A Current-Carrying Coil as a Magnetic Dipole — Question 1

Section 29-5 · Physics pages 4732–4733

Question 1 The axial dipole approximation has a measurable finite-radius error A thin circular N-turn coil of radius R lies in the xy plane. Current I gives moment⃗µ = NIπR2ˆz. Consider an on-axis point at z>0. Use the Biot–Savart law and pairwise cancellation of transverse components. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. z R P z Schematic; labels, not drawing lengths, specify the data. (a) Derive the exact axial field and its far-field dipole limit. (b) Find the minimum z/R for the dipole formula to overestimate the exact field by no more than 5.00 percent, using exact field in the error denominator.
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Question 1 - Solution The axial dipole approximation has a measurable finite-radius error z R P z (a) Method and calculation dBz = µ0NI 4π R dℓ (R2 + z2)3/2 , I dℓ= 2πR ⇒Bz = µ0NIR2 2(R2 + z2)3/2 . z ≫R : Bdip = µ0µ 2πz3 . (b) Interpretation and further calculation ϵ = Bdip Bz −1 = 1 + R2 z2 !3/2 −1 ≤.0500. z/R ≥ h (1.0500)2/3 −1 i−1/2 . Physical check The dipole expression preserves the moment but discards finite-radius corrections; its leading relative error is proportional to R squared over z squared.

Question 1

The axial dipole approximation has a measurable finite-radius error

A thin circular N-turn coil of radius R lies in the xy plane. Current I gives moment μ→=NIπR2ẑ\vec\mu=NI\pi R^2\hat z. Consider an on-axis point at z>0. Use the Biot–Savart law and pairwise cancellation of transverse components. Use vacuum permeability μ0≃4π×10−7Tm/A\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A} where numerical values are needed.

Diagram shown in the original SVG worksheet.

  1. Derive the exact axial field and its far-field dipole limit.

  2. Find the minimum z/R for the dipole formula to overestimate the exact field by no more than 5.00 percent, using exact field in the error denominator.

Question 1 - Solution

The axial dipole approximation has a measurable finite-radius error

Diagram shown in the original SVG worksheet.

(a) Method and calculation

dBz=μ0NI4πRdℓ(R2+z2)3/2,∮dℓ=2πR⇒Bz=μ0NIR22(R2+z2)3/2.dB_z=\frac{\mu_0NI}{4\pi}\frac{R\,d\ell}{(R^2+z^2)^{3/2}},\quad\oint d\ell=2\pi R\Rightarrow\boxed{B_z=\frac{\mu_0NIR^2}{2(R^2+z^2)^{3/2}}}.z≫R:Bdip=μ0μ2πz3.z\gg R:\quad\boxed{B_{\mathrm{dip}}=\frac{\mu_0\mu}{2\pi z^3}}.

(b) Interpretation and further calculation

ϵ=BdipBz−1=(1+R2z2)3/2−1≤.0500.\epsilon=\frac{B_{\mathrm{dip}}}{B_z}-1=\left(1+\frac{R^2}{z^2}\right)^{3/2}-1\le.0500.z/R≥[(1.0500)2/3−1]−1/2.\boxed{z/R\ge\left[(1.0500)^{2/3}-1\right]^{-1/2}}.

Physical check

The dipole expression preserves the moment but discards finite-radius corrections; its leading relative error is proportional to R squared over z squared.