Current Density — Question 2

Section 26-2 · Physics pages 4174–4175

Question 2 A parabolic radial profile concentrates current in the central core A cylindrical wire of radius R carries Jz(r) = J0[1 −(r/R)2] with J0 > 0. R Jz(r) Schematic; labels, not drawing lengths, specify the data. (a) Find total current and average current density. (b) Find the radius containing half the current and compare it with the half-area radius.
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Question 2 - Solution A parabolic radial profile concentrates current in the central core R Jz(r) (a) Method and calculation I = 2πJ0 ∫R 0 (1 −r2/R2)r dr = 1 2πJ0R2. Javg = I/(πR2) = J0/2. (b) Interpretation and further calculation I(r)/I = 2u −u2, u = r2/R2. 2u −u2 = 1/2 ⇒r1/2 = R √ 1 −1/ √ 2 = .541196R < R/ √ 2. Physical check The denser central current makes the half-current core smaller than the half-area core.

Question 2

A parabolic radial profile concentrates current in the central core

A cylindrical wire of radius R carries Jz(r)=J0[1−(r/R)2]J_z(r)=J_0[1-(r/R)^2] with J0>0J_0>0.

Diagram shown in the original SVG worksheet.

  1. Find total current and average current density.

  2. Find the radius containing half the current and compare it with the half-area radius.

Question 2 - Solution

A parabolic radial profile concentrates current in the central core

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I=2πJ0∫0R(1−r2/R2)rdr=12πJ0R2.I=2\pi J_0\int_0^R(1-r^2/R^2)r\,dr=\tfrac12\pi J_0R^2.Javg=I/(πR2)=J0/2.J_{\mathrm{avg}}=I/(\pi R^2)=J_0/2.

(b) Interpretation and further calculation

I(r)/I=2u−u2,u=r2/R2.I(r)/I=2u-u^2,\quad u=r^2/R^2.2u−u2=1/2⇒r1/2=R1−1/2=.541196R<R/2.2u-u^2=1/2\Rightarrow r_{1/2}=R\sqrt{1-1/\sqrt2}=.541196R<R/\sqrt2.

Physical check

The denser central current makes the half-current core smaller than the half-area core.