Solenoids and Toroids — Question 10

Section 29-4 · Physics pages 4730–4731

Question 10 A gap between two semi-infinite solenoids reduces their axial field Two thin coaxial solenoids of radius R and turn density n extend from z = −∞to −g/2 and from +g/2 to +∞. Their currents I both produce +z fields. For a turn at axial distance u, use dB = µ0nIR2du/[2(R2 + u2)3/2]. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. z midpoint g extends left extends right Schematic; labels, not drawing lengths, specify the data. (a) Integrate to find the field at the midpoint of the empty gap. (b) Find the gap g that reduces this field to half the uninterrupted infinite-solenoid value.
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Question 10 - Solution A gap between two semi-infinite solenoids reduces their axial field z midpoint g extends left extends right (a) Method and calculation B(0) = 2 ∫∞ g/2 µ0nIR2 du 2(R2 + u2)3/2 . B(0) = µ0nI ( 1 − g √ g2 + 4R2 ) . (b) Interpretation and further calculation B(0) = 1 2µ0nI ⇒ g √ g2 + 4R2 = 1 2 ⇒3g2 = 4R2. g = 2R/ √ 3 = 1.15470R . Physical check As g tends to zero the field approaches the uninterrupted solenoid value; for a very wide gap it tends to zero.

Question 10

A gap between two semi-infinite solenoids reduces their axial field

Two thin coaxial solenoids of radius R and turn density n extend from z=−∞z=-\infty to −g/2-g/2 and from +g/2+g/2 to +∞+\infty. Their currents I both produce +z fields. For a turn at axial distance u, use dB=μ0nIR2du/[2(R2+u2)3/2]dB=\mu_0nIR^2du/[2(R^2+u^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. Integrate to find the field at the midpoint of the empty gap.

  2. Find the gap g that reduces this field to half the uninterrupted infinite-solenoid value.

Question 10 - Solution

A gap between two semi-infinite solenoids reduces their axial field

Diagram shown in the original SVG worksheet.

(a) Method and calculation

B(0)=2∫g/2∞μ0nIR2du2(R2+u2)3/2.B(0)=2\int_{g/2}^{\infty}\frac{\mu_0nIR^2\,du}{2(R^2+u^2)^{3/2}}.B(0)=μ0nI(1−gg2+4R2).\boxed{B(0)=\mu_0nI\left(1-\frac g{\sqrt{g^2+4R^2}}\right)}.

(b) Interpretation and further calculation

B(0)=12μ0nI⇒gg2+4R2=12⇒3g2=4R2.B(0)=\tfrac12\mu_0nI\Rightarrow\frac g{\sqrt{g^2+4R^2}}=\frac12\Rightarrow3g^2=4R^2.g=2R/3=1.15470R.\boxed{g=2R/\sqrt3=1.15470R}.

Physical check

As g tends to zero the field approaches the uninterrupted solenoid value; for a very wide gap it tends to zero.