Irodov 3.333: Two Distant Loops Couple as Dipoles – Amit & Deepali Sinha Foundation

Problem Irodov 3.333

Two thin circular loops of the same radius $a$ have a common axis, and their centres are a distance $l$ apart, with $l \gg a$ . Find an approximate formula for their mutual inductance.

The two thin loops of radius a on a common axis, their centres a distance l apart (l a; drawn closer, schematically). — The two thin loops of radius $a$ on a common axis, their centres a distance $l$ apart ( $l \gg a$ ; drawn closer, schematically).

Given

two loops of radius $a$
common axis
centres $l$ apart, $l \gg a$
vacuum around them (assumed)

Find

the mutual inductance

L_{12}

, to leading order in

a/l

Solution

Step 1The field of one loop on its axis

A current I in the lower loop (sense chosen) and its field B_1 at the centre of the upper loop, along the common axis. — A current $I$ in the lower loop (sense chosen) and its field $\vect B_1$ at the centre of the upper loop, along the common axis.

Let a current $I$ flow in the lower loop. Every element $dl$ of it is at the same distance $\sqrt{a^2 + l^2}$ from a point of the axis at the height $l$ and perpendicular to the line joining them, so by the Biot and Savart law each element gives a field of the same magnitude:

dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl}{a^2 + l^2}

By symmetry the components across the axis cancel round the loop, and each element’s component along the axis is its field times $a/\sqrt{a^2 + l^2}$ . Adding round the circumference $2\pi a$ :

B_1 = \frac{\mu_0 I}{4\pi(a^2 + l^2)}\cdot\frac{a}{\sqrt{a^2 + l^2}}\cdot 2\pi a

B_1 = \frac{\mu_0 I a^2}{2\,(a^2 + l^2)^{3/2}}

(1)

Step 2The flux through the distant loop

For $l \gg a$ the field (1) is

B_1 \approx \frac{\mu_0 I a^2}{2\,l^3}

and over the upper loop, a disc of radius $a$ at the distance $l$ , it changes only by a fraction of order $(a/l)^2$ , which we drop. The flux through the upper loop is then this field times its area $\pi a^2$ :

\Phi_{21} = B_1\,\pi a^2 = \frac{\mu_0\pi a^4}{2\,l^3}\,I

Step 3The mutual inductance

The mutual inductance is the flux through one loop per unit current in the other:

L_{12} = \frac{\Phi_{21}}{I} \approx \frac{\mu_0\pi a^4}{2\,l^3}

(2)

It is positive when both loops are oriented the same way round the common axis.

Check

The same result from the two loops as magnetic dipoles. The lower loop is a dipole $p_m = I\pi a^2$ along the axis, whose field on its axis at the distance $l$ is $2\mu_0 p_m/(4\pi l^3) = \mu_0 I a^2/(2l^3)$ , the leading form above. Units: $\mu_0 a^4/l^3$ is $(\text{H/m})\cdot\text{m}$ , a henry. The value is symmetric in the two loops, as reciprocity requires.

Remark

The coupling falls as $1/l^3$ , the dipole law. Place the same two loops side by side in one plane, centres $l$ apart, and the first loop’s field at the second is the equatorial dipole field, half as strong and pointing back against the moment, so $L_{12} \approx -\mu_0\pi a^4/(4l^3)$ . Moving the second loop round from the axis to the side at the fixed distance $l$ , its plane kept parallel to the first, the coupling follows the dipole factor $3\cos^2\theta - 1$ , with $\theta$ the angle from the axis, and vanishes at $\theta \approx 54.7^\circ$ ; that is how coils that must not talk to each other can be placed.