Problem 4.1.8: Ring with a non-uniform surface current

Problem 4.1.8

Ring with a non-uniform surface current

There exists a nonuniform surface current density on the ring shown in the figure with an inner radius of $b$ and an outer radius of $2 b$ given as

J = I_{0} ρ^{'} sin (\frac{π}{b} ρ^{'}) a_{ϕ} [A/ m^{2}]

in vacuum where $I_{0}$ is a known constant. Find the vector potential $A$ and the magnetic field $H$ at the origin.

Try it yourself first.

Strategy

The source is a flat ring, so both $A$ and $H$ at the origin are surface integrals with the simplest possible geometry, $R = ρ^{'}$ . The point of the problem is that the direction of $J$ averages to zero round the ring, so $A$ vanishes, while the direction of $J \times a_{R}$ does not, so $H$ survives. Keep the two calculations apart.

1. Geometry

The source is the flat ring $b \leq ρ^{'} \leq 2 b$ , $z^{'} = 0$ , a sheet, so $J$ is a surface density and $I_{0}$ carries A/m $^{2}$ . The surface element is $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ , and for the field point at the origin

R = r - r^{'} = - ρ^{'} a_{ρ^{'}}, R = ρ^{'}, a_{R} = \frac{R}{R} = - a_{ρ^{'}} .

2. The vector potential

From Eq. (4.1.15) for a sheet,

A (0) = \frac{μ _{0}}{4 π} \int_{S} \frac{J}{R} d S^{'} = \frac{μ _{0} I _{0}}{4 π} \int_{0}^{2 π} \int_{b}^{2 b} ρ^{'} sin (\frac{π ρ ^{'}}{b}) a_{ϕ^{'}} d ρ^{'} d ϕ^{'} .

The radial and the angular integrals separate, and with $a_{ϕ^{'}} = - sin ϕ^{'} a_{x} + cos ϕ^{'} a_{y}$ (Section 2.3),

\int_{0}^{2 π} a_{ϕ^{'}} d ϕ^{'} = - a_{x} \int_{0}^{2 π} sin ϕ^{'} d ϕ^{'} + a_{y} \int_{0}^{2 π} cos ϕ^{'} d ϕ^{'} = 0 .

Answer

A (0) = 0

$A$ inherits the direction of $J$ : the element at $ϕ^{'}$ and the element at $ϕ^{'} + π$ have opposite $a_{ϕ^{'}}$ and the same $∣ J ∣/ R$ , so they cancel pair by pair, whatever the radial profile of $J$ (Figure 4.1.25a).

3. The magnetic field

$A = 0$ at one point says nothing about $H = \frac{1}{μ _{0}} \nabla \times A$ there, because the curl needs $A$ in a neighbourhood of the point. So compute $H$ directly from Biot–Savart's law for a surface distribution, Eq. (4.1.12):

H (0) = \frac{1}{4 π} \int_{S} \frac{J \times a _{R}}{R ^{2}} d S^{'}, a_{ϕ^{'}} \times (- a_{ρ^{'}}) = + a_{z} for every ϕ^{'} .

Now the direction does not cancel. With $R^{2} = ρ^{' 2}$ and $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ one power of $ρ^{'}$ is left over, and it cancels the $ρ^{'}$ in $J$ :

H (0) = \frac{a _{z}}{4 π} \int_{0}^{2 π} d ϕ^{'} \int_{b}^{2 b} I_{0} sin (\frac{π ρ ^{'}}{b}) d ρ^{'} = \frac{I _{0}}{2} [- \frac{b}{π} cos \frac{π ρ ^{'}}{b}]_{b}^{2 b} a_{z} = \frac{I _{0} b}{2 π} (- 1 - 1) a_{z} .

Answer

H (0) B (0) = - \frac{I _{0} b}{π} a_{z} [A/m], = - \frac{μ _{0} I _{0} b}{π} a_{z} [T]

(a) The ring seen from +z (darker = larger |\vec J|). — (a) The ring seen from $+ z$ (darker $=$ larger $∣ J ∣$ ).

$(b) The current profile and the contributions of the rings. Problem 4.1.8. (a) Why \vec A cancels but \vec H does not: \vec A inherits the direction of \vec J, which averages to zero round the ring, while \vec H inherits \vec J\times\vec a_R, whose direction \vec a_z is the same for every \phi'. The current circulates clockwise seen from +z because \sin(\pi\rho'/b)<0 on b<\rho'<2b. (b) Red: the outer rings carry more current. Blue: at the centre every ring counts the same, since dH_z=dI/2\rho' cancels the \rho' of \vec J. Both are negative on b<\rho'<2b, and the area under the blue curve, -2b/\pi, *is* the answer.$

Figure 4.1.25. (b) The current profile and the contributions of the rings. Problem 4.1.8. (a) Why

A

cancels but

H

does not:

A

inherits the direction of

J

, which averages to zero round the ring, while

H

inherits

J \times a_{R}

, whose direction

a_{z}

is the same for every

ϕ^{'}

. The current circulates clockwise seen from

+ z

because

sin (π ρ^{'} / b) < 0

b < ρ^{'} < 2 b

. (b) Red: the outer rings carry more current. Blue: at the centre every ring counts the same, since

d H_{z} = d I /2 ρ^{'}

cancels the

ρ^{'}

J

. Both are negative on

b < ρ^{'} < 2 b

, and the area under the blue curve,

- 2 b / π

, *is* the answer.

Check

Units: $I_{0}$ [A/m $^{2}$ ] times $b$ [m] gives A/m, correct for $H$ . Sign: on $b < ρ^{'} < 2 b$ the argument $π ρ^{'} / b$ runs from $π$ to $2 π$ , so $sin < 0$ and the current really circulates in the $- a_{ϕ}$ sense, and the right-hand rule then puts $H (0)$ along $- a_{z}$ . A second route, ring by ring: a ring of radius $ρ^{'}$ carries $d I = J_{ϕ} d ρ^{'}$ and gives $d H_{z} = d I /2 ρ^{'}$ at its centre (Problem 4.1.2 with $h = 0$ ), so

d H_{z} = \frac{I _{0} ρ ^{'} sin ( π ρ ^{'} / b ) d ρ ^{'}}{2 ρ ^{'}} = \frac{I _{0}}{2} sin (\frac{π ρ ^{'}}{b}) d ρ^{'},

the $ρ^{'}$ cancels, every ring counts the same, and $\int_{b}^{2 b} sin (π ρ^{'} / b) d ρ^{'} = - 2 b / π$ gives the same answer (Figure 4.1.25b). Had the ring extended from $0$ to $b$ , $sin$ would have been positive and $H (0)$ would point along $+ a_{z}$ . A flat spiral coil wound from $b$ to $2 b$ is, at distances large compared with the turn pitch, exactly this annular $a_{ϕ}$ sheet, the geometry of wireless-charging pads and induction hobs (Application 4.1.4).