The Magnetic Dipole Moment — Question 10

Section 28-9 · Physics pages 4650–4651

Question 10 Torque and potential energy together determine a dipole magnitude and polar angle A dipole in a known uniform .200 T field has torque magnitude .00800 newton metre and potential energy -.00600 J. The energy zero is defined by a perpendicular orientation. Let theta lie between zero and pi.⃗ µ θ⃗ B Schematic; labels, not drawing lengths, specify the data. (a) Find the dipole moment magnitude and theta without assuming the orientation in advance. (b) Find the quasistatic external work to reverse the moment to its opposite direction, and identify the orientation information still missing.
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Question 10 - Solution Torque and potential energy together determine a dipole magnitude and polar angle⃗ µ θ⃗ B (a) Method and calculation τ = µB sin θ, U = −µB cos θ ⇒(µB)2 = τ 2 + U2. µ = √ (.00800)2 + (.00600)2 .200 = .0500 A m2. (b) Interpretation and further calculation θ = tan−1(.00800/.00600) = 53.1301◦, cos θ > 0.⃗ µ →−⃗µ : Wext = Uf −Ui = .00600 −(−.00600) = .0120 J. The azimuth around B remains undetermined by these two scalar measurements. Physical check Energy fixes the sign of the field-parallel component, removing the acute-versus-obtuse ambiguity left by torque magnitude alone.

Question 10

Torque and potential energy together determine a dipole magnitude and polar angle

A dipole in a known uniform .200 T field has torque magnitude .00800 newton metre and potential energy -.00600 J. The energy zero is defined by a perpendicular orientation. Let theta lie between zero and pi.

Diagram shown in the original SVG worksheet.

  1. Find the dipole moment magnitude and theta without assuming the orientation in advance.

  2. Find the quasistatic external work to reverse the moment to its opposite direction, and identify the orientation information still missing.

Question 10 - Solution

Torque and potential energy together determine a dipole magnitude and polar angle

Diagram shown in the original SVG worksheet.

(a) Method and calculation

τ=μBsin⁡θ,U=−μBcos⁡θ⇒(μB)2=τ2+U2.\tau=\mu B\sin\theta,\quad U=-\mu B\cos\theta\Rightarrow(\mu B)^2=\tau^2+U^2.μ=(.00800)2+(.00600)2.200=.0500Am2.\mu=\frac{\sqrt{(.00800)^2+(.00600)^2}}{.200}=.0500\,\mathrm{A\,m^2}.

(b) Interpretation and further calculation

θ=tan⁡−1(.00800/.00600)=53.1301∘,cos⁡θ>0.\theta=\tan^{-1}(.00800/.00600)=53.1301^\circ,\quad\cos\theta>0.μ→→−μ→:Wext=Uf−Ui=.00600−(−.00600)=.0120J.\vec\mu\to-\vec\mu:\quad W_{\mathrm{ext}}=U_f-U_i=.00600-(-.00600)=.0120\,\mathrm J.The azimuth around B remains undetermined by these two scalar measurements.

Physical check

Energy fixes the sign of the field-parallel component, removing the acute-versus-obtuse ambiguity left by torque magnitude alone.