Current Density — Question 7

Section 26-2 · Physics pages 4184–4185

Question 7 A transverse density variation must be integrated across a rectangular contact A rectangular contact lies in the yz plane, −a ≤y ≤a, 0 ≤z ≤b, with normal +ˆx. Its current density is⃗J = J0(1 + y/a)2ˆx + Jtˆy. y −a a b Schematic; labels, not drawing lengths, specify the data. (a) Find the current through the contact. (b) Find the fraction crossing the half with y>0 and explain why Jt is absent.
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Question 7 - Solution A transverse density variation must be integrated across a rectangular contact y −a a b (a) Method and calculation I = ∫b 0 ∫a −a J0(1 + y/a)2 dy dz. I = 8 3J0ab. (b) Interpretation and further calculation Iy>0 = J0ab ∫1 0 (1 + u)2du = 7 3J0ab. Iy>0/I = 7/8, (Jtˆy) · ˆx = 0. Physical check Only the component normal to the chosen surface contributes to its flux.

Question 7

A transverse density variation must be integrated across a rectangular contact

A rectangular contact lies in the yz plane, −a≤y≤a-a\le y\le a, 0≤z≤b0\le z\le b, with normal +x̂+\hat x. Its current density is J→=J0(1+y/a)2x̂+Jtŷ\vec J=J_0(1+y/a)^2\hat x+J_t\hat y.

Diagram shown in the original SVG worksheet.

  1. Find the current through the contact.

  2. Find the fraction crossing the half with y>0 and explain why JtJ_t is absent.

Question 7 - Solution

A transverse density variation must be integrated across a rectangular contact

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I=∫0b∫−aaJ0(1+y/a)2dydz.I=\int_0^b\int_{-a}^a J_0(1+y/a)^2\,dy\,dz.I=83J0ab.I=\frac83J_0ab.

(b) Interpretation and further calculation

Iy>0=J0ab∫01(1+u)2du=73J0ab.I_{y>0}=J_0ab\int_0^1(1+u)^2du=\frac73J_0ab.Iy>0/I=7/8,(Jtŷ)⋅x̂=0.I_{y>0}/I=7/8,\quad(J_t\hat y)\cdot\hat x=0.

Physical check

Only the component normal to the chosen surface contributes to its flux.