A magnetometer is to sit at the centre of a thin closed spherical enclosure

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Magnetic materials · Electromagnetic induction and alternating-current circuits

The Boundary-Value Problem, Ch Magnetic diffusion, shielding and flux exclusion

Problem

A magnetometer is to sit at the centre of a thin closed spherical enclosure, mean radius a=4.0 cma = 4.0\,\mathrm{cm} and wall thickness d=1.2 mmd = 1.2\,\mathrm{mm}. Two wall materials are available: copper (σ=5.96×107 S m−1\sigma = 5.96\times10^{7}\,\mathrm{S\,m^{-1}}) and mu-metal (μr=5.0×104\mu_r = 5.0\times10^{4}, σ=1.8×106 S m−1\sigma = 1.8\times10^{6}\,\mathrm{S\,m^{-1}}). The sensor must be protected from the Earth's steady field, 50 μT50\,\mu\mathrm{T}, and from a 50 Hz50\,\mathrm{Hz} interference field of amplitude 3.0 μT3.0\,\mu\mathrm{T} from a nearby transformer. ((a)) Using the thin conducting shell result τs=μ0σad/3\tau_s = \mu_0\sigma ad/3 and ∣Bin/B0∣=1/1+ω2τs2|B_{\rm in}/B_0| = 1/\sqrt{1+\omega^2\tau_s^2}, find what fraction of the 50 Hz50\,\mathrm{Hz} field reaches the sensor through a copper enclosure, and state what happens to the steady Earth's field once enough time has passed. Decide whether copper alone is adequate. ((b)) Using the thin-wall static shielding factor Ssh≈1+2μrd/3aS_{\rm sh} \approx 1 + 2\mu_rd/3a, find the static shielding factor of a mu-metal enclosure of the same dimensions and the residual amplitude of both the Earth's field and the 50 Hz50\,\mathrm{Hz} field inside it (treat the static value as a floor for the 50 Hz50\,\mathrm{Hz} shielding, because eddy currents in the mu-metal wall add their own attenuation, provided its permeability at 50 Hz50\,\mathrm{Hz} stays near the static value). ((c)) Before the sensor is fitted, a small magnetised calibration bead (magnetic moment 0.020 A m20.020\,\mathrm{A\,m^2}, mass 0.50 g0.50\,\mathrm{g}) is accidentally dropped down a long vertical copper tube of the same mean radius and wall thickness, from a height of 30 cm30\,\mathrm{cm}. Using the eddy-current drag law F=βvF = \beta v with β=45μ02∣m∣2σd/1024a4\beta = 45\mu_0^2|\mathbf{m}|^2\sigma d/1024a^4, estimate whether the bead's fall is significantly slowed, and hence estimate the speed with which it strikes the bottom.

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