Question 26.11: What is the current in a wire of radius R = 3.40 mm if the m......
What is the current in a wire of radius R = 3.40 mm if the magnitude of the current density is given by (a) J_a=J_0 r / R and (b) J_b=J_0(1-r / R) , in which r is the radial distance and J_0 = 5.50 × 10^4 A/m²? (c) Which function maximizes the current density near the wire’s surface?
(a) The current resulting from this non-uniform current density is
\begin{aligned} i & =\int_{\text {cylinder }} J_a d A=\frac{J_0}{R} \int_0^R r \cdot 2 \pi r d r=\frac{2}{3} \pi R^2 J_0=\frac{2}{3} \pi\left(3.40 \times 10^{-3}\, m \right)^2\left(5.50 \times 10^4 \,A / m ^2\right) . \\ & =1.33 \,A . \end{aligned}
(b) In this case,
\begin{aligned} i & =\int_{\text {cylinder }} J_b d A=\int_0^R J_0\left(1-\frac{r}{R}\right) 2 \pi r d r=\frac{1}{3} \pi R^2 J_0=\frac{1}{3} \pi\left(3.40 \times 10^{-3} \,m \right)^2\left(5.50 \times 10^4 \,A / m ^2\right) \\ & =0.666\, A . \end{aligned}
(c) The result is different from that in part (a) because J_b is higher near the center of the cylinder (where the area is smaller for the same radial interval) and lower outward, resulting in a lower average current density over the cross section and consequently a lower current than that in part (a). So, J_a has its maximum value near the surface of the wire.