Question #43304

A small conducting loop of mass $m$, resistance $R$, and area $A$ is attached to a light rigid rod of length $L$ to form a simple pendulum. The pendulum oscillates in a vertical $xy$-plane, with $x$ being the horizontal displacement from equilibrium. A non-uniform magnetic field $\vec{B}$ is perpendicular to the oscillation plane, varying as $B(x) = B_0 + \beta x^2$ ($B_0, \beta > 0$). The loop is released from a small initial horizontal displacement $x_0$ ($x_0 \ll L$). Assume the loop's dimensions are small enough that $B$ is uniform over its area at any position $x$, its plane remains perpendicular to $\vec{B}$, and self-inductance is negligible. Due to electromagnetic damping, the mechanical energy decreases. Assuming the energy loss per cycle is very small, find the time $t_{1/2}$ required for the total mechanical energy to reduce to half of its initial value.