Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 4.1.6

Interactive: Ampère's Law Profiles

Problem 4.1.6

Hollow wire

An infinitely long conducting cylindrical wire of radius $b$ involves an inner cylindrical hole of radius $a$ as shown in the figure and carries a uniformly distributed constant current of $I$ flowing in the $a_{z}$ direction. What is the current density $J$ ? By using Ampère's formula, find the explicit expression of the magnetic field in the regions $0 < ρ < a$ , $a < ρ < b$ and $ρ > b$ .

Source: B. M. Notaroš, Electromagnetics, Problem 4.10.

Try it yourself first.

Strategy

Same symmetry, same circle, three regions: the hole, the metal and the outside. Only the enclosed current changes from region to region, and in the metal it is the current through the annulus between $a$ and $ρ$ .

1. Current density

The current is spread uniformly over the annular cross-section, whose area is $π (b^{2} - a^{2})$ :

Answer

J = \frac{I}{π ( b ^{2} - a ^{2} )} a_{z} (a < ρ < b), J = 0 elsewhere

2. Symmetry

Nothing depends on $z$ or $ϕ$ , so again $H = H_{ϕ} (ρ) a_{ϕ}$ and $2 π ρ H_{ϕ} = I_{enc} (ρ)$ on the circle of radius $ρ$ (Figure 4.1.23a).

3. Region (i), the hole

For $0 < ρ < a$ no current is enclosed:

I_{enc} = 0 ⟹ H = 0.

4. Region (ii), in the metal

For $a < ρ < b$ the circle encloses the annulus between $a$ and $ρ$ :

I_{enc} = \frac{I π ( ρ ^{2} - a ^{2} )}{π ( b ^{2} - a ^{2} )} = I \frac{ρ ^{2} - a ^{2}}{b ^{2} - a ^{2}} .

5. Region (iii), outside

For $ρ > b$ the circle encloses everything: $I_{enc} = I$ . Dividing each $I_{enc}$ by $2 π ρ$ :

Answer

H = ⎩ ⎨ ⎧ 0, \frac{I}{2 π ρ} \frac{ρ ^{2} - a ^{2}}{b ^{2} - a ^{2}} a_{ϕ}, \frac{I}{2 π ρ} a_{ϕ}, 0 < ρ < a a < ρ < b ρ > b

(in A/m)

$(b) The field profile across the tube. Problem 4.1.6. (a) The same circle in the three regions encloses nothing, part of I, and all of I. (b) All three curves share the *same* outside branch I/2\pi\rho and the *same* peak H_{\max}=I/2\pi b. The hole only changes the interior, dead space for the field, and as a\to b the rise is squeezed into a thin shell.$

Figure 4.1.23. (b) The field profile across the tube. Problem 4.1.6. (a) The same circle in the three regions encloses nothing, part of

I

, and all of

I

. (b) All three curves share the *same* outside branch

I /2 π ρ

and the *same* peak

H_{m a x} = I /2 π b

. The hole only changes the interior, dead space for the field, and as

a \to b

the rise is squeezed into a thin shell.

Shrink $a$ to zero to recover Problem 4.1.5, and push $a$ towards $b$ to see a thin shell.

InteractiveLoading simulation…

Check

Four checks. Continuity: the middle line gives $0$ at $ρ = a$ and $I /2 π b$ at $ρ = b$ , no jump, as it must be with no surface current. $a \to 0$ : it becomes $I ρ /2 π b^{2}$ , Problem 4.1.5. $a \to b$ : the thin shell, with the whole rise squeezed into the shell. Outside: $I /2 π ρ$ , from outside $a$ and $b$ are invisible (Figure 4.1.23b). The hole is field-free not because "the metal shields it": Ampère's law gives $H_{ϕ} = 0$ on every circle in the hole, and symmetry already forbade $H_{ρ}$ and $H_{z}$ . The peak $H_{m a x} = I /2 π b$ at the surface depends only on $I$ and the outer radius, whatever the hole: the middle of a conductor carries field-free dead metal, which is one reason why busbars are often tubes.

Problem 4.1.6: Hollow wire