Angular Momenta and Magnetic Dipole Moments — Question 4

Section 40-3 · Physics pages 6618–6619

Question 4 Magnetic torque produces precession without changing angular momentum magnitude An angular momentum J has moment µ = γJ in uniform B = Bˆz. Its dynamics obey dJ/dt = µ × B. The initial transverse component points along x. z x ˆn θ Schematic; labels, not drawing lengths, specify the data. (a) Derive the signed azimuthal angular velocity. (b) For an electron orbital moment, state the direction and period using γ = −e/(2m).
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Question 4 - Solution Magnetic torque produces precession without changing angular momentum magnitude z x ˆn θ (a) Method and calculation ˙Jx = γBJy, ˙Jy = −γBJx, ˙Jz = 0. Jx = J⊥cos ωt, Jy = J⊥sin ωt ⇒ω = −γB . (b) Interpretation and further calculation γ = −e/(2m) ⇒ω = eB/(2m) > 0 (x toward y). T = 2π/|ω| = 4πm/(eB) . Physical check The torque is perpendicular to J, so its magnitude remains fixed. Positive azimuth means counterclockwise viewed from positive z toward the origin.

Question 4

Magnetic torque produces precession without changing angular momentum magnitude

An angular momentum 𝑱\boldsymbol J has moment 𝝁=γ𝑱\boldsymbol\mu=\gamma\boldsymbol J in uniform 𝑩=Bẑ\boldsymbol B=B\hat z. Its dynamics obey d𝑱/dt=𝝁×𝑩d\boldsymbol J/dt=\boldsymbol\mu\times\boldsymbol B. The initial transverse component points along x.

Diagram shown in the original SVG worksheet.

  1. Derive the signed azimuthal angular velocity.

  2. For an electron orbital moment, state the direction and period using γ=−e/(2m)\gamma=-e/(2m).

Question 4 - Solution

Magnetic torque produces precession without changing angular momentum magnitude

Diagram shown in the original SVG worksheet.

(a) Method and calculation

J̇x=γBJy,J̇y=−γBJx,J̇z=0.\dot J_x=\gamma BJ_y,\quad\dot J_y=-\gamma BJ_x,\quad\dot J_z=0.Jx=J⟂cos⁡ωt,Jy=J⟂sin⁡ωt⇒ω=−γB.J_x=J_\perp\cos\omega t,\ J_y=J_\perp\sin\omega t\Rightarrow\boxed{\omega=-\gamma B}.

(b) Interpretation and further calculation

γ=−e/(2m)⇒ω=eB/(2m)>0(x toward y).\gamma=-e/(2m)\Rightarrow\boxed{\omega=eB/(2m)>0}\quad(x\text{ toward }y).T=2π/|ω|=4πm/(eB).\boxed{T=2\pi/|\omega|=4\pi m/(eB)}.

Physical check

The torque is perpendicular to J, so its magnitude remains fixed. Positive azimuth means counterclockwise viewed from positive z toward the origin.