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

# Total current through conical surface

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2067 Shrawan, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/current-conical-surface-h-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Given H = (3r²/sinθ) a_θ + 54r cosθ a_φ A/m in free space. Find the total current in the a_θ direction through the conical surface θ = 20°, 0 ≤ φ ≤ 2π, 0 ≤ r ≤ 5.

## Short answer

Current density is J = ∇×H; its a_θ component is −108 cosθ A/m². Integrating J·dS over the cone (θ=20°, r∈[0,5], φ∈[0,2π]) gives total current I = −2.73 kA.

## Answer

The current through a surface is I = ∫ J·dS with J = ∇×H. The surface is the cone θ = 20° with 0 ≤ r ≤ 5 and 0 ≤ φ ≤ 2π. Its normal is a_θ and dS = r sin θ dr dφ a_θ.

**Given:** H_r = 0, H_θ = 3r²/sin θ, H_φ = 54 r cos θ.

**Step 1: a_θ component of the curl** (spherical coordinates).

```text
(∇×H)_θ = (1/r) [ (1/sin θ) ∂H_r/∂φ − ∂(r H_φ)/∂r ]
```

H_r = 0, and r H_φ = 54 r² cos θ, so ∂(rH_φ)/∂r = 108 r cos θ:

```text
(∇×H)_θ = (1/r) [ 0 − 108 r cos θ ] = −108 cos θ   A/m²
```

**Step 2: integrate over the cone.**

```text
I = ∫_{φ=0}^{2π} ∫_{r=0}^{5} (−108 cos θ)(r sin θ) dr dφ
  = −108 cos θ sin θ × (r²/2)|₀⁵ × 2π
  = −108 × cos 20° × sin 20° × 12.5 × 2π
  = −108 × 0.9397 × 0.3420 × 12.5 × 6.2832
  = −2726 A
```

**Step 3: check by the line integral** (Stokes' theorem). The boundary of the cone is the circle r = 5, θ = 20°, with dl = r sin θ dφ a_φ.

```text
|∮ H·dl| = ∫₀^{2π} 54 r cos θ × r sin θ dφ
         = 54 × 25 × cos 20° sin 20° × 2π = 2726 A
```

The right-hand rule for the normal a_θ (pointing away from the z-axis and downward) makes the circulation run in the −a_φ direction, so the sign is negative. Both methods agree.

**Answer: I ≈ −2.73 kA** (magnitude 2726 A, flowing opposite to a_θ).

## Check yourself

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

1. For the cone theta = 20 degrees, what is dS?
   A. r dr d(theta) a_phi
   B. r^2 sin(theta) d(theta) d(phi) a_r
   C. r sin(theta) dr d(phi) a_theta
   D. r d(theta) dr a_z
   Answer: C. On a constant-theta surface, dS = r sin(theta) dr d(phi) a_theta. (Key point: Surface element)

2. The a_theta component of curl H depends on:
   A. H_theta only
   B. H_r and H_phi: (1/r)[(1/sin theta) dH_r/d(phi) - d(r H_phi)/dr]
   C. H_z only
   D. dH_theta/dr
   Answer: B. H_theta does not enter (curl H)_theta; only H_r and H_phi do. (Key point: Which components matter)

3. With H_r = 0 and H_phi = 54 r cos(theta), (curl H)_theta equals:
   A. 54 cos(theta)
   B. -54 cos(theta)
   C. -108 cos(theta)
   D. 108 r cos(theta)
   Answer: C. d(r H_phi)/dr = 108 r cos(theta), so (curl H)_theta = (1/r)(-108 r cos theta) = -108 cos(theta). (Key point: Value of the curl component)

4. What is the total current in the a_theta direction?
   A. -2726 A (about -2.73 kA)
   B. +2726 A
   C. -136 A
   D. 0
   Answer: A. I = -108 cos20 sin20 (12.5)(2 pi) = -2726 A; the Stokes line integral gives the same magnitude. (Key point: Total current)

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