 Solution

Use $N$ turns of mean radius $a\simeq 1.05\mathrm{mm}$, including the wire radius, on a light nonmagnetic rod. With wire radius $r_w$, length $l$ and density $\rho$, neglecting helical pitch givesThe last expression is the transverse central moment of inertia of the uniformly distributed turns. The magnetic torque is $-m_B{B}_h\sin \theta$, so the small-angle magnetic dipole oscillation obeys $J\ddot{\theta }+m_B{B}_h\theta =0$. HenceThe number of turns cancels. Taking the turn radius as $1.00\mathrm{mm}$ instead gives $4.04\mathrm{s}$, consistent at the accuracy of the thin-winding model. The rod's inertia, damping and lead-wire torque are neglected.