A Current-Carrying Coil as a Magnetic Dipole — Question 3

Section 29-5 · Physics pages 4736–4737

Question 3 Two axial measurements determine an unknown coil radius before its current is found A thin circular coil has N=100 turns and unknown radius R and current I. Its measured axial fields satisfy B(d)/B(2d) = (5/2)3/2, where d=.0400 m, and B(d)=20.0 microtesla. Use B(z) = µ0NIR2/[2(R2 + z2)3/2]. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. z R d d 2d Schematic; labels, not drawing lengths, specify the data. (a) Find R from the ratio without knowing I. (b) Find I and state why a far-field-only model would lose the radius information in these measurements.
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Question 3 - Solution Two axial measurements determine an unknown coil radius before its current is found z R d d 2d (a) Method and calculation s = [ B(d) B(2d) ]2/3 = R2 + 4d2 R2 + d2 = 5 2. R2 = 4 −s s −1d2 = d2 ⇒R = .0400 m . (b) Interpretation and further calculation I = 2B(d)(R2 + d2)3/2 µ0NR2 = 4 √ 2dB(d) µ0N = .0360127 A . Bdip(d)/Bdip(2d) = 8 independent of R. Only the finite-radius shape of the exact field distinguishes the coil radius from its moment. Physical check The measured ratio is smaller than eight, showing that these measurements are not both in the leading dipole regime.

Question 3

Two axial measurements determine an unknown coil radius before its current is found

A thin circular coil has N=100 turns and unknown radius R and current I. Its measured axial fields satisfy B(d)/B(2d)=(5/2)3/2B(d)/B(2d)=(5/2)^{3/2}, where d=.0400 m, and B(d)=20.0 microtesla. Use B(z)=μ0NIR2/[2(R2+z2)3/2]B(z)=\mu_0NIR^2/[2(R^2+z^2)^{3/2}]. 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. Find R from the ratio without knowing I.

  2. Find I and state why a far-field-only model would lose the radius information in these measurements.

Question 3 - Solution

Two axial measurements determine an unknown coil radius before its current is found

Diagram shown in the original SVG worksheet.

(a) Method and calculation

s=[B(d)B(2d)]2/3=R2+4d2R2+d2=52.s=\left[\frac{B(d)}{B(2d)}\right]^{2/3}=\frac{R^2+4d^2}{R^2+d^2}=\frac52.R2=4−ss−1d2=d2⇒R=.0400m.R^2=\frac{4-s}{s-1}d^2=d^2\Rightarrow\boxed{R=.0400\,\mathrm m}.

(b) Interpretation and further calculation

I=2B(d)(R2+d2)3/2μ0NR2=42dB(d)μ0N=.0360127A.I=\frac{2B(d)(R^2+d^2)^{3/2}}{\mu_0NR^2}=\frac{4\sqrt2dB(d)}{\mu_0N}=\boxed{.0360127\,\mathrm A}.Bdip(d)/Bdip(2d)=8independent of R.B_{\mathrm{dip}}(d)/B_{\mathrm{dip}}(2d)=8\quad\text{independent of }R.Only the finite-radius shape of the exact field distinguishes the coil radius from its moment.

Physical check

The measured ratio is smaller than eight, showing that these measurements are not both in the leading dipole regime.