---
title: "Total current through surface, H field"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:total-current-through-surface-h-rho"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/total-current-through-surface-h-rho
site: BCT Engineering
---

# Total current through surface, H field

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2070 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/total-current-through-surface-h-rho
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Given H = (2ρ + z) a_ρ + (2/z) a_z A/m, find the total current passing through the surface ρ = 2, π/4 < φ < π/2, 3 < z < 5, in the a_ρ direction.

## Short answer

Current is the flux of (∇×H)_ρ. For H = (2ρ+z)a_ρ + (2/z)a_z, (∇×H)_ρ = (1/ρ)∂H_z/∂φ - ∂H_φ/∂z = 0. Thus total current I = 0 A.

## Answer

The current through a surface equals the flux of the current density J = ∇×H, or equivalently the circulation of H around the boundary of the surface (Stokes' theorem). The surface is part of the cylinder ρ = 2, so we need the a_ρ component of J = ∇×H.

H = (2ρ + z) a_ρ + (2/z) a_z, i.e. H_ρ = 2ρ + z, H_φ = 0, H_z = 2/z.

**Step 1: a_ρ component of the curl.**

```text
(∇×H)_ρ = (1/ρ) ∂H_z/∂φ − ∂H_φ/∂z
        = (1/ρ)(0) − 0
        = 0
```

(H_z = 2/z does not depend on φ, and H_φ = 0.)

**Step 2: current.**

```text
I = ∬ J_ρ dS = ∫₃⁵ ∫_{π/4}^{π/2} 0 · ρ dφ dz = 0
```

**Step 3: check with the line integral ∮ H·dl** around the boundary of the patch (corners at φ = π/4, π/2 and z = 3, 5, all at ρ = 2):

```text
Along the φ-edges (z = 3 and z = 5): H_φ = 0 ⇒ contribution 0.
Along the z-edges (φ = π/4 and φ = π/2): H_z = 2/z, same function on both
edges but traversed in opposite directions ⇒ ∫₃⁵ (2/z) dz − ∫₃⁵ (2/z) dz = 0.
```

Both methods agree.

**Answer: I = 0 A**

## Check yourself

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

1. How can the current through a surface be found from H?
   A. I = integral of H . dS
   B. I = integral of (curl H) . dS
   C. I = integral of (div H) dv
   D. I = H x area
   Answer: B. J = curl H, so I = integral of J . dS, or equivalently the circulation of H around the boundary by Stokes' theorem. (Key point: Current from H)

2. The surface is part of the cylinder rho = 2 and the current is wanted in a_rho. Which component of curl H is needed?
   A. The rho component
   B. The phi component
   C. The z component
   D. The magnitude only
   Answer: A. The normal of the cylindrical patch is a_rho, so only (curl H)_rho matters. (Key point: Component needed)

3. For H_rho = 2 rho + z, H_phi = 0, H_z = 2/z, what is (curl H)_rho = (1/rho) dH_z/d phi - dH_phi/dz?
   A. 2/z^2
   B. 2
   C. 1/rho
   D. 0
   Answer: D. H_z does not depend on phi and H_phi = 0, so both terms vanish. (Key point: Evaluating (curl H)_rho)

4. Why does the line integral of H around the patch edge also give zero?
   A. H_phi = 0 on the phi-edges and the H_z contributions on the two z-edges cancel
   B. H_rho is zero everywhere
   C. The surface has zero area
   D. H is perpendicular to every edge
   Answer: A. The phi edges contribute nothing as H_phi = 0, and the 2/z integrals on the two z-edges are traversed in opposite directions and cancel. (Key point: Line-integral check)

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