---
title: "Find J from H in cylindrical"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:find-j-from-h-cylindrical"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/find-j-from-h-cylindrical
site: BCT Engineering
---

# Find J from H in cylindrical

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2076 Chaitra, 3 marks
- Video: [EMF || Curl of H  to find Current Density-J || Numerical problem](https://www.youtube.com/watch?v=Ygy2_d7w3ww) — KG Reddy College Of Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/find-j-from-h-cylindrical
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Calculate the value of the vector current density J in cylindrical coordinates at P_B(1.5, 90°, 0.5) if H = (2/ρ)(cos 0.2φ) a_ρ A/m.

## Short answer

J = ∇×H. For H = (2/ρ) cos(0.2φ) a_ρ, only the z-component is nonzero: J_z = 0.4 sin(0.2φ)/ρ² A/m². At P_B(1.5, 90°, 0.5), sin(0.2×90°) ≈ 0.3090, so J_z = 0.0549 A/m². J = 0.055 a_z A/m².

## Answer

J is found from Ampere's law in point form, J = ∇×H. Given H = (2/ρ) cos(0.2φ) a_ρ A/m, so H_ρ = (2/ρ) cos 0.2φ and H_φ = H_z = 0.

#### Curl in cylindrical coordinates
```text
(∇×H)_ρ = (1/ρ) ∂H_z/∂φ − ∂H_φ/∂z = 0
(∇×H)_φ = ∂H_ρ/∂z − ∂H_z/∂ρ = 0
(∇×H)_z = (1/ρ)[∂(ρH_φ)/∂ρ − ∂H_ρ/∂φ] = −(1/ρ) ∂H_ρ/∂φ
```
Only the z-component survives:

```text
∂H_ρ/∂φ = (2/ρ)(−0.2 sin 0.2φ) = −0.4 sin(0.2φ)/ρ
J_z = −(1/ρ)(−0.4 sin 0.2φ / ρ) = 0.4 sin(0.2φ)/ρ²   A/m²
```

#### At P_B(ρ = 1.5, φ = 90°, z = 0.5)
φ = 90° = π/2 rad, so 0.2φ = 0.3142 rad and sin(0.3142) = 0.3090.

```text
J_z = 0.4 × 0.3090 / (1.5)² = 0.1236/2.25 = 0.0549 A/m²
```

**Answer: J = 0.055 a_z A/m²** (J_ρ = J_φ = 0).

## Check yourself

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

1. How is J found from H?
   A. J = ∇·H
   B. J = ∇×H
   C. J = ∂H/∂t
   D. J = H/σ
   Answer: B. Ampere's law in point form gives J = ∇×H. (Key point: Method)

2. With only H_ρ(ρ,φ) present, which component of ∇×H survives?
   A. ρ
   B. φ
   C. z: −(1/ρ)∂H_ρ/∂φ
   D. None
   Answer: C. Terms with z-derivatives vanish; only (1/ρ)(∂(ρH_φ)/∂ρ − ∂H_ρ/∂φ) remains, giving J_z = 0.4 sin(0.2φ)/ρ². (Key point: Surviving component)

3. At ρ = 1.5, φ = 90°, J is:
   A. 0.055 a_z A/m²
   B. 0.55 a_z A/m²
   C. 0.0549 a_ρ A/m²
   D. 0
   Answer: A. 0.4(0.3090)/2.25 = 0.0549 A/m² in the a_z direction. (Key point: Value)

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