Irodov 3.286: Weighing a Ball's Pull to a Pole – Amit & Deepali Sinha Foundation

Problem Irodov 3.286

A small paramagnetic ball of volume $V = 41$ mm $^3$ hangs by a thread from the arm of a balance, on the axis of the pole shoe of an electromagnet $M$ below it, so that the balance measures the force with which the ball is attracted to the pole. On the axis the magnetic induction depends on the height $x$ above the pole as $B = B_0\exp(-ax^2)$ , with $B_0 = 1.50$ T and $a = 100$ m $^{-2}$ . Find:

the height $x_m$ at which the attraction is greatest;
the magnetic susceptibility of the material, if the greatest attraction is $F_{\max} = 160$ $\mu$ N.

The balance arm, the thread and the small paramagnetic ball hanging on the axis of the electromagnet M, whose field lines rise from the pole face and spread out round the ball. — The balance arm, the thread and the small paramagnetic ball hanging on the axis of the electromagnet $M$ , whose field lines rise from the pole face and spread out round the ball.

Given

$V = 41$ mm $^3$
$B = B_0 e^{-ax^2}$ on the axis
$B_0 = 1.50$ T
$a = 100$ m $^{-2}$
$F_{\max} = 160$ $\mu$ N
paramagnet, $\chi \ll 1$ (assumed)

Find

(a)

x_m

· (b)

\chi

Solution

Step 1The force on a small paramagnetic ball

For $\chi \ll 1$ (assumed) the ball barely changes the field it sits in, so the field strength inside it is $H = B/\mu_0$ and its magnetization is $J = \chi H$ . Its magnetic moment, along the field, is then

p_m = JV = \frac{\chi V B}{\mu_0} \tag{1}

On the axis the field varies only with height, and a moment along the field feels the force $p_m\,dB/dx$ along $x$ . With (1):

F_x = \frac{\chi V}{\mu_0}\,B\,\frac{dB}{dx} \tag{2}

Step 2The force as a function of height

Differentiating the given profile and substituting into (2):

\begin{aligned} \frac{dB}{dx} &= -2ax\,B_0e^{-ax^2}\\[2pt] F_x &= -\frac{2a\chi VB_0^2}{\mu_0}\,x\,e^{-2ax^2} \end{aligned}

The minus sign is the attraction toward the pole. Its magnitude is

F = \frac{2a\chi VB_0^2}{\mu_0}\,x\,e^{-2ax^2} \tag{3}

which vanishes at the pole face, where the field is flat, and far above it, where the field is gone.

The attracting force against the height $x$ for the given profile: zero at the pole face and far above it, greatest at $x_m$ .

Step 3(a) The height of greatest attraction

Setting the derivative of $x\,e^{-2ax^2}$ in (3) to zero:

\frac{d}{dx}\left(x\,e^{-2ax^2}\right) = \left(1 - 4ax^2\right)e^{-2ax^2} = 0

Numerically, $4a = 400$ m $^{-2}$ , and the positive root is

x_m = \frac{1}{\sqrt{4a}} \approx 5.0\text{ cm}

(4)

Step 4(b) The susceptibility

At $x_m$ the exponent is $-2ax_m^2 = -1/2$ , so (3) gives the greatest force:

F_{\max} = \frac{2a\chi VB_0^2}{\mu_0}\,\frac{1}{2\sqrt a}\,e^{-1/2} = \frac{\chi VB_0^2}{\mu_0}\sqrt{\frac{a}{e}}

Solving for $\chi$ , numerically,

\frac{\mu_0F_{\max}}{VB_0^2}\sqrt{\frac{e}{a}} = \frac{(4\pi\times10^{-7})(160\times10^{-6})}{(41\times10^{-9})(1.50)^2}\sqrt{\frac{2.718}{100}}

\chi = \frac{\mu_0F_{\max}}{VB_0^2}\sqrt{\frac{e}{a}} \approx 3.6\times10^{-4}

(5)

Check

The assumption holds: $\chi = 3.6\times10^{-4} \ll 1$ , so the ball’s own field is a fraction of a thousandth of $B$ . Units in (5): $B_0^2/\mu_0$ is an energy density, so $VB_0^2/\mu_0$ is an energy, in N m; dividing a force by it leaves m $^{-1}$ , and $\sqrt{e/a}$ is a length in m, so $\chi$ comes out a pure number, as it must.

Remark

The ball is weighed where the field is neither strongest nor weakest but where $B\,dB/dx$ peaks: at the pole face the field is large but flat, far away it is steep in relative terms but weak. Measuring at $x_m$ also makes the result insensitive to small errors in the ball’s height, since the force is stationary there. This is the Faraday method for susceptibilities, the small-sample cousin of the long-rod balance, and it gives the sign as well: a diamagnetic ball, $\chi < 0$ , would be pushed up, away from the pole.