Irodov 3.287: Work to Pull a Ball from a Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.287

A small ball of volume $V$ , made of a paramagnetic material with susceptibility $\chi$ , is moved slowly along the axis of a coil carrying a current, from a point where the magnetic induction is $B$ out to a region where there is practically no field. How much work is done in this process?

The coil in section with the small paramagnetic ball (susceptibility χ, volume V) on its axis at the point P, where the induction is B, moved slowly out along the axis at velocity v. — The coil in section with the small paramagnetic ball (susceptibility $\chi$ , volume $V$ ) on its axis at the point $P$ , where the induction is $B$ , moved slowly out along the axis at velocity $\vect v$ .

Given

volume $V$
susceptibility $\chi$
induction $B$ at the start
no field at the end
moved slowly
paramagnet, $\chi \ll 1$ (assumed)

Find

the work done

Solution

Step 1The magnetic force on the ball

Let $x$ run along the axis in the direction of the displacement, and $B(x)$ be the coil’s induction there. For $\chi \ll 1$ (assumed) the ball barely changes the field, so the field strength inside it is $H = B/\mu_0$ and its magnetization is $J = \chi H$ . Its moment $p_m = JV$ lies along the field, and on the axis it feels the force $p_m\,dB/dx$ :

F_x = \frac{\chi V}{\mu_0}\,B\,\frac{dB}{dx} \tag{1}

As $B$ falls along the way out, $dB/dx < 0$ and the force points back into the coil: the field holds the ball.

Step 2Slowly means the force is balanced

Moved slowly, the ball gains no kinetic energy, so the agent moving it applies at every point the force $-F_x$ , equal and opposite to (1). Its work over the displacement is

A = -\int F_x\,dx = -\frac{\chi V}{\mu_0}\int B\,\frac{dB}{dx}\,dx

Step 3Integrating from B to zero

With $B$ itself as the variable, running from $B$ at the start to $0$ at the end:

A = -\frac{\chi V}{\mu_0}\int_B^0 B'\,dB' = -\frac{\chi V}{\mu_0}\left(0 - \frac{B^2}{2}\right)

A = \frac{\chi VB^2}{2\mu_0}

(2)

The work done by the agent is positive for a paramagnet: the ball has to be pulled out against its attraction to the stronger field.

Check

Units: $B^2/(2\mu_0)$ is an energy density in J/m $^3$ , and times the volume $V$ it is an energy. Signs: the field’s force did the work $-A$ on the ball, negative, as it should for a body pulled out against an attraction. For a diamagnetic ball, $\chi < 0$ , the same formula gives $A < 0$ : the field itself pushes the ball out, and the agent only holds it back.

Remark

The result is the magnetic energy per unit volume $B^2/(2\mu_0)$ times $\chi V$ : the ball in the field behaves as if it carried a potential energy $-\chi VB^2/(2\mu_0)$ , lowest where the field is strongest. That potential is why paramagnetic oxygen collects at the poles of a strong magnet and why a diamagnetic body (a frog, in a famous experiment) can float in a steep field: there the energy $|\chi| VB^2/(2\mu_0)$ grows downward fast enough to carry its weight.