Induced Electric Fields — Question 3

Section 30-5 · Physics pages 4836–4837

Question 3 A nonuniform magnetic profile moves the strongest induced electric field inside its edge For r<a an axial field is Bz(r, t) = βt(1 −r2/a2), and it is zero outside. Assume axial symmetry and no added electrostatic field. a r axial B(t) Schematic; labels, not drawing lengths, specify the data. (a) Find Ephi inside and outside. (b) Locate the interior maximum of its magnitude for beta positive.
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Question 3 - Solution A nonuniform magnetic profile moves the strongest induced electric field inside its edge a r axial B(t) (a) Method and calculation ˙Φ(r) = 2πβ ∫r 0 (1 −s2/a2)s ds = πβ(r2 −r4/(2a2)). Eϕ = −β(r/2 −r3/(4a2)) (r < a); Eϕ = −βa2/(4r) (r > a) . (b) Interpretation and further calculation d dr|Eϕ| = β ( 1 2 −3r2 4a2 ) = 0 ⇒r = a √ 2/3 . |E|max = βa 3 √ 2/3 > βa/4 = |E(a)|. Physical check The radius of largest induced field is determined by the integrated flux profile, not solely by the field-region boundary.

Question 3

A nonuniform magnetic profile moves the strongest induced electric field inside its edge

For r<a an axial field is Bz(r,t)=βt(1−r2/a2)B_z(r,t)=\beta t(1-r^2/a^2), and it is zero outside. Assume axial symmetry and no added electrostatic field.

Diagram shown in the original SVG worksheet.

  1. Find Ephi inside and outside.

  2. Locate the interior maximum of its magnitude for beta positive.

Question 3 - Solution

A nonuniform magnetic profile moves the strongest induced electric field inside its edge

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Φ̇(r)=2πβ∫0r(1−s2/a2)sds=πβ(r2−r4/(2a2)).\dot\Phi(r)=2\pi\beta\int_0^r(1-s^2/a^2)s\,ds=\pi\beta(r^2-r^4/(2a^2)).Eϕ=−β(r/2−r3/(4a2))(r<a);Eϕ=−βa2/(4r)(r>a).\boxed{E_\phi=-\beta(r/2-r^3/(4a^2))\ (r<a);\quad E_\phi=-\beta a^2/(4r)\ (r>a)}.

(b) Interpretation and further calculation

ddr|Eϕ|=β(12−3r24a2)=0⇒r=a2/3.\frac d{dr}|E_\phi|=\beta\left(\frac12-\frac{3r^2}{4a^2}\right)=0\Rightarrow\boxed{r=a\sqrt{2/3}}.|E|max=βa32/3>βa/4=|E(a)|.\boxed{|E|_{\max}=\frac{\beta a}{3}\sqrt{2/3}}>\beta a/4=|E(a)|.

Physical check

The radius of largest induced field is determined by the integrated flux profile, not solely by the field-region boundary.