Irodov 3.270: A Small Coil Pulled Along an Axis – Amit & Deepali Sinha Foundation

Problem Irodov 3.270

A round loop of radius $R$ carries a current $I$ . A small coil with magnetic moment $\vect{p}_m$ sits on the loop’s axis at a distance $x$ from the loop’s centre, with $\vect{p}_m$ pointing along that axis. What is the magnitude of the force on the small coil?

The loop of radius R cut by the page (current I out at the top, in at the bottom) and, at distance x along its axis, the small coil with moment p_m along the axis, drawn in the sense of the loop's own field. — The loop of radius $R$ cut by the page (current $I$ out at the top, in at the bottom) and, at distance $x$ along its axis, the small coil with moment $\vect{p}_m$ along the axis, drawn in the sense of the loop’s own field.

Given

$\vect{p}_m$ along the axis
loop radius $R$
current $I$
distance $x$

Find

the magnitude

F

of the force on the coil

Solution

Step 1The loop’s field along its axis

By symmetry the parts of the Biot-Savart contributions transverse to the axis cancel round the loop, and every element adds the same axial part, so on the axis the field is axial with the standard value

B(x) = \frac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}} \tag{1}

It falls off on both sides of the loop’s plane, and that fall-off is what pushes or pulls a magnetic moment placed in it.

Step 2The force on a small moment

A small coil at fixed current has the mechanical energy $-\vect{p}_m\cdot\vect{B}$ , so the force on it is the gradient of $\vect{p}_m\cdot\vect{B}$ . Outside the loop’s wire the field is curl free, and with $\vect{p}_m$ constant that gradient becomes the derivative of $\vect{B}$ along $\vect{p}_m$ :

\vect{F} = \nabla(\vect{p}_m\cdot\vect{B}) = (\vect{p}_m\cdot\nabla)\,\vect{B} = p_m\,\frac{\partial \vect{B}}{\partial x}

On the axis the field has no transverse part at any $x$ , so the derivative along the axis has none either, and only the axial component survives:

F_x = p_m\,\frac{dB}{dx}

(2)

Step 3Differentiating the axial field

Differentiating (1) with respect to $x$ by the chain rule:

\begin{aligned} \frac{dB}{dx} &= \frac{\mu_0 I R^2}{2}\cdot\left(-\frac{3}{2}\right)(R^2 + x^2)^{-5/2}\cdot 2x\\[2pt] &= -\frac{3\mu_0 I R^2 x}{2\,(R^2 + x^2)^{5/2}} \end{aligned}

Substituting this into (2) gives the axial force,

F_x = -\frac{3\mu_0 I R^2 p_m x}{2\,(R^2 + x^2)^{5/2}}

and its magnitude is

F = \frac{3\mu_0}{2}\,\frac{R^2 I p_m x}{(R^2 + x^2)^{5/2}}

(3)

For $x > 0$ the axial component is negative: with $\vect{p}_m$ in the sense of the loop’s own field the coil is drawn toward the loop. With $\vect{p}_m$ reversed it is pushed away with the same magnitude.

The loop's B lines spread out with distance from the loop, so the field weakens across the coil; the force F on a coil whose p_m is along B points toward the loop. — The loop’s $\vect{B}$ lines spread out with distance from the loop, so the field weakens across the coil; the force $\vect{F}$ on a coil whose $\vect{p}_m$ is along $\vect{B}$ points toward the loop.

Check

Units: $\mu_0 I R^2 x/(R^2+x^2)^{5/2}$ is $\mu_0 I$ per square metre, which is tesla per metre, and tesla per metre times A m $^2$ is N. At $x = 0$ the force vanishes, as it must at the centre, where $B$ is at its maximum and has no slope. Far away, $x \gg R$ , (3) becomes $3\mu_0 R^2 I p_m/(2x^4)$ ; with the loop’s own moment $p_{1m} = \pi R^2 I$ this is $3\mu_0 p_{1m} p_m/(2\pi x^4)$ , the force between two coaxial dipoles.

Remark

The force is largest where $x/(R^2+x^2)^{5/2}$ peaks, which is where its derivative $(R^2 - 4x^2)/(R^2+x^2)^{7/2}$ vanishes, at $x = R/2$ . That is also the inflection point of $B(x)$ , where its second derivative is zero. Two equal coaxial loops a distance $R$ apart put their inflection points at the same midpoint: there each loop’s second derivative is zero and, by symmetry, the odd derivatives of the sum cancel, so the field between them is uniform to a high order. That is the Helmholtz pair.