Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 15C / ii

Table of contents
- Solution ii

Solution

An annulus of radius

s

and width

d s

has charge

d Q = 2 π s σ d s

, current

d I = \frac{ω}{2 π} d Q = σω s d s,

and dipole moment

d m = π s^{2} d I \overset{n}{^}

. Therefore the magnetic dipole moment of a rigidly rotating charge distribution is

m = πσω \int_{0}^{R} s^{3} d s \overset{n}{^} = \frac{πσω R ^{4}}{4} \overset{n}{^} .

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