Institutt for fysikk, NTNU TFY4155/FY1003 Elektrisitet og magnetisme Vår 2007

Løsningsforslag til øving 13

Veiledning uke 15.

Oppgave 2

a) Distance vector (from the origin) to a point in the ring:

r = sŝ + zẑ

Current density is:

j = \frac{I}{ab}ϕ̂

Hence:

r × j = \frac{I}{ab} (sẑ − zŝ)

Magnetic dipole moment becomes:

m = \frac{I}{2ab} ∫ (sẑ − zŝ) s ds dz dϕ

where ϕ runs from 0 to 2π, s from R − a/2 to R + a/2 and z from z_{0} to z_{0} + b. In the first term, i.e., the one which is proportional with ẑ, integration over z gives a factor b while the integration over ϕ gives a factor 2π. The integral over s becomes

∫_{R−a/2}^{R+a/2} s^{2}ds = \frac{1}{3} (R + a/2)^{3} − \frac{1}{3} (R − a/2)^{3}

The second term, i.e., the one which is proportional with ŝ, must, by symmetry arguments, disappear. We may also show this explicitly: The unit vector ŝ is not a constant, but depends on the angle ϕ in the following manner:

ŝ = (cos ϕ) x̂ + (sin ϕ) ŷ

Hence, this term will be proportional with

∫_{0}^{2π} ŝdϕ = ∫_{0}^{2π} (cos ϕ x̂ + sin ϕ ŷ) dϕ = 0

So, the magnetic dipole moment of the ring is:

m = \frac{I}{2ab} · b · 2π · [\frac{1}{3} (R + a/2)^{3} − \frac{1}{3} (R − a/2)^{3}] ẑ
= \frac{IπR^{3}}{3a} [(1 + \frac{a}{2R})^{3} − (1 − \frac{a}{2R})^{3}] ẑ

b) With a thin ring, a ≪ R, this expression for m may be expanded to lowest order in the ratio a/R:

(1 + \frac{a}{2R})^{3} ≃ 1 + \frac{3a}{2R}

(1 − \frac{a}{2R})^{3} ≃ 1 − \frac{3a}{2R}

Then, one has

m ≃ IπR^{2}ẑ

Alternatively, you may simply multiply out the two parentheses that are raised to the third power. This becomes no more than 4 terms from each. Then you will find:

(1 + \frac{a}{2R})^{3} − (1 − \frac{a}{2R})^{3} = \frac{3a}{R} + \frac{a^{3}}{4R^{3}} ≃ \frac{3a}{R}

where the final approximation could be performed because a ≪ R.