Irodov 3.279: A Magnetic Ball Gathers the Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.279

A ball of a uniform magnetic material with permeability $\mu$ is placed in an external uniform magnetic field of induction $\vect B_0$ , and it becomes uniformly magnetized. Find the magnetic induction $\vect B$ inside the ball. Use the fact that a uniformly magnetized ball, with magnetization $\vect J$ , produces inside itself a uniform field of strength $\vect H' = -\vect J/3$ .

A ball of permeability μ placed in the uniform external field B_0. — A ball of permeability $\mu$ placed in the uniform external field $\vect B_0$ .

Given

$\vect B_0$ uniform
permeability $\mu$
$\vect H' = -\vect J/3$ (own field)

Find

\vect B

inside the ball

Solution

Step 1The field strength inside: external plus the ball’s own

The field strength inside the ball is the external one, $\vect H_0 = \vect B_0/\mu_0$ , plus the field $\vect H'$ that the ball’s own magnetization produces inside it:

\vect H = \frac{\vect B_0}{\mu_0} - \frac{\vect J}{3} \tag{1}

Step 2Closing the loop with the material law

In a linear isotropic magnetic the magnetization is set by the field strength inside it,

\vect J = \chi\vect H = (\mu - 1)\vect H \tag{2}

Substituting (2) into (1) and collecting the $\vect H$ terms:

\begin{aligned} \vect H + \frac{\mu - 1}{3}\,\vect H &= \frac{\vect B_0}{\mu_0}\\[2pt] \frac{\mu + 2}{3}\,\vect H &= \frac{\vect B_0}{\mu_0} \end{aligned}

\vect H = \frac{3\vect B_0}{(\mu + 2)\,\mu_0}

(3)

The field strength inside is uniform, parallel to $\vect B_0$ , and smaller than the external $\vect B_0/\mu_0$ whenever $\mu > 1$ .

Step 3The induction inside

Inside the material $\vect B = \mu\mu_0\vect H$ , so multiplying (3) by $\mu\mu_0$ gives

\vect B = \frac{3\mu}{\mu + 2}\,\vect B_0

(4)

The B lines round and through the ball, drawn for μ = 4. The lines bend toward the ball and run straight and denser inside, where B is twice B_0. — The $\vect B$ lines round and through the ball, drawn for $\mu = 4$ . The lines bend toward the ball and run straight and denser inside, where $\vect B$ is twice $\vect B_0$ .

The induction inside points along $\vect B_0$ and is stronger than it for $\mu > 1$ : the lines crowd into the ball, as the figure shows.

Check

Three limits of (4). With $\mu = 1$ the ball is vacuum and $\vect B = \vect B_0$ . As $\mu \to \infty$ , $\vect B \to 3\vect B_0$ , a finite value although the material is ideal; then (3) gives $\vect H \to 0$ , as inside ideal iron. As $\mu \to 0$ (a body that expels the field), $\vect B \to 0$ .

Remark

However large $\mu$ , a ball can at most triple the field: its demagnetizing field $-\vect J/3$ grows with the magnetization and stops it. A long rod along the field has almost no demagnetizing field, so it gathers the field far more, by nearly the full factor $\mu$ ; the shape of a magnetic body matters as much as its material. The result is the twin of the dielectric ball, whose field inside is $3\vect E_0/(\varepsilon + 2)$ : there the bound charges weaken $\vect E$ , here the bound currents strengthen $\vect B$ .