Small Oscillations of a Magnetic Dipole
For small theta, I_rot theta_double_dot = -mB theta. Thus angular frequency is sqrt(mB/I_rot) and period is 2pi sqrt(I_rot/(mB)). The relevant field for a horizontal needle is Earth's horizontal component.
Why this shows up in the exam
Oscillating magnetometers · Compass-needle period questions · Comparing magnetic moments from measured periods
Learn the idea
A dipole slightly displaced from alignment performs SHM with period set by inertia, moment, and field. Near stable alignment, the magnetic torque is almost proportional to the angular displacement and points back toward equilibrium. This makes a compass needle oscillate like a torsional oscillator.
🧠 Memory hook: More mB restores faster; more inertia swings slower.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- T = 2 pi sqrt(I_rot/(m B)) — small-oscillation period in a uniform field
- omega = sqrt(m B/I_rot) — angular frequency of magnetic SHM
- T1/T2 = sqrt((I1 m2 B2)/(I2 m1 B1)) — comparison form for two oscillators
How to approach it
- 1Confirm small angular oscillations
- 2Identify the correct rotational inertia and field component
- 3Square period ratios before solving for moments or fields
Common slip-ups that cost marks
- •Using the formula for large angular amplitudes without qualification
- •Putting moment of inertia in the denominator inside the period
- •Using total Earth field instead of its horizontal component for a horizontal needle
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A charge of 2 microC moves perpendicular to a 3 T magnetic field at 4 x 10^5 m/s. Find the magnetic force.
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