Torque on a Current Loop — Question 8

Question 8

A freely pivoted current loop performs small angular oscillations about alignment

A rigid N-turn loop has area A, maintained current I and rotational inertia J about a pivot through its center. In uniform field B, let theta be the angle of its normal from B. The pivot constrains rotation to one plane and adds no torque.

1. Write the exact angular equation and derive the small-angle period near theta=0.

2. Find the effect of doubling I on the small-angle period and test the equilibrium at theta=pi with a small perturbation.

Question 8 - Solution

A freely pivoted current loop performs small angular oscillations about alignment

(a) Method and calculation

$$J\ddot\theta=-NIAB\sin\theta.$$$$|\theta|\ll1:\quad\ddot\theta+\Omega^2\theta=0,\quad\Omega=\sqrt{NIAB/J},\quad T=2\pi\sqrt{J/(NIAB)}.$$

(b) Interpretation and further calculation

$$I\prime=2I\Rightarrow T\prime=T/\sqrt2.$$$$\theta=\pi+\eta:\quad\ddot\eta\simeq+\Omega^2\eta\Rightarrow\eta\propto e^{\pm\Omega t};\quad\text{unstable}.$$

Physical check

Aligned and antialigned orientations both have zero instantaneous torque, but their responses to small disturbances differ.