 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, givesThe 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 isOn 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 givesFor 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.