Irodov 3.332: A Magnet Weighed by a Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.332

A small cylindrical magnet $M$ stands at the centre of a thin flat coil of radius $a$ with $N$ turns, the magnet’s axis along the coil’s axis. The coil is connected to a ballistic galvanometer, and the resistance of the whole circuit is $R$ . When the magnet is taken out of the coil, a charge $q$ flows through the galvanometer. Find the magnetic moment of the magnet.

The thin flat coil of N turns with the small cylindrical magnet M upright at its centre, its axis along the coil's axis, and the leads to the ballistic galvanometer. — The thin flat coil of $N$ turns with the small cylindrical magnet $M$ upright at its centre, its axis along the coil’s axis, and the leads to the ballistic galvanometer.

Given

coil radius $a$ , $N$ turns
circuit resistance $R$
charge $q$ on removing the magnet
magnet small, at the centre, axis along the coil’s
magnet taken far away (assumed)

Find

the magnet’s moment

p_m

Solution

Step 1The charge counts the change of flux

Let $\Phi$ be the magnet’s flux through one turn while it sits in the coil. As the magnet is withdrawn the flux changes and the emf $-N\,d\Phi/dt$ drives a current $I$ ; the coil’s own inductance $L$ adds $-L\,dI/dt$ :

RI = -N\,\frac{d\Phi}{dt} - L\,\frac{dI}{dt}

Integrating over the whole removal, from the magnet at the centre to the magnet far away (assumed), where its flux has gone to zero. The current is zero at both ends, so the inductance term integrates to nothing:

q = \int I\,dt = -\frac{N}{R}\int_{\Phi}^{0} d\Phi' - \frac{L}{R}\int_0^0 dI

q = \frac{N\Phi}{R}

(1)

Step 2The magnet’s flux through the coil

The section through the axis: the magnet's field lines and the coil's two cross-sections at the distance a. Lines leave the magnet upward through the plane; those crossing back down beyond the coil carry the flux that threads it. — The section through the axis: the magnet’s field lines and the coil’s two cross-sections at the distance $a$ . Lines leave the magnet upward through the plane; those crossing back down beyond the coil carry the flux that threads it.

The magnet is small compared with $a$ , so beyond the coil its field is that of a point dipole of moment $p_m$ along the axis. In the coil’s plane, the dipole’s equatorial plane, the field at the distance $r$ points against the moment with magnitude

B(r) = \frac{\mu_0 p_m}{4\pi r^3}

The field lines are closed, so the net flux through the whole plane is zero: the flux up through the disc of the coil equals the flux coming back down through the plane outside it. Summing that over rings of radius $r$ and width $dr$ , from $r = a$ outward:

\Phi = \int_a^\infty B(r)\,2\pi r\,dr = \frac{\mu_0 p_m}{2}\int_a^\infty\frac{dr}{r^2}

\Phi = \frac{\mu_0 p_m}{2a}

(2)

Step 3The magnetic moment

Substituting (2) into (1) and solving for $p_m$ :

q = \frac{N}{R}\,\frac{\mu_0 p_m}{2a}

p_m = \frac{2aRq}{\mu_0 N}

(3)

Check

Reciprocity gives the flux the other way round. The flux a small current loop at the coil’s centre sends through the whole coil equals the loop’s moment times the field the coil would make there per unit coil current. A current $I'$ in the coil makes at its centre the field $B' = \mu_0 N I'/(2a)$ along the axis, so the flux linkage is $p_m\mu_0 N/(2a)$ , which is $N\Phi$ with $\Phi$ from (2). Units: $aRq/\mu_0$ is $\text{m}\cdot\Omega\cdot\text{C}/(\text{H/m}) = \text{A}\cdot\text{m}^2$ , a magnetic moment.

Remark

This is the classic ballistic method of measuring a magnet, or with a known coil a field: a single throw of the galvanometer gives the flux change, whatever the speed of the hand that pulls the magnet out. Turning the magnet over in place instead of removing it reverses the flux and doubles the charge, a more sensitive variant of the same measurement.