Irodov 3.346: Small Loop Samples the Center Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.346

Two circular wire loops share the same center. The small one has radius $a$ and carries current $I_1$ ; the large one has radius $b$ , with $a \ll b$ , and carries current $I_2$ . The planes of the two loops make an angle $\theta$ with each other. Find the magnetic interaction energy of the two loops.

The two concentric loops: the large loop of radius b with current I_2, the small loop of radius a with current I_1, and the angle θ between their normals. — The two concentric loops: the large loop of radius $b$ with current $I_2$ , the small loop of radius $a$ with current $I_1$ , and the angle $\theta$ between their normals.

Given

$I_1$ , radius $a$
$I_2$ , radius $b$
$a \ll b$
common center
angle $\theta$ between the planes

Find

the interaction energy

W_{12}

Solution

Step 1Where the interaction energy lives

The magnetic field energy of two circuits carrying fixed currents splits into each circuit’s own energy and one cross term. The cross term is the interaction energy, and it is fixed by the mutual inductance $L_{12}$ .

W = \frac{L_1 I_1^2}{2} + \frac{L_2 I_2^2}{2} + L_{12} I_1 I_2 \tag{1}

The mutual inductance is the flux $\Phi_{12}$ through loop 1 produced by loop 2, per unit current in loop 2.

L_{12} = \frac{\Phi_{12}}{I_2} \tag{2}

Step 2Field of the large loop at its center

Every element $d\vect{l}$ of the large loop is a distance $b$ from the center and perpendicular to the line joining them, so the Biot and Savart law gives each element’s field along the loop’s normal with the same magnitude.

dB = \frac{\mu_0}{4\pi}\frac{I_2\,dl}{b^2}

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

B_2 = \frac{\mu_0 I_2}{4\pi b^2}\oint dl = \frac{\mu_0 I_2}{4\pi b^2}\,2\pi b

B_2 = \frac{\mu_0 I_2}{2b}

(3)

Step 3Flux through the small loop

The large loop's field B_2 at the common center lies along its normal; the small loop's normal, along its moment p_1, makes the angle θ with it. — The large loop’s field $\vect{B}_2$ at the common center lies along its normal; the small loop’s normal, along its moment $\vect{p}_1$ , makes the angle $\theta$ with it.

Take the small loop’s normal along its current by the right-hand rule, the direction of its moment $\vect{p}_1$ in the figure. With $\vect{B}_2$ uniform over the small disc of area $\pi a^2$ , the flux is the field times the area times the cosine of the angle between $\vect{B}_2$ and that normal, using (3).

\Phi_{12} = B_2\,\pi a^2\cos\theta = \frac{\mu_0 I_2}{2b}\,\pi a^2\cos\theta

Dividing by $I_2$ as in (2):

L_{12} = \frac{\mu_0\pi a^2}{2b}\cos\theta

(4)

Step 4The interaction energy

The cross term of (1) is the interaction energy. Substituting the mutual inductance:

W_{12} = L_{12} I_1 I_2 = \frac{\mu_0\pi a^2}{2b}\,I_1 I_2\cos\theta

(5)

Here $\theta$ is the angle between the two normals, each taken along its own current by the right-hand rule; reversing either current turns $\theta$ into $\pi - \theta$ and flips the sign of $W_{12}$ .

Check

Reciprocity, $L_{21} = L_{12}$ , checked the other way round. Seen from the large loop, the small one is a dipole $p_1 = I_1\pi a^2$ at its center, and only the component $p_1\cos\theta$ along the large loop’s normal sends net flux through it. The dipole’s field in its own equatorial plane is $\mu_0 p/(4\pi r^3)$ , pointing against the dipole, and the total flux through the whole plane is zero, so the flux inside radius $b$ equals the returning flux outside it:

\Phi_{21} = \int_b^{\infty}\frac{\mu_0 p_1\cos\theta}{4\pi r^3}\,2\pi r\,dr = \frac{\mu_0 I_1\pi a^2}{2b}\cos\theta

This is $L_{12}I_1$ with $L_{12}$ from (4), as reciprocity demands. Also, $\theta = 90^\circ$ gives zero: the large loop’s field then lies in the small loop’s plane and threads none of it.

Remark

At fixed currents the loops are pulled toward larger $W_{12}$ , not smaller: the sources supply twice the change in $W_{12}$ , so the mechanical potential energy is $-W_{12} = -\vect{p}_1\cdot\vect{B}_2$ . The torque on the small loop, $\partial W_{12}/\partial\theta = -p_1 B_2\sin\theta$ , turns it toward $\theta = 0$ , coaxial with currents circulating the same way.