Question #43022

A perfectly conducting U-shaped rail is fixed on a horizontal plane. A perfectly conducting rod of initial mass $m_0$ is placed at $x=0$, closing the circuit. The self-inductance of the closed loop is given by $L(x) = L_0 + \alpha x$, where $\alpha$ is a positive constant. The region $x > 0$ is covered with a uniform layer of stationary non-conducting dust. As the rod moves, it sweeps up and completely accumulates the dust. The linear mass density of the dust is $\rho = \frac{m_0 \alpha}{L_0}$. At $t=0$, an initial current $I_0$ is established in the loop, and the rod is released from rest. Assuming electrical resistance is zero and there is no mechanical friction other than the accretion of dust, what is the velocity of the rod when it reaches a position where the self-inductance of the loop has doubled ($L = 2L_0$)?