Irodov 3.285: The Coil Pulls a Paramagnetic Rod – Amit & Deepali Sinha Foundation

Problem Irodov 3.285

A long thin cylindrical rod of a paramagnetic material, with magnetic susceptibility $\chi$ and cross-sectional area $S$ , lies along the axis of a coil carrying a current. One end of the rod is at the centre of the coil, where the magnetic induction is $B$ ; the other end is far away, where there is practically no field. Find the force the coil exerts on the rod.

The coil in section with the thin rod (susceptibility χ, cross-section S) along its axis, from the centre O, where the induction is B, out to where the field has died away. — The coil in section with the thin rod (susceptibility $\chi$ , cross-section $S$ ) along its axis, from the centre $O$ , where the induction is $B$ , out to where the field has died away.

Given

susceptibility $\chi$
cross-section $S$
induction $B$ at the centre $O$
no field at the far end

Find

the force on the rod

Solution

Step 1The field inside the thin rod

Let $x$ run along the axis from $O$ out along the rod, and let $B(x)$ be the coil’s induction on the axis, with $B(0) = B$ and $B = 0$ at the far end. The rod’s long side runs along the field, and the tangential component of $\vect H$ is continuous across it; the rod is thin, so its end faces and their poles are negligible. The field strength inside the rod is therefore the coil’s own:

H(x) = \frac{B(x)}{\mu_0} \tag{1}

The rod is then magnetized along the axis with $J = \chi H$ .

Step 2The force on a slice of the rod

A slice of length $dx$ has the magnetic moment $dp_m = JS\,dx$ , along the field. A small magnetic moment along a field that changes along it feels the force $p_m\,\partial B/\partial x$ , so with (1)

dF_x = JS\,dx\,\frac{dB}{dx} = \frac{\chi S}{\mu_0}\,B\,\frac{dB}{dx}\,dx

Step 3Adding the slices

Integrating along the rod, the variable of integration becomes $B$ itself, running from $B$ at the centre to $0$ at the far end:

F_x = \frac{\chi S}{\mu_0}\int_0^{\infty} B\,\frac{dB}{dx}\,dx = \frac{\chi S}{\mu_0}\int_B^0 B'\,dB' = -\frac{\chi SB^2}{2\mu_0}

The minus sign says the force points toward the centre of the coil: the rod is drawn into it. Its magnitude is

F = \frac{\chi SB^2}{2\mu_0}

(2)

Check

Units: $B^2/\mu_0$ is an energy density, J/m $^3$ or N/m $^2$ , and times an area it is a force. An energy check gives the same result without any field profile: pushing the rod in by $dx$ adds a slice of volume $S\,dx$ at the centre, where the field is $B$ , and removes none where there is field (the far end is in zero field). At fixed current the field’s energy rises by the slice’s magnetic energy $\chi B^2/(2\mu_0)$ per unit volume times $S\,dx$ , and the force is that rise per unit length, $\chi SB^2/(2\mu_0)$ , pulling inward.

Remark

The force depends only on $B$ at the inner end, not on how the field falls off along the coil’s axis: every detail of $B(x)$ cancels in the integral. That is what makes this arrangement, a Gouy balance, a clean way to measure a susceptibility: hang the rod from a balance with one end in a strong uniform field and the other in none, and the pull $\chi SB^2/(2\mu_0)$ gives $\chi$ from $B$ and $S$ alone. It is the magnetic twin of a dielectric liquid drawn up between the plates of a capacitor. For a diamagnetic rod, $\chi < 0$ , the same formula pushes the rod out.