Solenoids and Toroids — Question 4

Section 29-4 · Physics pages 4718–4719

Question 4 Nested opposite solenoids can cancel a core field while leaving an annular field Two ideal infinitely long coaxial solenoids have radii a and b, with b>a. Their turn densities are n1 and n2. The inner solenoid current I1 produces field along +z; the outer current I2 produces field along -z. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. core annulus a b cross section of coaxial windings Schematic; labels, not drawing lengths, specify the data. (a) Find B in the core, the annulus and the exterior. (b) For n1=1500 per metre, n2=1000 per metre and I1=.400 A, choose I2 to cancel the core field, then find the annular field.
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Question 4 - Solution Nested opposite solenoids can cancel a core field while leaving an annular field core annulus a b cross section of coaxial windings (a) Method and calculation r < a :⃗B = µ0(n1I1 −n2I2)ˆz; a < r < b :⃗B = −µ0n2I2ˆz.⃗ B = 0 (r > b) for ideal infinite windings. (b) Interpretation and further calculation n1I1 = n2I2 ⇒I2 = .600 A .⃗ Bannulus = −(4π × 10−7)(600)ˆz = −.753982 mT ˆz . Physical check Canceling the field in one spatial region does not cancel every region because each winding has a different field support in the ideal model.

Question 4

Nested opposite solenoids can cancel a core field while leaving an annular field

Two ideal infinitely long coaxial solenoids have radii a and b, with b>a. Their turn densities are n1 and n2. The inner solenoid current I1 produces field along +z; the outer current I2 produces field along -z. 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 B in the core, the annulus and the exterior.

  2. For n1=1500 per metre, n2=1000 per metre and I1=.400 A, choose I2 to cancel the core field, then find the annular field.

Question 4 - Solution

Nested opposite solenoids can cancel a core field while leaving an annular field

Diagram shown in the original SVG worksheet.

(a) Method and calculation

r<a:B→=μ0(n1I1−n2I2)ẑ;a<r<b:B→=−μ0n2I2ẑ.r<a:\ \vec B=\mu_0(n_1I_1-n_2I_2)\hat z;\quad a<r<b:\ \vec B=-\mu_0n_2I_2\hat z.B→=0(r>b)for ideal infinite windings.\boxed{\vec B=0\quad(r>b)}\quad\text{for ideal infinite windings}.

(b) Interpretation and further calculation

n1I1=n2I2⇒I2=.600A.n_1I_1=n_2I_2\Rightarrow\boxed{I_2=.600\,\mathrm A}.B→annulus=−(4π×10−7)(600)ẑ=−.753982mTẑ.\boxed{\vec B_{\mathrm{annulus}}=-(4\pi\times10^{-7})(600)\hat z=-.753982\,\mathrm{mT}\,\hat z}.

Physical check

Canceling the field in one spatial region does not cancel every region because each winding has a different field support in the ideal model.