Current Density — Question 1

Question 1

An oblique cut carries the same current as a perpendicular cut

A straight wire has cross-sectional area $A_0$ perpendicular to a uniform current density $\vec J=J\hat x$. A plane cutting the whole wire has unit normal at angle $\theta<\pi/2$ to $+x$.

1. Find the oblique area and its current flux.

2. Explain the apparent divergence of area as the plane approaches parallel to the wire, and compare with reversing its normal.

Question 1 - Solution

An oblique cut carries the same current as a perpendicular cut

(a) Method and calculation

$$A_\theta\cos\theta=A_0\Rightarrow A_\theta=A_0/\cos\theta.$$$$I=\int\vec J\cdot d\vec A=JA_\theta\cos\theta=JA_0.$$

(b) Interpretation and further calculation

$$\theta\to\pi/2:\ A_\theta\to\infty,\quad J_n=J\cos\theta\to0.$$$$I(-\hat n)=-I(\hat n)=-JA_0.$$A finite wire eventually cannot contain the full ideal oblique cut.

Physical check

Current is an oriented scalar flux; changing the integration surface orientation changes its sign.