Induced Electric Fields — Question 8

Section 30-5 · Physics pages 4846–4847

Question 8 Two electric-field measurements determine a hidden changing-flux radius A uniform changing axial field is confined to unknown radius a. Clockwise induced electric-field magnitudes are .0300 V/m at r1=.0200 m inside and .0120 V/m at r2=.100 m outside. Assume axial symmetry with no added electrostatic field. a r axial B(t) Schematic; labels, not drawing lengths, specify the data. (a) Infer the positive rate beta of axial field increase. (b) Infer a and verify that the assumed inside/outside classification is consistent.
Show solutionHide solution
Question 8 - Solution Two electric-field measurements determine a hidden changing-flux radius a r axial B(t) (a) Method and calculation |E1| = βr1/2 ⇒β = 2(.0300)/.0200 = 3.00 T/s . |E2| = βa2/(2r2) ⇒a2 = 2r2|E2|/β. (b) Interpretation and further calculation a = √ 2(.100)(.0120)/3.00 = .0282843 m . .0200 < a < .100 as assumed. Physical check An exterior electric-field measurement measures the enclosed flux-change rate, while the interior measurement isolates the local field-change rate.

Question 8

Two electric-field measurements determine a hidden changing-flux radius

A uniform changing axial field is confined to unknown radius a. Clockwise induced electric-field magnitudes are .0300 V/m at r1=.0200 m inside and .0120 V/m at r2=.100 m outside. Assume axial symmetry with no added electrostatic field.

Diagram shown in the original SVG worksheet.

  1. Infer the positive rate beta of axial field increase.

  2. Infer a and verify that the assumed inside/outside classification is consistent.

Question 8 - Solution

Two electric-field measurements determine a hidden changing-flux radius

Diagram shown in the original SVG worksheet.

(a) Method and calculation

|E1|=βr1/2⇒β=2(.0300)/.0200=3.00T/s.|E_1|=\beta r_1/2\Rightarrow\boxed{\beta=2(.0300)/.0200=3.00\,\mathrm{T/s}}.|E2|=βa2/(2r2)⇒a2=2r2|E2|/β.|E_2|=\beta a^2/(2r_2)\Rightarrow a^2=2r_2|E_2|/\beta.

(b) Interpretation and further calculation

a=2(.100)(.0120)/3.00=.0282843m.a=\sqrt{2(.100)(.0120)/3.00}=\boxed{.0282843\,\mathrm m}..0200<a<.100as assumed.\boxed{.0200<a<.100}\quad\text{as assumed}.

Physical check

An exterior electric-field measurement measures the enclosed flux-change rate, while the interior measurement isolates the local field-change rate.