Magnetic Moment of Currents and Rotating Charge
A planar N-turn loop has m = NIA. A total charge q uniformly rotating around a circular loop with angular speed omega has I = q omega/(2pi) and m = q omega r^2/2 in magnitude.
Why this shows up in the exam
Rotating charged-ring questions · Relating coil geometry to torque · Gyromagnetic-ratio reasoning
Learn the idea
A current loop's magnetic strength is current times area, and rotating charge is an equivalent current. Charge circulating repeatedly around a loop constitutes current. Multiplying that current by the enclosed area gives the magnetic dipole moment.
🧠 Memory hook: Circulating charge makes current; current times area makes moment.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- m = N I A — magnetic moment magnitude of an N-turn planar loop
- I = q omega/(2 pi) — equivalent current of charge completing rotations
- m = q omega r²/2 — moment of uniformly rotating charge on a circular ring
How to approach it
- 1Convert rotation rate to current
- 2Find the enclosed area
- 3Use m = IA and assign direction by the right-hand rule
Common slip-ups that cost marks
- •Using circumference instead of enclosed area
- •Dropping the factor one-half for a rotating ring
- •Ignoring the direction set by charge sign and rotation
🌟 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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