Irodov 3.331: The Easy Way Round Through Reciprocity

Problem Irodov 3.331

Two thin circular wires of radii $a$ and $b$ , with $a \ll b$ , lie in one plane and have a common centre. Find:

the mutual inductance of the two wires;
the magnetic flux through the surface bounded by the outer wire when the inner wire carries a current $I$ .

The two concentric wires in one plane, radii a and b (a b, drawn schematically); the inner wire's current I of part (b), sense chosen. — The two concentric wires in one plane, radii $a$ and $b$ ( $a \ll b$ , drawn schematically); the inner wire’s current $I$ of part (b), sense chosen.

Given

radii $a$ and $b$ , $a \ll b$
concentric, coplanar
current $I$ in the inner wire (part b)
vacuum around them (assumed)

Find

(a) the mutual inductance

L_{12}

; (b) the flux

\Phi

through the outer wire’s surface

Solution

Step 1The choice of which wire carries the current

The mutual inductance can be found from either wire, since $L_{12} = L_{21}$ . The flux of the small loop through the large one is hard to compute directly: its field over the large disc is far from uniform and changes sign outside the small loop. So send a current $I'$ round the large wire instead and take its flux through the small one, where $a \ll b$ makes the field nearly uniform.

Step 2The large wire’s field at the centre

A current I' in the outer wire (sense chosen) and its field B at the common centre, normal to the plane, nearly uniform over the small disc. — A current $I'$ in the outer wire (sense chosen) and its field $\vect B$ at the common centre, normal to the plane, nearly uniform over the small disc.

Every element $dl$ of the outer wire is at the distance $b$ from the centre and perpendicular to the line joining them, so the Biot and Savart law gives each element’s field normal to the plane with the same magnitude:

dB = \frac{\mu_0}{4\pi}\,\frac{I'\,dl}{b^2}

Adding the elements round the whole circumference $2\pi b$ :

B = \frac{\mu_0 I'}{4\pi b^2}\oint dl = \frac{\mu_0 I'}{4\pi b^2}\,2\pi b

B = \frac{\mu_0 I'}{2b}

(1)

Step 3(a) The mutual inductance

With the field (1) uniform over the small disc of area $\pi a^2$ and normal to it, the flux through the small wire is

\Phi_{12} = B\,\pi a^2 = \frac{\mu_0\pi a^2}{2b}\,I'

Per unit current in the large wire:

L_{12} = \frac{\Phi_{12}}{I'} = \frac{\mu_0\pi a^2}{2b}

(2)

The flux is positive when both wires are oriented the same way round, as drawn.

Step 4(b) The flux of the small loop through the large one

Now the inner wire carries $I$ . By reciprocity the flux it sends through the outer wire’s surface is $L_{21}I$ with $L_{21} = L_{12}$ , so from (2):

\Phi = L_{12}I = \frac{\mu_0\pi a^2 I}{2b}

(3)

It threads the outer wire in the same sense as the inner current, out of the page for the counter-clockwise current drawn.

Check

Part (b) directly, with no reciprocity. Far from its wire the small loop is a dipole of moment $p_m = I\pi a^2$ , whose field in its own plane at the distance $r$ is $\mu_0 p_m/(4\pi r^3)$ , pointing against $\vect p_m$ . All the flux that goes up through the small loop comes back down through the plane, so the flux up through the disc of radius $b$ equals the flux down through the rest of the plane:

\Phi = \int_b^{\infty}\frac{\mu_0 p_m}{4\pi r^3}\,2\pi r\,dr = \frac{\mu_0 p_m}{2b} = \frac{\mu_0\pi a^2 I}{2b},

which is (3). Units: $\mu_0 a^2/b$ is $(\text{H/m})\cdot\text{m}$ , a henry.

Remark

The flux in (b) is only a small part of the flux that rises through the small loop itself: almost all of that closes back down inside the large circle, between the two wires. Only the part that returns beyond the radius $b$ counts, and the dipole’s field there falls as $1/r^3$ , which is why the result goes as $1/b$ : the small loop’s strong field near its own wire never enters, only the weak return field beyond the outer wire.