Irodov 3.342: Magnetization Work Carries a Half – Amit & Deepali Sinha Foundation

Problem Irodov 3.342

Starting from the expression for the energy per unit volume of a magnetic field, show that the work spent on magnetizing a unit volume of a paramagnetic or diamagnetic substance is $A = -\vect J\cdot\vect B/2$ , where $\vect J$ is the magnetization and $\vect B$ the induction.

Given

para- or diamagnet
magnetization $\vect J$ , induction $\vect B$
$\vect J$ proportional to $\vect B$ (linear medium)

Find

the work

A

per unit volume spent on magnetization

Solution

Step 1The field strength in terms of B and J

The field strength is defined from the induction and the magnetization:

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

In a para- or diamagnet $\vect J = \chi\vect H$ with a constant susceptibility $\chi$ , so $\vect J$ , $\vect H$ and $\vect B$ are all parallel and proportional to one another.

Step 2Splitting the energy density

The energy density of the magnetic field in a medium is $\vect B\cdot\vect H/2$ . Substituting (1) splits it into two terms:

w = \frac{\vect B\cdot\vect H}{2} = \frac{\vect B}{2}\cdot\left(\frac{\vect B}{\mu_0} - \vect J\right) = \frac{B^2}{2\mu_0} - \frac{\vect J\cdot\vect B}{2} \tag{2}

The first term is the energy density the same induction would have in vacuum. The second exists only because the substance is magnetized.

Energy density against B: in vacuum, w_0 = B^2/2μ_0 (sky), and in a paramagnet, w = B^2/2μμ_0 (gold), drawn for μ = 1.6 (schematic). At each B the gap between them is J· B/2. — Energy density against $B$ : in vacuum, $w_0 = B^2/2\mu_0$ (sky), and in a paramagnet, $w = B^2/2\mu\mu_0$ (gold), drawn for $\mu = 1.6$ (schematic). At each $B$ the gap between them is $\vect J\cdot\vect B/2$ .

Step 3The same split from the work done

To see that the second term is work spent on the substance, build the field up from zero. Raising the induction by $d\vect B$ costs the sources $\vect H\cdot d\vect B$ per unit volume, and by (1)

\vect H\cdot d\vect B = \frac{\vect B\cdot d\vect B}{\mu_0} - \vect J\cdot d\vect B

The first part would be spent in vacuum too; the second is spent on magnetization. Because $\vect J$ is along $\vect B$ and proportional to it, its two terms in the differential of the product are equal:

d(\vect J\cdot\vect B) = \vect J\cdot d\vect B + \vect B\cdot d\vect J = 2\,\vect J\cdot d\vect B

Integrating the magnetization part from zero up to the final state:

A = -\int_0^{B}\vect J\cdot d\vect B = -\frac{1}{2}\int_0^{B} d(\vect J\cdot\vect B)

A = -\frac{\vect J\cdot\vect B}{2}

(3)

which is the second term of (2). In a paramagnet $\vect J$ is along $\vect B$ and $A < 0$ ; in a diamagnet it is opposite and $A > 0$ .

Check

With $\vect J = \chi\vect H$ and $\vect B = \mu\mu_0\vect H$ , $\mu = 1 + \chi$ , the split (2) should reproduce $B^2/2\mu\mu_0$ . Substituting $J = \chi B/\mu\mu_0$ :

\frac{B^2}{2\mu_0} - \frac{\chi B^2}{2\mu\mu_0} = \frac{B^2}{2\mu\mu_0}\,(\mu - \chi) = \frac{B^2}{2\mu\mu_0},

as it must. With $\chi = 0$ the work vanishes and $w$ is the vacuum value.

Remark

The factor $1/2$ is the mark of an induced moment. A permanent moment $\vect p_m$ in a field has the energy $-\vect p_m\cdot\vect B$ , but a moment that grows in proportion to the field is built up while the field rises, and on average it meets only half the final field, giving $-\vect J\cdot\vect B/2$ per unit volume. The same half separates an induced electric dipole, whose energy in a field is $-\vect p\cdot\vect E/2$ , from a permanent one with $-\vect p\cdot\vect E$ .