= Solution
\b[<Magnetic field> and <force>.] A steady <electric current> produces a <magnetic field>. In <vacuum> the <Biot-Savart law> gives
$$
\mathbf B_1(\mathbf r)=\frac{\mu_0I_1}{4\pi}
\int_{C_1}\frac{d\boldsymbol\ell_1\times(\mathbf r-\mathbf r_1)}{|\mathbf r-\mathbf r_1|^3}.
$$
The <Lorentz force> on the charge carriers transfers <force> to the material carrying them. The resulting <force> on a current element of another circuit is $d\mathbf F_2=I_2\,d\boldsymbol\ell_2\times\mathbf B_1$, or, for a continuous current distribution, the magnetic <force> per unit <volume (thermodynamics)> is $\mathbf f=\mathbf J\times\mathbf B$.
For two long parallel straight wires a distance $r$ apart, <Ampère's law> gives $B_1=\mu_0 I_1/(2\pi r)$. Therefore
$$
\boxed{\frac{F}{L}=\frac{\mu_0 I_1I_2}{2\pi r}.}
$$
The <force> is \b[attractive for parallel currents and repulsive for opposite currents], by the <cross product> rule. Its units are $\mathrm{N\,m^{-1}}$. This approximation neglects wire thickness and end effects; a complete circuit includes its return current. Element-by-element <forces> need not form an isolated Newton's-third-law pair, but complete stationary circuits have balanced mutual <forces> when the entire system is included.
\b[<Energy> description.] For two linear coupled circuits the <magnetic energy> is
$$
W_B=\frac12L_1I_1^2+\frac12L_2I_2^2+M(q)I_1I_2.
$$
Here $q$ is a mechanical coordinate and $M$ the signed <mutual inductance>. At prescribed currents, the generalized <force> is $I_1I_2\,\partial M/\partial q$ if the self-inductances do not vary. This positive <derivative> is obtained from <magnetic coenergy>, or by including the electrical sources' <work>; differentiating $W_B$ with a minus sign at fixed currents would omit that source <work>. At prescribed flux linkages the <force> instead equals the negative <derivative> of stored <magnetic energy> with those linkages fixed. The geometry thus determines both <inductance> and mechanical <force>.
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