Question

Derive this equation from an energy standpoint. Start from the total energy of an oscillating rigid system in a magnetic field and obtain the same equation as above. Note that this derivation is cleaner, as you don’t have to put in the minus sign by hand.

4*1сом MB TSHMB

Answer 1

The torque of a magnetic dipole in a magnetic field B is given by

$\tau = \mu_B*B sin(\theta)$

Since the restoring force is proportional to the angle (the wave equation of a SHM)

$I *\frac{\partial^2 \theta}{\partial t^2} =- \mu_B*B sin(\theta) \approx- \mu_BB*\theta$

So, the angular frequency of the osculillator is given by

$\omega ^2 = \mu_B*B/I_{com}$

So, the time pertiod of the simple harmonic motion is given by

$T = 2\pi/\omega = 2\pi/\sqrt{I_{com}/\mu_BB}$

Squaring both sides and rearranging,

$\mu_B = \frac{4\pi^2 I_{com}}{BT^2_{SHM}}$