Irodov 3.345: Interaction Energy Lives in the Overlap

Problem Irodov 3.345

Two loops carrying currents are in vacuum. Let $\vect B_1$ and $\vect B_2$ be the magnetic fields that the first loop’s current and the second loop’s current would each produce alone at the location of a volume element $dV$ . Show that the magnetic interaction energy of the two loops can be written as $W_{ia} = (1/\mu_0)\int \vect B_1\cdot\vect B_2\,dV$ , the integral taken over all space.

Two current loops in vacuum, with currents I_1 and I_2 (senses chosen), and a volume element dV in the space round them. — Two current loops in vacuum, with currents $I_1$ and $I_2$ (senses chosen), and a volume element $dV$ in the space round them.

Given

two loops with currents $I_1$ , $I_2$ , in vacuum
$\vect B_1$ , $\vect B_2$ : each loop’s field alone

Find

show

W_{ia} = (1/\mu_0)\int \vect B_1\cdot\vect B_2\,dV

Solution

Step 1The field energy of the pair

In vacuum the magnetic energy is stored in the field with density $B^2/(2\mu_0)$ . The fields superpose, so with both currents on the field is $\vect B = \vect B_1 + \vect B_2$ at every point, and the total energy is

W = \frac{1}{2\mu_0}\int (\vect B_1 + \vect B_2)^2\,dV \tag{1}

Step 2Splitting off each loop’s own energy

Expanding the square in (1):

W = \frac{1}{2\mu_0}\int B_1^2\,dV + \frac{1}{2\mu_0}\int B_2^2\,dV + \frac{1}{\mu_0}\int \vect B_1\cdot\vect B_2\,dV

The first term is the energy the first loop has with the second absent, $L_1 I_1^2/2$ , and the second term likewise $L_2 I_2^2/2$ . Neither depends on where the loops are relative to each other; what is left over is the energy the pair has because both are present, which is the interaction energy:

W_{ia} = \frac{1}{\mu_0}\int \vect B_1\cdot\vect B_2\,dV

(2)

Step 3The same energy as a mutual inductance

To see that this cross term is the familiar $L_{12}I_1I_2$ , write $\vect B_1 = \nabla\times\vect A_1$ with $\vect A_1$ the vector potential of the first loop, and use the identity

\vect B_2\cdot(\nabla\times\vect A_1) = \vect A_1\cdot(\nabla\times\vect B_2) + \nabla\cdot(\vect A_1\times\vect B_2)

Integrated over all space the divergence becomes a surface integral at infinity, which vanishes because $A_1 B_2$ falls off as $1/r^5$ while the surface grows as $r^2$ . With $\nabla\times\vect B_2 = \mu_0\vect j_2$ , nonzero only in the second wire, where $\vect j_2\,dV = I_2\,d\vect l_2$ :

\frac{1}{\mu_0}\int \vect B_1\cdot\vect B_2\,dV = \int \vect A_1\cdot\vect j_2\,dV = I_2\oint \vect A_1\cdot d\vect l_2

By Stokes’s theorem the last circulation is the flux $\Phi_{21}$ of $\vect B_1$ through the second loop, which is $L_{21} I_1$ :

W_{ia} = I_2\,\Phi_{21} = L_{21} I_1 I_2 \tag{3}

the interaction energy of two circuits in terms of their mutual inductance.

Check

Symmetry: (2) is unchanged when the labels 1 and 2 are swapped, so (3) read the other way gives $L_{12}I_1I_2$ and hence $L_{12} = L_{21}$ , the reciprocity of mutual inductance. Units: $B^2/\mu_0$ is $\text{T}^2/(\text{H/m}) = \text{J/m}^3$ , an energy density, so the integral is an energy.

Remark

Field energies do not add; field amplitudes do. Two equal coaxial loops brought together with parallel currents double the field near them and so carry four times one loop’s field energy there, not twice: the extra is $W_{ia}$ , which is positive for parallel currents. As the loops attract at fixed currents, the sources supply twice the increase of $W_{ia}$ : half goes into this extra field energy, half into the work of attraction.