∇×E_s = −jωμH_s | Faraday's law |
| ∇×H_s = jωεE_s | Ampere's law (with displacement current) |
| ∇·D_s = ρ_vs  (∇·E_s = ρ_vs/ε) | Gauss's law |
| ∇·B_s = 0  (∇·H_s = 0) | No magnetic monopoles |

In a source-free region, ρ_vs = 0 so ∇·E_s = 0. For a conducting medium, the Ampere equation becomes ∇×H_s = (σ + jωε)E_s.

Here E_s and H_s are the phasors (functions of position only), related to the instantaneous fields by E = Re[E_s e^(jωt)].

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In phasor form for e^(jωt) fields, ∂/∂t is replaced by:
   A. jω
   B. −ω
   C. 1/(jω)
   D. ω²
   Answer: A. Differentiation in time becomes multiplication by jω. (Key point: Phasor substitution)

2. Faraday's law in phasor form is:
   A. ∇×E_s = jωμH_s
   B. ∇×E_s = −jωμH_s
   C. ∇×E_s = −jωεH_s
   D. ∇×E_s = 0
   Answer: B. ∇×E = −μ∂H/∂t becomes ∇×E_s = −jωμH_s. (Key point: Faraday's law)

3. In a conducting medium, Ampere's law becomes:
   A. ∇×H_s = jωεE_s
   B. ∇×H_s = σE_s
   C. ∇×H_s = (σ + jωε)E_s
   D. ∇×H_s = −jωμE_s
   Answer: C. Conduction current σE adds to displacement current jωεE. (Key point: Ampere's law in conductor)

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