Solenoids and Toroids — Question 3

Section 29-4 · Physics pages 4716–4717

Question 3 A center-field tolerance places a quantitative limit on the long-solenoid approximation For a thin finite solenoid, the exact center field is Bc = µ0nI/ √ 1 + (2R/ℓ)2. An engineer uses B∞= µ0nI. Define relative error as (B∞−Bc)/Bc. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. z ℓ N turns, current I Schematic; labels, not drawing lengths, specify the data. (a) Find the minimum ell/R for at most 1.00 percent error. (b) For ell=10R, calculate the actual error and the multiplicative current correction that produces the target field predicted by the ideal model.
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Question 3 - Solution A center-field tolerance places a quantitative limit on the long-solenoid approximation z ℓ N turns, current I (a) Method and calculation ϵ = √ 1 + 4R2/ℓ2 −1 ≤.0100. ℓ/R ≥ 2 √ (1.01)2 −1 = 14.1069 . (b) Interpretation and further calculation ℓ= 10R : ϵ = √ 1.04 −1 = 1.98039% . Icorrected/Iideal = √ 1.04 = 1.019804 . Physical check The ideal model overestimates the center field at fixed current, so the compensating current must be larger.

Question 3

A center-field tolerance places a quantitative limit on the long-solenoid approximation

For a thin finite solenoid, the exact center field is Bc=μ0nI/1+(2R/ℓ)2B_c=\mu_0nI/\sqrt{1+(2R/\ell)^2}. An engineer uses B∞=μ0nIB_\infty=\mu_0nI. Define relative error as (B∞−Bc)/Bc(B_\infty-B_c)/B_c. 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 the minimum ell/R for at most 1.00 percent error.

  2. For ell=10R, calculate the actual error and the multiplicative current correction that produces the target field predicted by the ideal model.

Question 3 - Solution

A center-field tolerance places a quantitative limit on the long-solenoid approximation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ϵ=1+4R2/ℓ2−1≤.0100.\epsilon=\sqrt{1+4R^2/\ell^2}-1\le.0100.ℓ/R≥2(1.01)2−1=14.1069.\boxed{\ell/R\ge\frac2{\sqrt{(1.01)^2-1}}=14.1069}.

(b) Interpretation and further calculation

ℓ=10R:ϵ=1.04−1=1.98039%.\ell=10R:\quad\epsilon=\sqrt{1.04}-1=\boxed{1.98039\%}.Icorrected/Iideal=1.04=1.019804.\boxed{I_{\mathrm{corrected}}/I_{\mathrm{ideal}}=\sqrt{1.04}=1.019804}.

Physical check

The ideal model overestimates the center field at fixed current, so the compensating current must be larger.