= Solution
In <magnetostatics> <Ampère's circuital law> is $\oint\mathbf B\cdot d\mathbf l=\mu_0I_{\rm enc}$. Its general vacuum version adds $\mu_0\epsilon_0\,d\Phi_E/dt$ on the right. <Faraday's law> is $\mathcal E=\oint\mathbf E\cdot d\mathbf l=-d\Phi_B/dt$ for a fixed loop. A circular Amperian path around a long straight wire gives
$$
\boxed{B(r)=\frac{\mu_0 I}{2\pi r},}
$$
with direction given by the right-hand rule. The <force> on a wire element is $d\mathbf F=I\,d\mathbf l\times\mathbf B$; its magnitude is $I B\,dl\sin\chi$ and its direction follows this cross product.
Choose the straight-wire current rightward and the square below it, as printed. Its positive <magnetic field> there is into the page. With an into-page loop normal, the <magnetic flux linkage> is
$$
\Lambda=N\int_s^{s+a}\frac{\mu_0I}{2\pi r}a\,dr=\mathcal M I,
\qquad
\boxed{\mathcal M=\frac{\mu_0Na}{2\pi}\ln\frac{s+a}{s}.}
$$
Here $\mathcal M$ denotes <mutual inductance>, not a <mass>. Choose positive coil current to produce an out-of-page field. Then its induced current in the resistance-dominated approximation is
$$
\boxed{i(t)=\frac{\mathcal M}{R}\frac{dI}{dt}
=\frac{\mu_0\omega NaI_0}{2\pi R}\ln\frac{s+a}{s}\cos(\omega t).}
$$
This reproduces the specified sign with the declared orientation. It neglects coil <self-inductance>, requiring $\omega L_{\rm coil}\ll R$, as implicit in the requested expression.
The nearest side contains N parallel current segments and is at a uniform distance $s$, so its <force> magnitude is
$$
\boxed{F_{\rm near}=\frac{\mu_0Na}{2\pi s}|I(t)i(t)|.}
$$
The two sides perpendicular to the wire have opposite currents and equal <magnetic fields> at corresponding distances; their <forces> cancel pairwise. The far side opposes the near-side <force> and has the smaller field. Thus
$$
\boxed{F_{\rm total}=\frac{\mu_0Na}{2\pi}|I(t)i(t)|
\left(\frac1s-\frac1{s+a}\right).}
$$
When $I$ and $i$ have the same sign under the chosen conventions, the near-side coil current opposes the straight-wire current and the total <force> is away from the wire; otherwise it is toward it. In particular the signed <force> alternates with $\sin(\omega t)\cos(\omega t)$. The coil is fixed, so these <forces> do no mechanical work on it. The mean <Joule heating> is
$$
\boxed{\overline P=R\overline{i^2}=\frac{\mathcal M^2\omega^2I_0^2}{2R}.}
$$
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