Question #43263

A rigid, non-conducting thin rod of mass $M$ and length $L$ is hinged frictionlessly at the origin $(0,0)$ and is free to rotate in the $xy$-plane. The rod has a non-uniform linear mass density $\lambda(x) = \frac{2M}{L^2}x$, where $x$ is the distance from the hinge. It possesses a permanent magnetic dipole moment of magnitude $\mu$, directed along its length from the hinge to the free end. The rod initially hangs at rest vertically downwards along the $-\hat{j}$ axis. A uniform, constant magnetic field $\vec{B}_0 = B_0 \hat{j}$ exists in the region, and gravity acts along $-\hat{j}$. A short magnetic field pulse $\vec{B}_{pulse}(t) = B_p(t) \hat{i}$ is applied for a negligible duration $\Delta t$, such that $\int_0^{\Delta t} B_p(t) dt = \beta$. If the rod swings upwards and comes to instantaneous rest exactly at the horizontal position ($+\hat{i}$ axis), which of the following gives the value of $\beta^2$? (Assume $\frac{2}{3} M g L > \mu B_0$)