(Samuel Li) \[\,\] (a) Since \(\mu_r \gg 1\), to a good approximation the \(\vec{H}\) field is confined to within the ring. The reluctance of the ring is \[ \mathcal{R}_{\text{ring}} = \frac{2\pi R}{\mu_r \mu_0 \pi a^2}, \] and the reluctance of the gap is \[ \mathcal{R}_{\text{gap}} = \frac{2s}{\mu_0 \pi a^2}. \] The magneto-motive force is \(NI = 640 \text{A turns}\), so the magnetic flux through the circuit is \[ \Phi_B = \frac{NI}{\mathcal{R}_{\text{tot}}} = \pi a^2 \mu_0 NI \left(\frac{2\pi R}{\mu_r} + 2s\right)^{-1} = 39 \, \mu \text{Wb}. \] Thus we have \[ |\vec{B}| = \frac{\Phi_B}{A} = \mu_0 N I \left(\frac{2\pi R}{\mu_r} + 2s\right)^{-1} = 58 \text{mT} \] everywhere throughout the circuit. Within the ring, we have \[ |\vec{H}| = \frac{|\vec{B}|}{\mu_0 \mu_r} = 115 \text{ A/m}, \] and within the gap, we have \[ |\vec{H}| = \frac{|\vec{B}|}{\mu_0} = 46 \text{ kA/m}. \] (b) The total magnetic energy is \[ U = \frac{1}{2} \int \vec{B} \cdot \vec{H} \text{d}V = \frac{\pi a^2 \mu_0}{2} N^2 I^2 \left(\frac{2\pi R}{\mu_r} + 2s\right)^{-1} = 13 \text{ mJ}. \] (c) The coil's self-inductance \(L\) satisfies \[ U = \frac{1}{2} L I^2, \] yielding \[ L = \pi a^2 \mu_0 N^2 \, \left(\frac{2\pi R}{\mu_r} + 2s\right)^{-1} = 41 \text{ mH}. \] (d) The total force on the coil is \[ F = \frac{\partial U}{\partial s} = \pi a^2 \mu_0 N^2 I^2 \left(\frac{2\pi R}{\mu_r} + 2s\right)^{-2} = 70 \text{ nN}, \] corresponding to a mass \[ m = \frac{F}{g} = \boxed{7.1 \, \mu\text{g}}. \]