Induced Electric Fields — Question 2

Section 30-5 · Physics pages 4834–4835

Question 2 An annular changing field leaves the central hole free of induced circulation A changing axial field has uniform derivative beta in the annulus a<r<b and zero derivative elsewhere. The system is axially symmetric and has no additional electric field. ˙B = β hole b a Schematic; labels, not drawing lengths, specify the data. (a) Find Ephi in all three radial regions. (b) Show continuity at both boundaries and locate the maximum magnitude.
Show solutionHide solution
Question 2 - Solution An annular changing field leaves the central hole free of induced circulation ˙B = β hole b a (a) Method and calculation r < a : ˙Φ = 0 ⇒Eϕ = 0 . a < r < b : Eϕ = −β(r2 −a2) 2r ; r > b : Eϕ = −β(b2 −a2) 2r . (b) Interpretation and further calculation Eϕ(a) = 0, Eϕ(b) = −β(b2 −a2)/(2b). d dr ( r −a2 r ) = 1 + a2 r2 > 0 ⇒|E|max = |β|(b2 −a2) 2b . Physical check The hole result uses symmetry in addition to zero circulation; an unrelated electrostatic field could change the local field.

Question 2

An annular changing field leaves the central hole free of induced circulation

A changing axial field has uniform derivative beta in the annulus a<r<b and zero derivative elsewhere. The system is axially symmetric and has no additional electric field.

Diagram shown in the original SVG worksheet.

  1. Find Ephi in all three radial regions.

  2. Show continuity at both boundaries and locate the maximum magnitude.

Question 2 - Solution

An annular changing field leaves the central hole free of induced circulation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

r<a:Φ̇=0⇒Eϕ=0.r<a:\quad\dot\Phi=0\Rightarrow\boxed{E_\phi=0}.a<r<b:Eϕ=−β(r2−a2)2r;r>b:Eϕ=−β(b2−a2)2r.a<r<b:\quad\boxed{E_\phi=-\frac{\beta(r^2-a^2)}{2r}};\quad r>b:\ E_\phi=-\frac{\beta(b^2-a^2)}{2r}.

(b) Interpretation and further calculation

Eϕ(a)=0,Eϕ(b)=−β(b2−a2)/(2b).E_\phi(a)=0,\quad E_\phi(b)=-\beta(b^2-a^2)/(2b).ddr(r−a2r)=1+a2r2>0⇒|E|max=|β|(b2−a2)2b.\frac d{dr}\left(r-\frac{a^2}r\right)=1+\frac{a^2}{r^2}>0\Rightarrow\boxed{|E|_{\max}=\frac{|\beta|(b^2-a^2)}{2b}}.

Physical check

The hole result uses symmetry in addition to zero circulation; an unrelated electrostatic field could change the local field.