The Magnetic Dipole Moment — Question 6

Section 28-9 · Physics pages 4642–4643

Question 6 A fixed magnetic moment feels a translational force in a prescribed field gradient A small permanent dipole is constrained to keep⃗µ = .0400ˆz ampere square metre as it moves along x. Along its path, the applied field is⃗B = (B0 + Gx)ˆz, with G=.300 tesla per metre. Use U=-mu dot B and ignore the dipole effect on the field source.⃗ µ x Bz = B0 + Gx Schematic; labels, not drawing lengths, specify the data. (a) Find the force along x from the potential-energy derivative. (b) Find magnetic work and potential-energy change over a displacement delta x=.200 m.
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Question 6 - Solution A fixed magnetic moment feels a translational force in a prescribed field gradient⃗ µ x Bz = B0 + Gx (a) Method and calculation U(x) = −µ(B0 + Gx), Fx = −dU/dx = µG. Fx = (.0400)(.300) = .0120 N. (b) Interpretation and further calculation WB = Fx∆x = (.0120)(.200) = 2.40 × 10−3 J. ∆U = −WB = −2.40 × 10−3 J. Physical check An aligned fixed moment is pulled toward stronger field in this one-dimensional constrained model.

Question 6

A fixed magnetic moment feels a translational force in a prescribed field gradient

A small permanent dipole is constrained to keep μ→=.0400ẑ\vec\mu=.0400\hat z ampere square metre as it moves along x. Along its path, the applied field is B→=(B0+Gx)ẑ\vec B=(B_0+Gx)\hat z, with G=.300 tesla per metre. Use U=-mu dot B and ignore the dipole effect on the field source.

Diagram shown in the original SVG worksheet.

  1. Find the force along x from the potential-energy derivative.

  2. Find magnetic work and potential-energy change over a displacement delta x=.200 m.

Question 6 - Solution

A fixed magnetic moment feels a translational force in a prescribed field gradient

Diagram shown in the original SVG worksheet.

(a) Method and calculation

U(x)=−μ(B0+Gx),Fx=−dU/dx=μG.U(x)=-\mu(B_0+Gx),\quad F_x=-dU/dx=\mu G.Fx=(.0400)(.300)=.0120N.F_x=(.0400)(.300)=.0120\,\mathrm N.

(b) Interpretation and further calculation

WB=FxΔx=(.0120)(.200)=2.40×10−3J.W_B=F_x\Delta x=(.0120)(.200)=2.40\times10^{-3}\,\mathrm J.ΔU=−WB=−2.40×10−3J.\Delta U=-W_B=-2.40\times10^{-3}\,\mathrm J.

Physical check

An aligned fixed moment is pulled toward stronger field in this one-dimensional constrained model.