Induced Magnetic Fields — Question 2

Section 32-2 · Physics pages 5194–5195

Question 2 An annular electric-flux change leaves a field-free central hole A quasistatic vacuum field changes at uniform rate beta along +z only in the annulus a < r < b. No conduction current is present. Use H B dl = c−2 ˙ΦE. hole b a Schematic; labels, not drawing lengths, specify the data. (a) Find Bphi in all three radial regions. (b) Locate its greatest magnitude for beta positive.
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Question 2 - Solution An annular electric-flux change leaves a field-free central hole hole b a (a) Method and calculation r < a : Bϕ = 0 . a < r < b : Bϕ = β(r2 −a2) 2c2r ; r > b : Bϕ = β(b2 −a2) 2c2r . (b) Interpretation and further calculation d dr(r −a2/r) = 1 + a2/r2 > 0, d dr(1/r) < 0. Bmax = β(b2 −a2) 2c2b at r = b . Physical check The field is continuous at both radial boundaries even though the changing electric-field region has sharp edges.

Question 2

An annular electric-flux change leaves a field-free central hole

A quasistatic vacuum field changes at uniform rate beta along +z only in the annulus a<r<ba<r<b. No conduction current is present. Use ∮Bdl=c−2Φ̇E\oint B\,dl=c^{-2}\dot\Phi_E.

Diagram shown in the original SVG worksheet.

  1. Find Bphi in all three radial regions.

  2. Locate its greatest magnitude for beta positive.

Question 2 - Solution

An annular electric-flux change leaves a field-free central hole

Diagram shown in the original SVG worksheet.

(a) Method and calculation

r<a:Bϕ=0.r<a:\quad\boxed{B_\phi=0}.a<r<b:Bϕ=β(r2−a2)2c2r;r>b:Bϕ=β(b2−a2)2c2r.a<r<b:\quad\boxed{B_\phi=\frac{\beta(r^2-a^2)}{2c^2r}};\quad r>b:\quad\boxed{B_\phi=\frac{\beta(b^2-a^2)}{2c^2r}}.

(b) Interpretation and further calculation

ddr(r−a2/r)=1+a2/r2>0,ddr(1/r)<0.\frac d{dr}(r-a^2/r)=1+a^2/r^2>0,\quad\frac d{dr}(1/r)<0.Bmax=β(b2−a2)2c2bat r=b.\boxed{B_{\max}=\frac{\beta(b^2-a^2)}{2c^2b}\quad\text{at }r=b}.

Physical check

The field is continuous at both radial boundaries even though the changing electric-field region has sharp edges.