= Solution
Choose positive circulation anticlockwise as viewed from positive $z$. The linked magnetic flux is $\Phi=2\ell XB$, so <Faraday law>, including the motional contribution, gives
$$
\boxed{\mathcal E=-\frac{d\Phi}{dt}=-2\ell\frac{d(XB)}{dt}.}
$$
The conducting circuit consists of two arms of length $X$ and two transverse pieces of length $2\ell$. Its resistance is $\mathcal R=2R(X+2\ell)$, where $R$ is resistance per unit length. Neglecting self-inductance as in the quasistatic circuit model, the current is
$$
I=\frac{\mathcal E}{\mathcal R}=-\frac{\ell}{R(X+2\ell)}\frac{d(XB)}{dt}.
$$
On the moving rod this positive current points along $+y$, so the <Lorentz force> is $I(2\ell\mathbf e_y)\times(B\mathbf e_z)=2\ell IB\mathbf e_x$. Its mass is $2\ell M$, not the mass of the entire circuit. Newton's equation consequently gives
$$
\boxed{M\ddot X=IB=-\frac{B}{R(X+2\ell)}\frac{d(X\ell B)}{dt}.}
$$
For constant $B$, the force opposes $\dot X$, as expected from <Lenz's law>. The prescribed time dependence of the external field can also supply energy to the rod.
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