Diamagnetism — Question 4

Section 32-8 · Physics pages 5318–5319

Question 4 The vertical field-gradient product sets a levitation balance A small weakly diamagnetic sample of density rho has susceptibility chi less than zero. The applied field magnitude varies with upward coordinate z. Neglect air buoyancy and demagnetization. Use Fz = χV B(dB/dz)/µ0. Fmag ρV g z Schematic; labels, not drawing lengths, specify the data. (a) Derive the field-gradient product required to balance gravity. (b) Evaluate it for ρ = 1000 kg/m3, χ = −9.0 × 10−6, g = 9.81 m/s2, and µ0 = 4π × 10−7 H/m. Determine the sign of dB/dz if B is positive.
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Question 4 - Solution The vertical field-gradient product sets a levitation balance Fmag ρV g z (a) Method and calculation Fz −ρV g = 0 ⇒B dB dz = µ0ρg χ . V cancels: the required product is independent of sample volume. (b) Interpretation and further calculation B dB dz = (4π × 10−7)(1000)(9.81) −9.0 × 10−6 = −1.370 × 103 T2/m . B > 0 ⇒dB/dz < 0 . Physical check Force balance alone does not establish stability; the spatial derivative of the net force must also be checked.

Question 4

The vertical field-gradient product sets a levitation balance

A small weakly diamagnetic sample of density rho has susceptibility chi less than zero. The applied field magnitude varies with upward coordinate z. Neglect air buoyancy and demagnetization. Use Fz=χVB(dB/dz)/μ0F_z=\chi V B(dB/dz)/\mu_0.

Diagram shown in the original SVG worksheet.

  1. Derive the field-gradient product required to balance gravity.

  2. Evaluate it for ρ=1000kg/m3\rho=1000\,\mathrm{kg/m^3}, χ=−9.0×10−6\chi=-9.0\times10^{-6}, g=9.81m/s2g=9.81\,\mathrm{m/s^2}, and μ0=4π×10−7H/m\mu_0=4\pi\times10^{-7}\,\mathrm{H/m}. Determine the sign of dB/dz if B is positive.

Question 4 - Solution

The vertical field-gradient product sets a levitation balance

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Fz−ρVg=0⇒BdBdz=μ0ρgχ.F_z-\rho Vg=0\Rightarrow\boxed{B\frac{dB}{dz}=\frac{\mu_0\rho g}{\chi}}.V cancels: the required product is independent of sample volume.V\text{ cancels: the required product is independent of sample volume.}

(b) Interpretation and further calculation

BdBdz=(4π×10−7)(1000)(9.81)−9.0×10−6=−1.370×103T2/m.B\frac{dB}{dz}=\frac{(4\pi\times10^{-7})(1000)(9.81)}{-9.0\times10^{-6}}=\boxed{-1.370\times10^3\,\mathrm{T^2/m}}.B>0⇒dB/dz<0.B>0\Rightarrow\boxed{dB/dz<0}.

Physical check

Force balance alone does not establish stability; the spatial derivative of the net force must also be checked.