Irodov 3.271: Two Coaxial Dipoles – Amit & Deepali Sinha Foundation

Problem Irodov 3.271

Two small coils have magnetic moments $p_{1m} = 4.0$ mA $\cdot$ m $^2$ and $p_{2m} = 6.0$ mA $\cdot$ m $^2$ , and their axes lie along one common line. They are $l = 20$ cm apart, a distance much larger than the size of either coil. Find the force with which they interact.

The two small coils on their common axis, moments p_1m and p_2m along it, separated by l. Drawn with the moments parallel; reversing one leaves the magnitude of the force unchanged. — The two small coils on their common axis, moments $\vect{p}_{1m}$ and $\vect{p}_{2m}$ along it, separated by $l$ . Drawn with the moments parallel; reversing one leaves the magnitude of the force unchanged.

Given

$p_{1m} = 4.0$ mA $\cdot$ m $^2$
$p_{2m} = 6.0$ mA $\cdot$ m $^2$
common axis
$l = 20$ cm, much larger than the coils

Find

the interaction force

F

Solution

Step 1The first coil’s field at the second

Seen from a distance much larger than its size, the first coil is a point magnetic dipole. On its axis, at distance $r$ , the dipole field points along the axis and has the standard value

B_1(r) = \frac{\mu_0}{4\pi}\,\frac{2p_{1m}}{r^3} \tag{1}

(the far-field form of the loop’s axial field $\mu_0 IR^2/2(R^2+r^2)^{3/2}$ with $p_{1m} = \pi R^2 I$ ).

Step 2The force from the field’s gradient

The second coil sits on the axis of the first with its moment along that axis, so the force on it is axial and equals its moment times the rate at which the first coil’s field changes along the axis:

F = p_{2m}\,\frac{dB_1}{dr}\bigg|_{r=l}

Differentiating (1) and setting $r = l$ ,

\frac{dB_1}{dr}\bigg|_{r=l} = \frac{\mu_0}{4\pi}\,2p_{1m}\cdot\left(-\frac{3}{l^4}\right) = -\frac{3\mu_0\,p_{1m}}{2\pi\,l^4}

The magnitude of the force is $p_{2m}$ times the magnitude of this slope. Numerically, that product is

\frac{3\,(4\pi\times10^{-7})(4.0\times10^{-3})(6.0\times10^{-3})}{2\pi\,(0.20)^4} = \frac{(6.0\times10^{-7})(2.4\times10^{-5})}{1.6\times10^{-3}}\text{ N}

F = \frac{3\mu_0\,p_{1m}\,p_{2m}}{2\pi\,l^4} \approx 9.0\text{ nN}

(2)

With the moments parallel, as drawn, the force on the second coil points toward the first, so the coils attract; with one moment reversed they repel with the same magnitude. The force on the first coil is equal and opposite.

Check

The result is symmetric in the two moments, as Newton’s third law requires: computing the force on the first coil in the field of the second gives the same expression. Units: $\mu_0$ in N/A $^2$ times (A m $^2$ ) $^2$ over m $^4$ is N. It also agrees with the force on a small coil on the axis of a loop of radius $R$ , $3\mu_0R^2Ip_mx/2(R^2+x^2)^{5/2}$ , which for $x \gg R$ and $\pi R^2 I = p_{1m}$ is exactly this expression.

Remark

The force falls as $1/l^4$ , one power faster than the $1/l^3$ dipole field, because a dipole feels only the gradient of a field. Doubling the separation cuts the force sixteen times: at $l = 40$ cm it would be about $0.56$ nN. That steep fall is why small magnets seem to snap together only at the last moment.